Re: [R] cbind question, please
Hello, I am not sure what you mean by a matrix. If you want to have a matrix, use the function matrix, (matrix(c(dog,cat,tree),3)) but I have the feeling you really want a data frame as you are talking about variables. In that case simply use mydataframe - data.frame(dog,cat,tree) If you are not sure what you want, please read the intro to R pdf which is included in your installed R library. Best regards, daniel Feladó: R-help [r-help-boun...@r-project.org] ; meghatalmaz#243;: Erin Hodgess [erinm.hodg...@gmail.com] Küldve: 2015. április 24. 0:41 To: R help Tárgy: [R] cbind question, please Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
On 24/04/15 10:41, Erin Hodgess wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 I don't understand how you managed to get *that*. When I did the obvious thing --- do.call(cbind,as.list(big.char)) --- I got (as expected :-) ) [,1] [,2] [,3] [1,] dog cat tree Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. do.call(cbind,lapply(big.char,get,envir=.GlobalEnv)) Note: I had to throw in the specification of envir otherwise get() got the cat() function from base rather than your vector cat. Probably not a problem for your real application; shouldn't hurt, but. Another salutary example of why it's not a good idea to give data sets names that are names of R built-ins. cheers, Rolf -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
What are you expecting? dog - 1:3 cat - 2:4 tree - 5:7 big.char - c(dog,cat,tree) xx - cbind(dog, cat, tree, big.char) gives me xx1 - structure(c(1, 2, 3, 2, 3, 4, 5, 6, 7, dog, cat, tree), .Dim = 3:4, .Dimnames = list(NULL, c(dog, cat, tree, big.char))) John Kane Kingston ON Canada -Original Message- From: erinm.hodg...@gmail.com Sent: Thu, 23 Apr 2015 18:41:05 -0400 To: r-h...@stat.math.ethz.ch Subject: [R] cbind question, please Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Steve Taylor steve.tay...@aut.ac.nz on Thu, 23 Apr 2015 23:32:00 + writes: This works for me... get0 = function(x) get(x,pos=1) sapply(big.char, get0) Note that get0() is a _ somewhat important for efficient code _ new function since R 3.2.0 so you'd rather call your functions differently... The extra step seems necessary because without it, get() gets base::cat() instead of cat. cheers, Steve -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Erin Hodgess Sent: Friday, 24 April 2015 10:41a To: R help Subject: [R] cbind question, please Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Here is the big picture. I have a character vector with all of the names of the variables in it. I want to cbind all of the variables to create a matrix. Doing 3 is straightforward, but many, not so much. Hence my question. Thanks so much for your answers! Sincerely, Erin On Thu, Apr 23, 2015 at 7:44 PM, John Kane jrkrid...@inbox.com wrote: What are you expecting? dog - 1:3 cat - 2:4 tree - 5:7 big.char - c(dog,cat,tree) xx - cbind(dog, cat, tree, big.char) gives me xx1 - structure(c(1, 2, 3, 2, 3, 4, 5, 6, 7, dog, cat, tree), .Dim = 3:4, .Dimnames = list(NULL, c(dog, cat, tree, big.char))) John Kane Kingston ON Canada -Original Message- From: erinm.hodg...@gmail.com Sent: Thu, 23 Apr 2015 18:41:05 -0400 To: r-h...@stat.math.ethz.ch Subject: [R] cbind question, please Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! Check it out at http://www.inbox.com/marineaquarium -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Is this what you're looking for? dog - 1:3 bat - 2:4 tree - 5:7 big.char - c(dog,bat,tree) do.call(cbind,lapply(big.char, get)) [,1] [,2] [,3] [1,]125 [2,]236 [3,]347 On Thu, Apr 23, 2015 at 3:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
G'day Erin, On Thu, 23 Apr 2015 20:51:18 -0400 Erin Hodgess erinm.hodg...@gmail.com wrote: Here is the big picture. I have a character vector with all of the names of the variables in it. I want to cbind all of the variables to create a matrix. Doing 3 is straightforward, but many, not so much. So I guess you want something like: R do.call(cbind, sapply(big.char, as.name)) dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 Which does not seem to have the problem of confusing the numeric vector `cat' with the function 'cat'. HTH. Cheers, Berwin __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
I am amazed at the number of rather obtuse misunderstandings of the actual nature of Erin's question. The suggestion that Erin should read the intro to R made me smile. Erin is a long time and highly sophisticated user of R; she has no need to read the intro. The person who made that suggestion should have read her question more thoughtfully. Also I liked Steve Taylor's and Marc Swartz's nifty solutions that use sapply(); much sexier than my rather kludgy effort using do.call(). Berwin Turlach's combination of do.call() and sapply() is pretty sexy too. cheers, Rolf Turner -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Hello Erin, I think you have explain your goal more detailed. Maybe I am completely lost but as far as I understand now you only need the command cbind: m1 - cbind(dog, dat, tree) dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 But I can't imagine that is the solution you are looking for. Cheers David On 24.04.2015 00:41, Erin Hodgess wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- David Kienle Department of Biogeography University of Bayreuth GEO II, Rm 003 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
You could do something tricky like do.call(cbind, lapply(big.char, as.name)) dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 but you are usually better off creating these things as part of a list and passing that to do.call(cbind, list). There is a slight danger of using do.call with cbind. If your list has a component with the unlikely name 'deparse.level', then that will be taken as cbind's deparse.level argument, not as a column of the matrix to be made. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Apr 23, 2015 at 3:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Perhaps: dog - 1:3 cat - 2:4 tree - 5:7 big.char - cbind(dog,cat,tree) big.char dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 colnames(big.char)-c(dog,cat,tree) big.char dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 Clint BowmanINTERNET: cl...@ecy.wa.gov Air Quality Modeler INTERNET: cl...@math.utah.edu Department of Ecology VOICE: (360) 407-6815 PO Box 47600FAX:(360) 407-7534 Olympia, WA 98504-7600 USPS: PO Box 47600, Olympia, WA 98504-7600 Parcels:300 Desmond Drive, Lacey, WA 98503-1274 On Thu, 23 Apr 2015, Erin Hodgess wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
Hi Erin, Well, if I do this: dog - 1:3 cat - 2:4 tree - 5:7 dct-cbind(dog,cat,tree) I get this: dct dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 If I assume that you want to include the character vector as well: rownames(dct)-big.char dct Jim On Fri, Apr 24, 2015 at 8:41 AM, Erin Hodgess erinm.hodg...@gmail.com wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
On Apr 23, 2015, at 5:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Hello!
I have a cbind type question, please: Suppose I have the following:
dog - 1:3
cat - 2:4
tree - 5:7
and a character vector
big.char - c(dog,cat,tree)
I want to end up with a matrix that is a cbind of dog, cat, and tree.
This is a toy example. There will be a bunch of variables.
I experimented with do.call, but all I got was
1
2
3
Any suggestions would be much appreciated. I still think that do.call
might be the key, but I'm not sure.
R Version 3-1.3, Windows 7.
Thanks,
Erin
Hi Erin,
One approach could be:
sapply(big.char, get, mode = integer)
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
or
sapply(big.char, get, mode = numeric)
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
Note that I used the ‘mode' argument to get(). You used ‘cat’ as the name of
one of the objects and of course, there is an R function cat(). By default for
get(), mode = “any”, which would otherwise result in:
sapply(big.char, get)
$dog
[1] 1 2 3
$cat
function (..., file = , sep = , fill = FALSE, labels = NULL,
append = FALSE)
{
if (is.character(file))
if (file == )
file - stdout()
else if (substring(file, 1L, 1L) == |) {
file - pipe(substring(file, 2L), w)
on.exit(close(file))
}
else {
file - file(file, ifelse(append, a, w))
on.exit(close(file))
}
.Internal(cat(list(...), file, sep, fill, labels, append))
}
bytecode: 0x7fe942d78f78
environment: namespace:base
$tree
[1] 5 6 7
In the above, the cat() function body is returned, instead of the vector cat.
So just need to be cautious.
An alternative approach, depending upon where your vectors are stored, might be:
sapply(big.char, get, pos = 1)
dog cat tree
[1,] 1 25
[2,] 2 36
[3,] 3 47
which specifies which environment to search for the named objects and the cat()
function is not returned since it is in namespace:base.
See ?get
Regards,
Marc Schwartz
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
This works for me... get0 = function(x) get(x,pos=1) sapply(big.char, get0) The extra step seems necessary because without it, get() gets base::cat() instead of cat. cheers, Steve -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Erin Hodgess Sent: Friday, 24 April 2015 10:41a To: R help Subject: [R] cbind question, please Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind question, please
These are great! Thank you! On Thu, Apr 23, 2015 at 7:14 PM, William Dunlap wdun...@tibco.com wrote: You could do something tricky like do.call(cbind, lapply(big.char, as.name)) dog cat tree [1,] 1 25 [2,] 2 36 [3,] 3 47 but you are usually better off creating these things as part of a list and passing that to do.call(cbind, list). There is a slight danger of using do.call with cbind. If your list has a component with the unlikely name 'deparse.level', then that will be taken as cbind's deparse.level argument, not as a column of the matrix to be made. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Apr 23, 2015 at 3:41 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind question, please
Hello! I have a cbind type question, please: Suppose I have the following: dog - 1:3 cat - 2:4 tree - 5:7 and a character vector big.char - c(dog,cat,tree) I want to end up with a matrix that is a cbind of dog, cat, and tree. This is a toy example. There will be a bunch of variables. I experimented with do.call, but all I got was 1 2 3 Any suggestions would be much appreciated. I still think that do.call might be the key, but I'm not sure. R Version 3-1.3, Windows 7. Thanks, Erin -- Erin Hodgess Associate Professor Department of Mathematical and Statistics University of Houston - Downtown mailto: erinm.hodg...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x))
Both work fine. Thanks to Jeff Laake (2) + David Carlson (1) for their
suggestions.
On 10/8/2014 3:12 PM, Evan Cooch wrote:
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e.,
given an (m x n) matrix x, this produces the vector Y of length m*n
that contains the columns of the matrix x, stacked below each other),
and then (3) cbinds them together.
Here is an example using the case where I know how many matrices I
need to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way? | summary
Actually Jeff Laake's can be made even shorter with
sapply(mat_list, as.vector)
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Thursday, October 9, 2014 7:37 AM
To: Evan Cooch; r-help@r-project.org
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x))
Both work fine. Thanks to Jeff Laake (2) + David Carlson (1) for their
suggestions.
On 10/8/2014 3:12 PM, Evan Cooch wrote:
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e.,
given an (m x n) matrix x, this produces the vector Y of length m*n
that contains the columns of the matrix x, stacked below each other),
and then (3) cbinds them together.
Here is an example using the case where I know how many matrices I
need to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way? | summary
How about foreach() run in parallel?
On Oct 9, 2014 1:54 PM, David L Carlson dcarl...@tamu.edu wrote:
Actually Jeff Laake's can be made even shorter with
sapply(mat_list, as.vector)
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Evan Cooch
Sent: Thursday, October 9, 2014 7:37 AM
To: Evan Cooch; r-help@r-project.org
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x))
Both work fine. Thanks to Jeff Laake (2) + David Carlson (1) for their
suggestions.
On 10/8/2014 3:12 PM, Evan Cooch wrote:
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e.,
given an (m x n) matrix x, this produces the vector Y of length m*n
that contains the columns of the matrix x, stacked below each other),
and then (3) cbinds them together.
Here is an example using the case where I know how many matrices I
need to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way? | summary
Thanks!
On 10/9/2014 1:52 PM, David L Carlson wrote:
Actually Jeff Laake's can be made even shorter with
sapply(mat_list, as.vector)
David C
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Thursday, October 9, 2014 7:37 AM
To: Evan Cooch; r-help@r-project.org
Subject: Re: [R] cbind in a loop...better way? | summary
Two solutions proposed -- not entirely orthogonal, but both do the
trick. Instead of nesting cbin in a loop (as I did originally -- OP,
below),
1\ do.call(cbind, lapply(mat_list, as.vector))
or
2\ sapply(mat_list,function(x) as.vector(x))
Both work fine. Thanks to Jeff Laake (2) + David Carlson (1) for their
suggestions.
On 10/8/2014 3:12 PM, Evan Cooch wrote:
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e.,
given an (m x n) matrix x, this produces the vector Y of length m*n
that contains the columns of the matrix x, stacked below each other),
and then (3) cbinds them together.
Here is an example using the case where I know how many matrices I
need to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind in a loop...better way?
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e., given
an (m x n) matrix x, this produces the vector Y of length m*n that
contains the columns of the matrix x, stacked below each other), and
then (3) cbinds them together.
Here is an example using the case where I know how many matrices I need
to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way?
How about
do.call(cbind, lapply(env, as.vector))
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Wednesday, October 8, 2014 2:13 PM
To: r-help@r-project.org
Subject: [R] cbind in a loop...better way?
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e., given
an (m x n) matrix x, this produces the vector Y of length m*n that
contains the columns of the matrix x, stacked below each other), and
then (3) cbinds them together.
Here is an example using the case where I know how many matrices I need
to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in a loop...better way?
That works as well. I'll collate your response and a couple of others,
and post tomorrow.
On 10/8/2014 4:17 PM, David L Carlson wrote:
How about
do.call(cbind, lapply(env, as.vector))
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Evan Cooch
Sent: Wednesday, October 8, 2014 2:13 PM
To: r-help@r-project.org
Subject: [R] cbind in a loop...better way?
...or some such. I'm trying to work up a function wherein the user
passes a list of matrices to the function, which then (1) takes each
matrix, (2) performs an operation to 'vectorize' the matrix (i.e., given
an (m x n) matrix x, this produces the vector Y of length m*n that
contains the columns of the matrix x, stacked below each other), and
then (3) cbinds them together.
Here is an example using the case where I know how many matrices I need
to cbind together. For this example, 2 square (3x3) matrices:
a - matrix(c,0,20,50,0.05,0,0,0,0.1,0),3,3,byrow=T)
b - matrix(c(0,15,45,0.15,0,0,0,0.2,0),3,3,byrow=T)
I want to vec them, and then cbind them together. So,
result - cbind(matrix(a,nr=9), matrix(b,nr=9))
which yields the following:
[,1] [,2]
[1,] 0.00 0.00
[2,] 0.05 0.15
[3,] 0.00 0.00
[4,] 20.00 15.00
[5,] 0.00 0.00
[6,] 0.10 0.20
[7,] 50.00 45.00
[8,] 0.00 0.00
[9,] 0.00 0.00
Easy enough. But, I want to put it in a function, where the number and
dimensions of the matrices is not specified. Something like
Using matrices (a) and (b) from above, let
env - list(a,b).
Now, a function (or attempt at same) to perform the desired operations:
vec=function(matlist) {
n_mat=length(matlist);
size_mat=dim(matlist[[1]])[1];
result=cbind()
for (i in 1:n_mat) {
result=cbind(result,matrix(matlist[[i]],nr=size_mat^2))
}
return(result)
}
When I run vec(env), I get the *right answer*, but I am wondering if
there is a *better* way to get there from here than the approach I use
(above). I'm not so much interested in 'computational efficiency' as I
am in stability, and flexibility.
Thanks...
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind with row names to serveral files in R
Hi,
The rownames part is not clear as your expected output and input files didn't
show them as rownames.
##Suppose you have all the files in a folder
##here I am creating the names of those files
files1 - paste0(sample, rep(1:777, each=29),chr,1:29,.txt)
length(files1)
#[1] 22533
lst1 - split(files1,as.numeric(gsub(.*chr(\\d+)\\..*,\\1,files1)))
##use list.files()
##in your case, ##not tested
lst1 - split(list.files(pattern=sample\\d+),
as.numeric(gsub(.*chr(\\d+)\\..*,\\1,list.files(pattern=sample\\d+
##in case other files are also in the folder
library(plyr)
lst2 - lapply(lst1,function(x) {x1 -join_all(lapply(x,function(y)
read.table(y,header=TRUE,stringsAsFactors=FALSE,sep=)),c(Name,Chr,Position))
})
lapply(seq_along(lst2),function(i)
write.table(lst2[[i]],paste0(LRRrawallchr,i,.txt),row.names=FALSE,quote=FALSE))
###if you need to create rownames using the first three columns:
lapply(seq_along(lst2),function(i) {x1 - lst2[[i]]; row.names(x1) -
as.character(interaction(x1[,1:3],sep=_)); x2 - x1[,-(1:3)]; write.table(x2,
paste0(LRRrawallchr,i,.txt), quote=FALSE)})
A.K.
I would like to use the `cbind` in a list of files. However each file are
splited in a specific chromosome (chr) `(k in 1:29)`, and specific sample `(i
in 1:777)`. The files are like:
sample1chr1.txt, sample1chr2.txt ... sample1chr29.txt, sample2chr1.txt ...
sample777chr29.txt
All files have exactly the same rows names (3 first collumns represent my row
names). I would like to get a final file to each chr merging to all sample
files, with and do not repeat the row names in the final file (the first 3
collumns representing my rows names).
I tried this:
`#Creating file with row names (3 first collumns) to each Chr
{
{for(k in 1:29){
infile - paste0(sample1chr,k,.txt)
outfile - paste0(LRRrawallchr,k,.txt)
rows - read.table(infile, header=TRUE, sep=\t)
rows - rows[, -grep(Log.R.Ratio, colnames(rows))]
write.table(rows, outfile, sep=;)}}`
`#Cbind in one file per Chr
{ for(i in 1:777)
for(k in 1:29){
base - paste0(LRRrawallchr,k,.txt)
chr - read.table(base, header=TRUE, sep=;)
infile - paste0(sample,i,chr,k,.txt)
chr2 - read.table(infile, header=TRUE, sep=\t)
outfile - paste0(LRRrawallchr,k,.txt)
chr2 - chr2[, -grep(Name, colnames(chr2))]
chr2 - chr2[, -grep(Chr, colnames(chr2))]
chr2 - chr2[, -grep(Position, colnames(chr2))]
chr - cbind(chr, chr2)
write.table(chr, outfile, sep=;, row.names=FALSE, col.names=FALSE)}
}`
Input example (sample1chr1.txt):
Name Chr Position sample1value
BAC-11034 1 128 0.302
BAC-11044 1 129 -0.56
BAC-11057 1 134 0.0840
Input example (sample2chr1.txt):
Name Chr Position sample2value
BAC-11034 1 128 0.25
BAC-11044 1 129 0.41
BAC-11057 1 134 -0.14
Expected output (LRRrawallchr1):
Name Chr Position sample1value sample2value
BAC-11034 1 128 0.302 0.25
BAC-11044 1 129 -0.56 0.41
BAC-11057 1 134 0.0840 -0.14
I have 22553 diferent .txt files (29 files (one per chr) to each of 777
samples). All 22553 files (sample1chr1.txt, sample1chr2.txt ...
sample1chr29.txt, sample2chr1.txt ... sample777chr29.txt) are like above
example.
I wanna 29 files like (LRRrawallchr1), one per Chr. The LRRrawallchr,k, files
have to be with 777+3 (800 collumns). The 3 row names and one collumn per
sample.
Cheers!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind() function : Not able to display columns
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Problem: when we print the matrix k,
k
a b c
[1,] 1 a ee
[2,] 2 b tt
[3,] 3 c rr
we can see that rows are represented by [1,] , [2,] and [3,]. Similarly,
the columns are denoted by [a], [b] and [c]. When we try to print the
corresponding columns, we are able to print for k[a], i.e., the first
column but not able to correctly print the second and third columns.
k[a]
[1] 1 2 3
k[b]
[1] NA NA NA
k[c]
[1] NA NA NA
Please let me know what am I doing wrong.
--
Thanks and Regards,
Vivek Kumar Singh
Research Assistant,
School of Computing,
National University of Singapore
Mobile:(0065) 82721535
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind() function : Not able to display columns
Hi,
Try:
k[,a]
#[1] 1 2 3
k[,b]
#[1] a b c
k[,c]
#[1] ee tt rr
A.K.
On Tuesday, October 22, 2013 11:37 PM, Vivek Singh vksingh.ii...@gmail.com
wrote:
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Problem: when we print the matrix k,
k
a b c
[1,] 1 a ee
[2,] 2 b tt
[3,] 3 c rr
we can see that rows are represented by [1,] , [2,] and [3,]. Similarly,
the columns are denoted by [a], [b] and [c]. When we try to print the
corresponding columns, we are able to print for k[a], i.e., the first
column but not able to correctly print the second and third columns.
k[a]
[1] 1 2 3
k[b]
[1] NA NA NA
k[c]
[1] NA NA NA
Please let me know what am I doing wrong.
--
Thanks and Regards,
Vivek Kumar Singh
Research Assistant,
School of Computing,
National University of Singapore
Mobile:(0065) 82721535
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind() function : Not able to display columns
Hard to say, not sure what you want to do. But the columns are not denoted by
[a], [b] or [c]. You should learn to use the str function to understand what
various expressions really are, and return to the Introduction to R document
that comes with the software. There is a distinct difference between a and a
in R, and square brackets are not at all like quotes. See help([) and the
ItoR section on indexing.
You might get what you want by k[,b] for example.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Vivek Singh vksingh.ii...@gmail.com wrote:
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Problem: when we print the matrix k,
k
a b c
[1,] 1 a ee
[2,] 2 b tt
[3,] 3 c rr
we can see that rows are represented by [1,] , [2,] and [3,].
Similarly,
the columns are denoted by [a], [b] and [c]. When we try to print the
corresponding columns, we are able to print for k[a], i.e., the first
column but not able to correctly print the second and third columns.
k[a]
[1] 1 2 3
k[b]
[1] NA NA NA
k[c]
[1] NA NA NA
Please let me know what am I doing wrong.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind() function : Not able to display columns
In addition to what the others have told you, it looks like you might
be confusing
matrices with data.frames. Please see
?data.frame
I think what you are looking for is
b - c('a','b','c')
c - c(ee,tt,rr)
k - cbind(a,b,c)
K - data.frame(a, b, c)
K
a b c
1 1 a ee
2 2 b tt
3 3 c rr
I recommend using - for assignment (not the spaces on both sides), not =.
On Wed, Oct 23, 2013 at 12:01 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.us wrote:
Hard to say, not sure what you want to do. But the columns are not denoted by
[a], [b] or [c]. You should learn to use the str function to understand what
various expressions really are, and return to the Introduction to R
document that comes with the software. There is a distinct difference between
a and a in R, and square brackets are not at all like quotes. See help([)
and the ItoR section on indexing.
You might get what you want by k[,b] for example.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Vivek Singh vksingh.ii...@gmail.com wrote:
Hi All,
I have create a matrix using cbind() function as follows:
a=c(1,2,3)
b=c('a','b','c')
c=c(ee,tt,rr)
k=cbind(a,b,c)
Problem: when we print the matrix k,
k
a b c
[1,] 1 a ee
[2,] 2 b tt
[3,] 3 c rr
we can see that rows are represented by [1,] , [2,] and [3,].
Similarly,
the columns are denoted by [a], [b] and [c]. When we try to print the
corresponding columns, we are able to print for k[a], i.e., the first
column but not able to correctly print the second and third columns.
k[a]
[1] 1 2 3
k[b]
[1] NA NA NA
k[c]
[1] NA NA NA
Please let me know what am I doing wrong.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind with headers
Hi, You can save it in file. I copy and paste: Subtype,Gender,Expression A,m,-0.54 A,f,-0.8 B,f,-1.03 C,m,-0.41 on the gedit and save it as data1.csv. You might be able to do the same with notepad. x - read.csv(data1.csv,header=T,sep=,) x2 - read.csv(data2N.csv,header=T,sep=,) x3 - cbind(x,x2) x3 # Subtype Gender Expression Age City #1 A m -0.54 32 New York #2 A f -0.80 21 Houston #3 B f -1.03 34 Seattle #4 C m -0.41 67 Houston #or if the dataset is small as in the example x- read.table(text= Subtype,Gender,Expression A,m,-0.54 A,f,-0.8 B,f,-1.03 C,m,-0.41 ,sep=,,header=TRUE,stringsAsFactors=FALSE) x2- read.table(text= Age,City 32,New York 21,Houston 34,Seattle 67,Houston ,sep=,,header=TRUE,stringsAsFactors=FALSE) cbind(x,x2) # Subtype Gender Expression Age City #1 A m -0.54 32 New York #2 A f -0.80 21 Houston #3 B f -1.03 34 Seattle #4 C m -0.41 67 Houston A.K. Hi, I can't seem to get this to work: http://www.endmemo.com/program/R/cbind.php Do I save the data as data1.csv in note pad and pull in the file? Do I type data1.csv-Subtype,Gender,Expression,A,m,-0.54,A,f,-0.8,B,f,-1.03,C,m,-0.41?? I can do a simple matrix. But, I want to have headers and data to combine. Simple Matrix I combined. m-as.data.frame(matrix(c(1:6),ncol=2)) n-as.data.frame(matrix(c(7:12),ncol=2)) m V1 V2 1 1 4 2 2 5 3 3 6 n V1 V2 1 7 10 2 8 11 3 9 12 Mary-cbind(m,n) Mary V1 V2 V1 V2 1 1 4 7 10 2 2 5 8 11 3 3 6 9 12 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind with headers
Hi, I can't seem to get this to work: http://www.endmemo.com/program/R/cbind.php Do I save the data as data1.csv in note pad and pull in the file? Do I type data1.csv-Subtype,Gender,Expression,A,m,-0.54,A,f,-0.8,B,f,-1.03,C,m,-0.41?? I can do a simple matrix. But, I want to have headers and data to combine. Simple Matrix I combined. m-as.data.frame(matrix(c(1:6),ncol=2)) n-as.data.frame(matrix(c(7:12),ncol=2)) m V1 V2 1 1 4 2 2 5 3 3 6 n V1 V2 1 7 10 2 8 11 3 9 12 Mary-cbind(m,n) Mary V1 V2 V1 V2 1 1 4 7 10 2 2 5 8 11 3 3 6 9 12 -- View this message in context: http://r.789695.n4.nabble.com/cbind-with-headers-tp4673354.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind error with check.names
Hello all, I posted a question about cbind and an error when specifying the check.names argument here: http://stackoverflow.com/questions/17810470/cbind-error-with-check-names I was advised to take this to R-devel, but before that, I'd like to get the opinions of people on this list. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
Thanks for the explanations.
Wouldn't the following bit of checking in do.call() make it easier to figure
such things out in the future?
my.call - function(what,args,...) {
## Get the name of the function to call.
if (!is.character(what))
whatStr - deparse(substitute(what))
else
whatStr - what
callFctn - paste0(gsub(paste0('.',class(args[[1]])),'',whatStr), '.',
class(args[[1]]))
## Check whether list names are arguments to the function.
if ( any( names(args) %in% names(formals(callFctn)) ) )
warning( 'Element of args is also an argument for ', callFctn, '.' )
## Do the actual call.
do.call( what, args, ... )
}
Sent from my iPhone
On Apr 8, 2013, at 7:23 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote:
On Mon, Apr 8, 2013 at 3:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
all= is an argument to cbind.zoo so it cannot be used as a column name.
args(zoo:::cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
On Tue, Apr 9, 2013 at 7:31 AM, Harry Mamaysky h.mamay...@gmail.com wrote:
Thanks for the explanations.
Wouldn't the following bit of checking in do.call() make it easier to figure
such things out in the future?
Sure, it would have helped you figure out your issue, but you don't
want a *warning* when you purposefully try to set named function
arguments. I.e., what if you want to set a function argument to
something other than its default value? Do you really want a warning
when you do that?
my.call - function(what,args,...) {
## Get the name of the function to call.
if (!is.character(what))
whatStr - deparse(substitute(what))
else
whatStr - what
callFctn - paste0(gsub(paste0('.',class(args[[1]])),'',whatStr), '.',
class(args[[1]]))
## Check whether list names are arguments to the function.
if ( any( names(args) %in% names(formals(callFctn)) ) )
warning( 'Element of args is also an argument for ', callFctn, '.' )
## Do the actual call.
do.call( what, args, ... )
}
Sent from my iPhone
On Apr 8, 2013, at 7:23 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
On Mon, Apr 8, 2013 at 3:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
all= is an argument to cbind.zoo so it cannot be used as a column name.
args(zoo:::cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
R/Finance 2013: Applied Finance with R | www.RinFinance.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
That's true. So perhaps there should be a flag that turns on this error
checking. Often args is just a list that gets generated automatically and you
don't know what all of its elements are. It just leads to a bit of
non-deterministic behavior. It would be useful to have the option of flagging
when one of those list elements (inadvertently) has the same name as an
argument of what.
Sent from my iPhone
On Apr 9, 2013, at 9:07 AM, Joshua Ulrich josh.m.ulr...@gmail.com wrote:
On Tue, Apr 9, 2013 at 7:31 AM, Harry Mamaysky h.mamay...@gmail.com wrote:
Thanks for the explanations.
Wouldn't the following bit of checking in do.call() make it easier to figure
such things out in the future?
Sure, it would have helped you figure out your issue, but you don't
want a *warning* when you purposefully try to set named function
arguments. I.e., what if you want to set a function argument to
something other than its default value? Do you really want a warning
when you do that?
my.call - function(what,args,...) {
## Get the name of the function to call.
if (!is.character(what))
whatStr - deparse(substitute(what))
else
whatStr - what
callFctn - paste0(gsub(paste0('.',class(args[[1]])),'',whatStr), '.',
class(args[[1]]))
## Check whether list names are arguments to the function.
if ( any( names(args) %in% names(formals(callFctn)) ) )
warning( 'Element of args is also an argument for ', callFctn, '.' )
## Do the actual call.
do.call( what, args, ... )
}
Sent from my iPhone
On Apr 8, 2013, at 7:23 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
On Mon, Apr 8, 2013 at 3:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
all= is an argument to cbind.zoo so it cannot be used as a column name.
args(zoo:::cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
R/Finance 2013: Applied Finance with R | www.RinFinance.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
On Tue, Apr 9, 2013 at 9:15 AM, Harry Mamaysky h.mamay...@gmail.com wrote: That's true. So perhaps there should be a flag that turns on this error checking. Often args is just a list that gets generated automatically and you don't know what all of its elements are. It just leads to a bit of non-deterministic behavior. It would be useful to have the option of flagging when one of those list elements (inadvertently) has the same name as an argument of what. You can force an error by specifying all the default arguments. Using the version of zz with an all component it will encounter all= twice: do.call(cbind, c(zz, all = TRUE, fill = NA, suffixes = list(NULL), drop = FALSE)) Error in cbind(deparse.level, ...) : formal argument all matched by multiple actual arguments or more compactly: do.call(cbind, c(zz, formals(zoo:::cbind.zoo)[-1])) Error in cbind(deparse.level, ...) : formal argument all matched by multiple actual arguments -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
That's a nice solution. Thanks. Sent from my iPhone On Apr 9, 2013, at 10:00 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Tue, Apr 9, 2013 at 9:15 AM, Harry Mamaysky h.mamay...@gmail.com wrote: That's true. So perhaps there should be a flag that turns on this error checking. Often args is just a list that gets generated automatically and you don't know what all of its elements are. It just leads to a bit of non-deterministic behavior. It would be useful to have the option of flagging when one of those list elements (inadvertently) has the same name as an argument of what. You can force an error by specifying all the default arguments. Using the version of zz with an all component it will encounter all= twice: do.call(cbind, c(zz, all = TRUE, fill = NA, suffixes = list(NULL), drop = FALSE)) Error in cbind(deparse.level, ...) : formal argument all matched by multiple actual arguments or more compactly: do.call(cbind, c(zz, formals(zoo:::cbind.zoo)[-1])) Error in cbind(deparse.level, ...) : formal argument all matched by multiple actual arguments -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind for list of zoo objects
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
Because 'all' is the name of one of the arguments to cbind.zoo:
R args(cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
do.call constructs a call somewhat like:
R cbind(test=zz$test, all=zz$all, foo=zz$foo)
The same thing would happen for list elements named 'fill',
'suffixes', or 'drop'.
Best,
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
R/Finance 2013: Applied Finance with R | www.RinFinance.com
On Mon, Apr 8, 2013 at 2:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind for list of zoo objects
On Mon, Apr 8, 2013 at 3:54 PM, Harry Mamaysky h.mamay...@gmail.com wrote:
Can someone explain why this happens when one of the list elements is named
'all'?
zz - list( zoo(1:10,1:10), zoo(101:110,1:10), zoo(201:210,1:10) )
names(zz)-c('test','bar','foo')
do.call(cbind,zz)
test bar foo
1 1 101 201
2 2 102 202
3 3 103 203
4 4 104 204
5 5 105 205
6 6 106 206
7 7 107 207
8 8 108 208
9 9 109 209
10 10 110 210
names(zz)-c('test','all','foo')
do.call(cbind,zz)
test foo
1 1 201
2 2 202
3 3 203
4 4 204
5 5 205
6 6 206
7 7 207
8 8 208
9 9 209
10 10 210
all= is an argument to cbind.zoo so it cannot be used as a column name.
args(zoo:::cbind.zoo)
function (..., all = TRUE, fill = NA, suffixes = NULL, drop = FALSE)
NULL
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind and cbind.data.frame
## Hi, I'm having trouble understanding how the cbind function decides what method to apply to its arguments. Easy cut and paste code below. ## create two columns c1 - c(A,A,B,B) c2 - 1:4 ## cbind outputs a matrix with elements that are characters, seems reasonable out - cbind(c1,c2) out c1 c2 [1,] A 1 [2,] A 2 [3,] B 3 [4,] B 4 class(out) [1] matrix mode(out) [1] character ## there are two methods associated with cbind. so i assume that ## cbind is calling the data.frame method, because neither of my ## arguments are of ts class methods(cbind) [1] cbind.data.frame cbind.ts* Non-visible functions are asterisked ## now i explicitly tell cbind to use the data.frame method ## (which is what i assumed it was doing automatically in the last example) ## but this produces something very different from before, a data.frame out2 - cbind.data.frame(c1,c2) out2 c1 c2 1 A 1 2 A 2 3 B 3 4 B 4 class(out2) [1] data.frame mode(out2) [1] list ## can someone explain why these outputs are different. Thanks, James. c1 - c(A,A,B,B) c2 - 1:4 out - cbind(c1,c2) out class(out) mode(out) methods(cbind) out2 - cbind.data.frame(c1,c2) out2 class(out2) mode(out2) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind and cbind.data.frame
Short answer: there is a default method used here which returns a matrix. It's defined at the C-level for speed so you don't see it with methods() Longer: cbind() isn't a regular S3 generic since it has no UseMethod(). Look at WRE and R-Internals (internal generics) for more info. Best (and good question!), Michael On Jul 9, 2012, at 10:21 PM, James Long jpl2...@gmail.com wrote: ## Hi, I'm having trouble understanding how the cbind function decides what method to apply to its arguments. Easy cut and paste code below. ## create two columns c1 - c(A,A,B,B) c2 - 1:4 ## cbind outputs a matrix with elements that are characters, seems reasonable out - cbind(c1,c2) out c1 c2 [1,] A 1 [2,] A 2 [3,] B 3 [4,] B 4 class(out) [1] matrix mode(out) [1] character ## there are two methods associated with cbind. so i assume that ## cbind is calling the data.frame method, because neither of my ## arguments are of ts class methods(cbind) [1] cbind.data.frame cbind.ts* Non-visible functions are asterisked ## now i explicitly tell cbind to use the data.frame method ## (which is what i assumed it was doing automatically in the last example) ## but this produces something very different from before, a data.frame out2 - cbind.data.frame(c1,c2) out2 c1 c2 1 A 1 2 A 2 3 B 3 4 B 4 class(out2) [1] data.frame mode(out2) [1] list ## can someone explain why these outputs are different. Thanks, James. c1 - c(A,A,B,B) c2 - 1:4 out - cbind(c1,c2) out class(out) mode(out) methods(cbind) out2 - cbind.data.frame(c1,c2) out2 class(out2) mode(out2) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cbind help
Hi, we have a problem concerning the use and forecasting on the GARCH (1, 1) on
financial data.
We would like to make predictions on future returns in a loop which moves the
window of estimation forward one unit for each iteration. Our code is as
follows:
BJORN - read.table(C:/Users/Osvald/Desktop/DATATILLUPPSATS.txt,
+ header=TRUE, sep=\t, na.strings=NA, dec=,, strip.white=TRUE)
t - as.timeSeries(BJORN)
start=length(t[2:1834])
end=length(t)
fore = array(0, dim=(end-start))
for (i in start:end) {
+fit - garchFit(formula =garch(1,1), data = t[1:i])
+pred - predict(fit, n.ahead=1)
+fore[i-start+1] = pred[,1]
+ }
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind, data.frame | numeric to string?
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on elements of
a and b columns are failing. If I look at the structure of the data.frame,
I see that in fact *all* the variables are being entered as factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a column of
character variables change everything else? And, more to the point, what do
I need to do to 'correct' the problem (i.e., stop this from happening).
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen anser.c...@gmail.com wrote:
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on elements of
a and b columns are failing. If I look at the structure of the data.frame,
I see that in fact *all* the variables are being entered as factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a column of
character variables change everything else? And, more to the point, what do
I need to do to 'correct' the problem (i.e., stop this from happening).
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
Still didn't work for me without cbind , although you really don't need it ;)
worked after i set options(stringsAsFactors=F).
options(stringsAsFactors=F)
df-data.frame(intVec,chaVec)
df
intVec chaVec
1 1 a
2 2 b
3 3 c
df$chaVec
[1] a b c
documentation of data.frame says the option is true by default.
Am 10.04.2012 um 17:38 schrieb R. Michael Weylandt:
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen anser.c...@gmail.com wrote:
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on elements of
a and b columns are failing. If I look at the structure of the data.frame,
I see that in fact *all* the variables are being entered as factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a column of
character variables change everything else? And, more to the point, what do
I need to do to 'correct' the problem (i.e., stop this from happening).
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
cbind() works as well, but only if c is attached to the existing test variable:
tst - cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
str( tst )
'data.frame': 5 obs. of 3 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
Not saying it is a good idea, though...
Rainer
On Tuesday 10 April 2012 11:38:51 R. Michael Weylandt wrote:
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen anser.c...@gmail.com wrote:
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on elements of
a and b columns are failing. If I look at the structure of the data.frame,
I see that in fact *all* the variables are being entered as factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a column of
character variables change everything else? And, more to the point, what do
I need to do to 'correct' the problem (i.e., stop this from happening).
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
Sorry, I missed that the OP's real question was in character/factor,
not in the why are these all factors bit...good catch.
Rant about cbind() still stands though. :-) [Your way with cbind()
would give him all characters, not some characters and some numerics
since cbind() gives a matrix by default -- note that in Rainer's
construction, it doesn't coerce because it's giving a data.frame (as
cbind.data.frame does but cdbind.default doesn't) so no coercion is
necessary ]
Best,
Michael
On Tue, Apr 10, 2012 at 11:55 AM, Jessica Streicher
j.streic...@micromata.de wrote:
Still didn't work for me without cbind , although you really don't need it ;)
worked after i set options(stringsAsFactors=F).
options(stringsAsFactors=F)
df-data.frame(intVec,chaVec)
df
intVec chaVec
1 1 a
2 2 b
3 3 c
df$chaVec
[1] a b c
documentation of data.frame says the option is true by default.
Am 10.04.2012 um 17:38 schrieb R. Michael Weylandt:
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen anser.c...@gmail.com wrote:
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on elements of
a and b columns are failing. If I look at the structure of the data.frame,
I see that in fact *all* the variables are being entered as factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a column of
character variables change everything else? And, more to the point, what do
I need to do to 'correct' the problem (i.e., stop this from happening).
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst - cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
str( tst )
'data.frame': 5 obs. of 3 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
Not saying it is a good idea, though...
To be somewhat more expansive ... 'cbind' is not just one function,
but rather a set of functions, since it is generic. The one that is
chosen by the interpreter will depend on whether the first argument
has a class. If it does have a class as in the example above having a
class of data.frame, then the cbind.data.frame function will be
dispatched to process the list of arguments. If the first argument
doesn't have a class as in the OP's second example below, then the
internal cbind function will be used and returns a matrics which
strips off all but a few attributes and forces a lowest common
denominator mode. If only one of the arguments were logical, then
cbind would return a a matrix of all TRUEs and FALSEs.
(This all assumes that the typos in the OP's original example that
created 'c' as an incomplete expression and a and b with unequal
lengths were fixed.)
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
test - data.frame(cbind(a,b))
Warning message:
In cbind(a, b) :
number of rows of result is not a multiple of vector length (arg 1)
c - c(y1,y2,y3,y4,y5)
cbind(c, test)
Error in data.frame(..., check.names = FALSE) :
arguments imply differing number of rows: 5, 6
--
David.
Rainer
On Tuesday 10 April 2012 11:38:51 R. Michael Weylandt wrote:
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen anser.c...@gmail.com
wrote:
Complete newbie to R -- struggling with something which should be
pretty
basic. Trying to create a simple data set (which I gather R refers
to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of
characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character
vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on
elements of
a and b columns are failing. If I look at the structure of the
data.frame,
I see that in fact *all* the variables are being entered as
factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a
column of
character variables change everything else? And, more to the
point, what do
I need to do to 'correct' the problem (i.e., stop this from
happening).
[[alternative HTML version deleted]]
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, data.frame | numeric to string?
On Apr 10, 2012, at 12:19 PM, David Winsemius wrote:
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst - cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
str( tst )
'data.frame': 5 obs. of 3 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
Not saying it is a good idea, though...
To be somewhat more expansive ... 'cbind' is not just one function,
but rather a set of functions, since it is generic. The one that
is chosen by the interpreter will depend on whether the first
argument has a class.
That was just my erroneous impression. If _any_ of the objects in the
argument list is a data.frame then cbind.data.frame appears to get
used. There is a Dispatch section on the help page for cbind that
appears to cover this adequately.
If it does have a class as in the example above having a class of
data.frame, then the cbind.data.frame function will be dispatched
to process the list of arguments. If the first argument doesn't have
a class as in the OP's second example below, then the internal cbind
function will be used and returns a matrics which strips off all but
a few attributes and forces a lowest common denominator mode. If
only one of the arguments were logical, then cbind would return a a
matrix of all TRUEs and FALSEs.
(This all assumes that the typos in the OP's original example that
created 'c' as an incomplete expression and a and b with unequal
lengths were fixed.)
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
test - data.frame(cbind(a,b))
Warning message:
In cbind(a, b) :
number of rows of result is not a multiple of vector length (arg 1)
c - c(y1,y2,y3,y4,y5)
cbind(c, test)
Error in data.frame(..., check.names = FALSE) :
arguments imply differing number of rows: 5, 6
--
David.
Rainer
On Tuesday 10 April 2012 11:38:51 R. Michael Weylandt wrote:
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very frequent over the last few weeks)
Michael
On Tue, Apr 10, 2012 at 10:33 AM, Anser Chen
anser.c...@gmail.com wrote:
Complete newbie to R -- struggling with something which should be
pretty
basic. Trying to create a simple data set (which I gather R
refers to as a
data.frame). So
a - c(1,2,3,4,5);
b - c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
test - data.frame(cbind(a,b))
Seems to do the trick:
test
a b
1 1 0.3
2 2 0.4
3 3 0.5
4 4 0.6
5 5 0.7
Confirm that each variable is numeric:
is.numeric(test$a)
[1] TRUE
is.numeric(test$b)
[1] TRUE
OK, so far so good. But, now I want to merge in a vector of
characters:
c - c('y1,y2,y3,y4,y5)
Confirm that this is string:
is.numeric(c);
[1] FALSE
cbind c into the data frame:
test - data.frame(cbind(a,b,c))
Looks like everything is in place:
test
a b c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
Except that it seems as if the moment I cbind in a character
vector, it
changes numeric data to string:
is.numeric(test$a)
[1] FALSE
is.numeric(test$b)
[1] FALSE
which would explain why the operations I'm trying to perform on
elements of
a and b columns are failing. If I look at the structure of the
data.frame,
I see that in fact *all* the variables are being entered as
factors.
str(test)
'data.frame': 5 obs. of 3 variables:
$ a: Factor w/ 5 levels 1,2,3,4,..: 1 2 3 4 5
$ b: Factor w/ 5 levels 0.3,0.4,0.5,..: 1 2 3 4 5
$ c: Factor w/ 5 levels y1,y2,y3,..: 1 2 3 4 5
But, if I try
test - data.frame(cbind(a,b))
str(test)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: num 0.3 0.4 0.5 0.6 0.7
a and b are coming back as numeric. So, why does cbind'ing a
column of
character variables change everything else? And, more to the
point, what do
I need to do to 'correct' the problem (i.e., stop this from
happening).
[[alternative HTML version deleted]]
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind alternate
I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
You could break the data into chunks, so you cbind and save 50,000
observations at a time. That should be less taxing on your machine and
memory.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mary Kindall
Sent: Friday, January 06, 2012 12:43 PM
To: r-help@r-project.org
Subject: [R] cbind alternate
I have two one dimensional list of elements and want to perform cbind
and then write into a file. The number of entries are more than a
million in both lists. R is taking a lot of time performing this
operation.
Is there any alternate way to perform cbind?
x = table1[1:100,1]
y = table2[1:100,5]
z = cbind(x,y) //hanging the machine
write.table(z,'out.txt)
--
-
Mary Kindall
Yorktown Heights, NY
USA
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
***
This message is for the named person's use only. It may\...{{dropped:11}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) The issue is not the use of cbind(), but that write.table() can be slow with data frames, where each column may be a different class (data type) and requires separate formatting for output. This is referenced in the Note section of ?write.table: write.table can be slow for data frames with large numbers (hundreds or more) of columns: this is inevitable as each column could be of a different class and so must be handled separately. If they are all of the same class, consider using a matrix instead. I suspect in this case, while you don't have a large number of columns, you do have a large number of rows, so that there is a tradeoff. If all of the columns in your source tables are of the same type (eg. all numeric), coerce 'z' to a matrix and then try using write.table(). z - matrix(rnorm(100 * 6), ncol = 6) str(z) num [1:100, 1:6] -0.713 0.79 -0.538 0.945 1.621 ... system.time(write.table(z, file = test.txt)) user system elapsed 12.664 0.292 13.029 The resultant file is about 118 Mb on my system. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
On Jan 6, 2012, at 12:43 PM, Mary Kindall wrote:
I have two one dimensional list of elements and want to perform
cbind and
then write into a file. The number of entries are more than a
million in
both lists. R is taking a lot of time performing this operation.
Is there any alternate way to perform cbind?
x =
y = table2[1:100,5]
z = cbind(x,y) //hanging the machine
You should have been able to bypass the intermediate steps with just:
z = cbind( table1[1:100,1] table2[1:100,5])
Whether you will have sufficient contiguous memory for that object at
the moment or even after rm(x), rm(y) is in doubt, but had you not
created the unneeded x and y, you _might_ have succeeded in your
limited environment. (Real answer: Buy more RAM.)
I speculate that you are on Windows and so refer your to the R-Win FAQ
for further reading about memory limits.
write.table(z,'out.txt)
I do not know of a way to bypass the requirement of a named object to
pass to write.table, but testing suggests that you could try:
write( t(cbind( table1[1:100,1] table2[1:100,5])).
test.txt, 2)
write() does not require a named object but is less inquisitive than
write table and will give you a transposed matrix with 5 columns by
default which will really mess up things, so you need to transpose and
specify the number of columns. (And that may not save any space over
creating a z object.)
So there is another thread today to which master R programmer Bill
Dunlap has offered this strategy (with minor modifications to your
situation by me):
###
f1 - function (n, fileName) {
unlink(fileName)
system.time({
fileConn - file(fileName, wt)
on.exit(close(fileConn))
for (i in seq_len(n)) cat( table1[i, 1], ,
table2[i, 5],
\n, file = fileConn)
})
}
f1(100, 'out.txt')
#
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
On Jan 6, 2012, at 12:39 PM, Marc Schwartz wrote: On Jan 6, 2012, at 11:43 AM, Mary Kindall wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) Apologies, I mis-read where the hang up was. It is in the use of cbind() prior to calling write.table(), not in write.table() itself. Not sure why that part is taking a long time, unless as already mentioned, you are short on memory available. This runs quickly for me: x - matrix(rnorm(100 * 3), ncol = 3) y - matrix(rnorm(100 * 3), ncol = 3) system.time(z - cbind(x, y)) user system elapsed 0.039 0.025 0.065 str(z) num [1:100, 1:6] -0.5102 1.8776 2.4635 0.2982 0.0901 ... To give an example with two data frames containing differing data types, let's use the built-in 'iris' data set, which has 5 columns and 150 rows by default. Let's create a new version with over a million rows: iris.new - iris[rep(seq(nrow(iris)), 7000), ] str(iris.new) 'data.frame': 105 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... system.time(iris.new2 - cbind(iris.new, iris.new)) user system elapsed 5.289 0.282 5.658 str(iris.new2) 'data.frame': 105 obs. of 10 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1 1 1 1 ... You might verify the structures of your 'x' and 'y' to be sure that there is not something amiss with either one. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
Hello,
I believe this function can handle a problem of that size, or bigger.
It does NOT create the full matrix, just writes it to a file, a certain
number of lines at a time.
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf - file(outfile, at) # or wt - write text
on.exit(close(testf))
step - nmax # how many at a time
inx - seq(1, length(x), by=step) # index into 'x' and 'y'
mat - matrix(0, nrow=step, ncol=2) # create a work matrix
# do it 'nmax' rows per iteration
for(i in inx){
mat - cbind(x[i:(i+step-1)], y[i:(i+step-1)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# and now the remainder
mat - NULL
mat - cbind(x[(i+1):length(x)], y[(i+1):length(y)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
# return the output filename
outfile
}
x - 1:1e6 # a numeric vector
y - sample(letters, 1e6, replace=TRUE) # and a character vector
length(x);length(y) # of the same length
fl - test.txt# output file
system.time(write.big.matrix(x, y, outfile=fl))
On my system it takes (sample output)
user system elapsed
1.590.041.65
and can handle different types of data. In the example, numeric and
character.
If you also need the matrix, try to use 'cbind' first, without writing to a
file.
If it's still slow, adapt the code above to keep inserting chunks in an
output matrix.
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-alternate-tp4270188p4270444.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
Sorry Mary,
My function would write the remainder twice, I had only tested it
with multiples of the chunk size.
(And without looking at the lenghty output correctly.)
Now checked:
write.big.matrix - function(x, y, outfile, nmax=1000){
if(file.exists(outfile)) unlink(outfile)
testf - file(outfile, at) # or wt - write text
on.exit(close(testf))
step - nmax # how many at a time
inx - seq(1, length(x)-step, by=step) # index into 'x' and 'y'
mat - matrix(0, nrow=step, ncol=2) # create a work matrix
# do it 'nmax' rows per iteration
for(i in inx){
mat - cbind(x[i:(i+step-1)], y[i:(i+step-1)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# and now the remainder
if(i+step length(x)){
mat - NULL
mat - cbind(x[(i+step):length(x)], y[(i+step):length(y)])
write.table(mat, file=testf, quote=FALSE, row.names=FALSE,
col.names=FALSE)
}
# return the output filename
outfile
}
x - 1:(1e6 + 1234) # a numeric vector
y - sample(letters, 1e6 + 1234, replace=TRUE) # and a character vector
length(x);length(y) # of the same length
fl - test.txt# output file
system.time(write.big.matrix(x, y, outfile=fl, nmax=100))
user system elapsed
3.040.063.09
system.time(write.big.matrix(x, y, outfile=fl))
user system elapsed
1.640.121.76
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-alternate-tp4270188p4270687.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind alternate
What is it you want to do with the data after you save it? Are you just going to read it back into R? If so, consider using save/load. On Fri, Jan 6, 2012 at 12:43 PM, Mary Kindall mary.kind...@gmail.com wrote: I have two one dimensional list of elements and want to perform cbind and then write into a file. The number of entries are more than a million in both lists. R is taking a lot of time performing this operation. Is there any alternate way to perform cbind? x = table1[1:100,1] y = table2[1:100,5] z = cbind(x,y) //hanging the machine write.table(z,'out.txt) -- - Mary Kindall Yorktown Heights, NY USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
Note that this issue was raised on StackOverflow recently. http://stackoverflow.com/questions/7678090/xts-merge-odd-behaviour Here's the solution: index(a) - index(a) index(b) - index(b) merge(a,b) ZWD.UGX SCHB.Close 2010-03-31 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 28.30 2010-04-05 7.4040 28.38 2010-04-06 7.3317 28.21 2010-04-07 NA 28.31 2010-04-08 NA 28.47 Jeff's answer on StackOverflow explains why this works. Best, -- Joshua Ulrich | FOSS Trading: www.fosstrading.com On Fri, Aug 26, 2011 at 8:09 AM, R. Michael Weylandt michael.weyla...@gmail.com wrote: This seems to be the easiest way to handle the problem: a = xts(coredata(a), time(a)) b = xts(coredata(b), time(b)) merge(a,b) ZWD.UGX SCHB.Close 2010-03-31 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 28.30 2010-04-05 7.4040 28.38 2010-04-06 7.3317 28.21 2010-04-07 NA 28.31 2010-04-08 NA 28.47 I can't be sure where the problem was coming from, but if you look closely into the mumbo jumbo of the dput() code you'll see that the time indices don't actually match anywhere as POSIXct/POSIXt objects. Even though they map to the same day, something or other is keeping them as internally different and it's probably inherited from the data sources -- one set maps to times captured at 2300 EDT on my machine and the other to 0800 EDT. If you want, it might be interesting to track down the difference, but the fix I gave above, which basically pulls the time data out through the xts class and only keeps the date in creating a new xts that's exactly the same should handle it. In short, it's all black magic derived from R's various time/date mechanisms. If you don't really want to see whats under the hood, this will also work in this case. a = as.xts(a) b = as.xts(b) merge(a, b) But this can lead to some strange code if you don't take a moment to think about it is.xts(a) TRUE identical(a, as.xts(a)) FALSE Hope this helps and thanks for the fun problem! Michael Weylandt On Fri, Aug 26, 2011 at 8:08 AM, Petr PIKAL petr.pi...@precheza.cz wrote: I was rather too quick It has probably something to do with versions of zoo and xts after updating to zoo 1.7.4 and xts 0.8.2 I got with your examples merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 but when I did a-as.zoo(a) b-as.zoo(b) I got merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 7.6343 28.02 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 7.4040 28.30 2010-04-06 7.3317 28.38 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 which is probably what you want. Regards Petr Hi On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote: If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Ok... dput(a) structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts, zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200, 1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, ZWD.UGX)) dput(b) structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS = Date, .indexTZ = , src = yahoo, updated = structure(1314356091.21457, class = c(POSIXct, POSIXt)), class = c(xts, zoo), index = structure(c(1270072800, 1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, SCHB.Close)) merge(a,b) merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 q() I get slightly different result xx-(merge(a,b)) xx ZWD.UGX SCHB.Close 1270072800 NA 28.02 1270105200 7.6343 NA 1270191600 7.6343 NA 1270278000 7.5458 NA
[R] cbind giving NA's?
I have two xts objects, call them a and b, and am trying to merge them... class(a) [1] xts zoo class(b) [1] xts zoo head(a) 2010-04-01 7.6343 2010-04-02 7.6343 2010-04-03 7.5458 2010-04-04 7.4532 2010-04-05 7.4040 2010-04-06 7.3317 head(b) 2010-04-01 568.80 2010-04-05 571.01 2010-04-06 568.22 2010-04-07 563.54 2010-04-08 567.49 2010-04-09 566.22 head(cbind(a,b)) 2010-04-01 568.80 NA 2010-04-01 NA 0.3998 2010-04-02 NA 0.3998 2010-04-03 NA 0.3999 2010-04-04 NA 0.3994 2010-04-05 571.01 NA q() I'd like the cbind'ed version to look like: 2010-04-01 568.800.3998 2010-04-02 NA 0.3998 2010-04-03 NA 0.3999 2010-04-04 NA 0.3994 2010-04-05 571.01 0.3991 What gives? Many thanks, in advance. -- H -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
Hmm, that's quite puzzling. I don't know but I'd willing to guess the time/date stamps on a,b are more different than the output is leading us to believe. My experience is that there's always more to a time/date object to trick us up than one would expect. If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Michael Weylandt On Fri, Aug 26, 2011 at 6:05 AM, Hasan Diwan hasan.di...@gmail.com wrote: I have two xts objects, call them a and b, and am trying to merge them... class(a) [1] xts zoo class(b) [1] xts zoo head(a) 2010-04-01 7.6343 2010-04-02 7.6343 2010-04-03 7.5458 2010-04-04 7.4532 2010-04-05 7.4040 2010-04-06 7.3317 head(b) 2010-04-01 568.80 2010-04-05 571.01 2010-04-06 568.22 2010-04-07 563.54 2010-04-08 567.49 2010-04-09 566.22 head(cbind(a,b)) 2010-04-01 568.80 NA 2010-04-01 NA 0.3998 2010-04-02 NA 0.3998 2010-04-03 NA 0.3999 2010-04-04 NA 0.3994 2010-04-05 571.01 NA q() I'd like the cbind'ed version to look like: 2010-04-01 568.800.3998 2010-04-02 NA 0.3998 2010-04-03 NA 0.3999 2010-04-04 NA 0.3994 2010-04-05 571.01 0.3991 What gives? Many thanks, in advance. -- H -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote: If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Ok... dput(a) structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts, zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200, 1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, ZWD.UGX)) dput(b) structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS = Date, .indexTZ = , src = yahoo, updated = structure(1314356091.21457, class = c(POSIXct, POSIXt)), class = c(xts, zoo), index = structure(c(1270072800, 1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, SCHB.Close)) merge(a,b) merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 q() So, no, merge doesn't work... -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
Hi On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote: If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Ok... dput(a) structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts, zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200, 1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, ZWD.UGX)) dput(b) structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS = Date, .indexTZ = , src = yahoo, updated = structure(1314356091.21457, class = c(POSIXct, POSIXt)), class = c(xts, zoo), index = structure(c(1270072800, 1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, SCHB.Close)) merge(a,b) merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 q() I get slightly different result xx-(merge(a,b)) xx ZWD.UGX SCHB.Close 1270072800 NA 28.02 1270105200 7.6343 NA 1270191600 7.6343 NA 1270278000 7.5458 NA 1270364400 7.4532 NA 1270418400 NA 28.30 1270450800 7.4040 NA 1270504800 NA 28.38 1270537200 7.3317 NA 1270591200 NA 28.21 1270677600 NA 28.31 1270764000 NA 28.47 but from what you want it seems to me that aggregate could be suitable second step. something like aggregate(xx, list(those dates you have but I don't), mean, na.rm=T) could do what you want. Regards Petr So, no, merge doesn't work... -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
Hi:
Try this:
require('xts')
merge.zoo(zoo(a), zoo(b), all = c(TRUE, TRUE))
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7.4532 28.30
2010-04-05 7.4040 28.38
2010-04-06 7.3317 28.21
2010-04-07 NA 28.31
2010-04-08 NA 28.47
HTH,
Dennis
On Fri, Aug 26, 2011 at 3:58 AM, Hasan Diwan hasan.di...@gmail.com wrote:
On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
If you could, dput() them so we can see everything about them. You also
might see if merge() gives you more expected behavior
Ok...
dput(a)
structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts,
zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200,
1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone =
, tclass = Date), .Dim = c(6L,
1L), .Dimnames = list(NULL, ZWD.UGX))
dput(b)
structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS =
Date, .indexTZ = , src = yahoo, updated =
structure(1314356091.21457, class = c(POSIXct,
POSIXt)), class = c(xts, zoo), index = structure(c(1270072800,
1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone =
, tclass = Date), .Dim = c(6L,
1L), .Dimnames = list(NULL, SCHB.Close))
merge(a,b)
merge(a,b)
ZWD.UGX SCHB.Close
2010-04-01 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7.4532 NA
2010-04-05 NA 28.30
2010-04-05 7.4040 NA
2010-04-06 NA 28.38
2010-04-06 7.3317 NA
2010-04-07 NA 28.21
2010-04-08 NA 28.31
2010-04-09 NA 28.47
q()
So, no, merge doesn't work...
--
Sent from my mobile device
Envoyait de mon telephone mobil
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
I was rather too quick It has probably something to do with versions of zoo and xts after updating to zoo 1.7.4 and xts 0.8.2 I got with your examples merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 but when I did a-as.zoo(a) b-as.zoo(b) I got merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 7.6343 28.02 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 7.4040 28.30 2010-04-06 7.3317 28.38 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 which is probably what you want. Regards Petr Hi On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote: If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Ok... dput(a) structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts, zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200, 1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, ZWD.UGX)) dput(b) structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS = Date, .indexTZ = , src = yahoo, updated = structure(1314356091.21457, class = c(POSIXct, POSIXt)), class = c(xts, zoo), index = structure(c(1270072800, 1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, SCHB.Close)) merge(a,b) merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 q() I get slightly different result xx-(merge(a,b)) xx ZWD.UGX SCHB.Close 1270072800 NA 28.02 1270105200 7.6343 NA 1270191600 7.6343 NA 1270278000 7.5458 NA 1270364400 7.4532 NA 1270418400 NA 28.30 1270450800 7.4040 NA 1270504800 NA 28.38 1270537200 7.3317 NA 1270591200 NA 28.21 1270677600 NA 28.31 1270764000 NA 28.47 but from what you want it seems to me that aggregate could be suitable second step. something like aggregate(xx, list(those dates you have but I don't), mean, na.rm=T) could do what you want. Regards Petr So, no, merge doesn't work... -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind giving NA's?
This seems to be the easiest way to handle the problem: a = xts(coredata(a), time(a)) b = xts(coredata(b), time(b)) merge(a,b) ZWD.UGX SCHB.Close 2010-03-31 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 28.30 2010-04-05 7.4040 28.38 2010-04-06 7.3317 28.21 2010-04-07 NA 28.31 2010-04-08 NA 28.47 I can't be sure where the problem was coming from, but if you look closely into the mumbo jumbo of the dput() code you'll see that the time indices don't actually match anywhere as POSIXct/POSIXt objects. Even though they map to the same day, something or other is keeping them as internally different and it's probably inherited from the data sources -- one set maps to times captured at 2300 EDT on my machine and the other to 0800 EDT. If you want, it might be interesting to track down the difference, but the fix I gave above, which basically pulls the time data out through the xts class and only keeps the date in creating a new xts that's exactly the same should handle it. In short, it's all black magic derived from R's various time/date mechanisms. If you don't really want to see whats under the hood, this will also work in this case. a = as.xts(a) b = as.xts(b) merge(a, b) But this can lead to some strange code if you don't take a moment to think about it is.xts(a) TRUE identical(a, as.xts(a)) FALSE Hope this helps and thanks for the fun problem! Michael Weylandt On Fri, Aug 26, 2011 at 8:08 AM, Petr PIKAL petr.pi...@precheza.cz wrote: I was rather too quick It has probably something to do with versions of zoo and xts after updating to zoo 1.7.4 and xts 0.8.2 I got with your examples merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 but when I did a-as.zoo(a) b-as.zoo(b) I got merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 7.6343 28.02 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 7.4040 28.30 2010-04-06 7.3317 28.38 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 which is probably what you want. Regards Petr Hi On 26 August 2011 03:37, R. Michael Weylandt michael.weyla...@gmail.com wrote: If you could, dput() them so we can see everything about them. You also might see if merge() gives you more expected behavior Ok... dput(a) structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c(xts, zoo), .indexCLASS = Date, .indexTZ = , index = structure(c(1270105200, 1270191600, 1270278000, 1270364400, 1270450800, 1270537200), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, ZWD.UGX)) dput(b) structure(c(28.02, 28.3, 28.38, 28.21, 28.31, 28.47), .indexCLASS = Date, .indexTZ = , src = yahoo, updated = structure(1314356091.21457, class = c(POSIXct, POSIXt)), class = c(xts, zoo), index = structure(c(1270072800, 1270418400, 1270504800, 1270591200, 1270677600, 1270764000), tzone = , tclass = Date), .Dim = c(6L, 1L), .Dimnames = list(NULL, SCHB.Close)) merge(a,b) merge(a,b) ZWD.UGX SCHB.Close 2010-04-01 NA 28.02 2010-04-01 7.6343 NA 2010-04-02 7.6343 NA 2010-04-03 7.5458 NA 2010-04-04 7.4532 NA 2010-04-05 NA 28.30 2010-04-05 7.4040 NA 2010-04-06 NA 28.38 2010-04-06 7.3317 NA 2010-04-07 NA 28.21 2010-04-08 NA 28.31 2010-04-09 NA 28.47 q() I get slightly different result xx-(merge(a,b)) xx ZWD.UGX SCHB.Close 1270072800 NA 28.02 1270105200 7.6343 NA 1270191600 7.6343 NA 1270278000 7.5458 NA 1270364400 7.4532 NA 1270418400 NA 28.30 1270450800 7.4040 NA 1270504800 NA 28.38 1270537200 7.3317 NA 1270591200 NA 28.21 1270677600 NA 28.31 1270764000 NA 28.47 but from what you want it seems to me that aggregate could be suitable second step. something like aggregate(xx, list(those dates you have but I don't), mean, na.rm=T) could do what you want. Regards Petr So, no, merge doesn't work... -- Sent from my mobile device Envoyait de mon telephone mobil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do
Re: [R] cbind in aggregate formula - based on an existing object (vector)
For a little lateral thinking, consider the use of . on the LHS. That could
play out as follows:
myvars - c(Ozone,Wind)
f - . ~ Month
j - union(all.vars(f[[3]]), myvars)
aggregate(. ~ Month, data=airquality[j], mean, na.rm=T)
MonthOzone Wind
1 5 23.61538 11.457692
2 6 29.4 12.18
3 7 59.11538 8.523077
4 8 59.96154 8.565385
5 9 31.44828 10.075862
(and of course, when you play with something unusual, a buglet pops up: it
doesn't work with f instead of the explicit formula in the call to aggregate.)
On Jul 15, 2011, at 00:10 , Dennis Murphy wrote:
Hi:
I think Bill's got the right idea for your problem, but for the fun of
it, here's how Bert's suggestion would play out:
# Kind of works, but only for the first variable in myvars...
aggregate(get(myvars) ~ group + mydate, FUN = sum, data = example)
group mydate get(myvars)
1 group1 2008-12-01 4
2 group2 2008-12-01 6
3 group1 2009-01-01 40
4 group2 2009-01-01 60
5 group1 2009-02-01 400
6 group2 2009-02-01 600
# Maybe sapply() with get as the function will work...
aggregate(sapply(myvars, get) ~ group + mydate, FUN = sum, data = example)
group mydate myvars get
1 group1 2008-12-01 4 4.2
2 group2 2008-12-01 6 6.2
3 group1 2009-01-01 40 40.2
4 group2 2009-01-01 60 60.2
5 group1 2009-02-01400 400.2
6 group2 2009-02-01600 600.2
Apart from the variable names, it matches example.agg1. OTOH, Bill's
suggestion matches example.agg1 exactly and has an advantage in terms
of code clarity:
byVars - c('group', 'mydate')
aggregate(example[myvars], by = example[byVars], FUN = sum)
group mydate value1 value2
1 group1 2008-12-01 44.2
2 group2 2008-12-01 66.2
3 group1 2009-01-01 40 40.2
4 group2 2009-01-01 60 60.2
5 group1 2009-02-01400 400.2
6 group2 2009-02-01600 600.2
FWIW,
Dennis
On Thu, Jul 14, 2011 at 12:05 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
I am aggregating using a formula in aggregate - of the type:
aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata)
However, I actually have an object (vector of my variables to be aggregated):
myvars-c(var1,var2,var3)
I'd like my aggregate formula (its cbind part) to be able to use my
myvars object. Is it possible?
Thanks for your help!
Dimitri
Reproducible example:
mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4)
value1=c(1,10,100,2,20,200,3,30,300,4,40,400)
value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1)
example-data.frame(mydate=mydate,value1=value1,value2=value2)
example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3))
example$group-as.factor(example$group)
(example);str(example)
example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example)
# this works
(example.agg1)
### Building my object (vector of 2 names - in reality, many more):
myvars-c(value1,value2)
example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example)
### does not work
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
THAT'S IT, Bill - exactly what I was looking for! Thanks a lot for the input, everyone. I find the by method the most straigtfoward and clear. Dimitri On Thu, Jul 14, 2011 at 5:12 PM, William Dunlap wdun...@tibco.com wrote: You may find it easier to use the data.frame method for aggregate instead of the formula method when you are using vectors of column names. E.g., responseVars - c(mpg, wt) byVars - c(cyl, gear) aggregate(mtcars[responseVars], by=mtcars[byVars], FUN=median) gives the same result as aggregate(cbind(mpg, wt) ~ cyl + gear, FUN=median, data=mtcars) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Liakhovitski Sent: Thursday, July 14, 2011 1:45 PM To: David Winsemius Cc: r-help Subject: Re: [R] cbind in aggregate formula - based on an existing object (vector) Thanks a lot! actually, what I tried to do is very simple - just passing tons of variable names into the formula. Maybe that get thing suggested by Bert would work... Dimitri On Thu, Jul 14, 2011 at 4:01 PM, David Winsemius dwinsem...@comcast.net wrote: Dmitri: as.matrix makes a matrix out of the dataframe that is passed to it. As a further note I attempted and failed for reasons that are unclear to me to construct a formula that would (I hoped) preserve the column names which are being mangle in the posted effort: form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) myvars-c(value1,value2) example.agg1-aggregate(formula=form,data=example, FUN=sum) Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable traceback() 2: aggregate.formula(formula = form, data = example, FUN = sum) 1: aggregate(formula = form, data = example, FUN = sum) form cbind(value1, value2) ~ group + mydate parse(text=form) expression(~ cbind(value1, value2), group + mydate) So it seems to be correctly dispatched to aggregate.formula but not passing some check or another. Also tried with formula() rather than as.formula with identical error message. Also tried including without naming the argument. -- David On Jul 14, 2011, at 3:32 PM, Dimitri Liakhovitski wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self
Re: [R] cbind in aggregate formula - based on an existing object (vector)
On Jul 15, 2011, at 15:06 , peter dalgaard wrote: For a little lateral thinking, consider the use of . on the LHS. That could play out as follows: myvars - c(Ozone,Wind) f - . ~ Month j - union(all.vars(f[[3]]), myvars) aggregate(. ~ Month, data=airquality[j], mean, na.rm=T) MonthOzone Wind 1 5 23.61538 11.457692 2 6 29.4 12.18 3 7 59.11538 8.523077 4 8 59.96154 8.565385 5 9 31.44828 10.075862 (and of course, when you play with something unusual, a buglet pops up: it doesn't work with f instead of the explicit formula in the call to aggregate.) ...however, once you go down that road, you might as well construct the LHS directly: lhs - as.call(lapply(c(cbind, myvars), as.name)) eval(bquote(aggregate(.(lhs) ~ Month, data=airquality, mean, na.rm=T))) MonthOzone Wind 1 5 23.61538 11.457692 2 6 29.4 12.18 3 7 59.11538 8.523077 4 8 59.96154 8.565385 5 9 31.44828 10.075862 -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind in aggregate formula - based on an existing object (vector)
Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group +mydate,sum,data=example) example.agg1 group mydate example[myvars]NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1, 3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group +mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
Dmitri: Look at my vars from myvars-c(value1,value2) It's just a character vector of length 2! You can't cbind a character vector of length 2! These are not references/pointers. It's not at all clear to me what you ultimately want to do, but IF it's: pass a character vector of names to be used as the LHS of the aggregate .formula call, then something like: (untested) MyVars - do.call(cbind, lapply(myvars,get)) and then aggregate(MyVars ~ ...) might do. But there are so many potential scoping problems here that I would not be surprised if it failed. The usual advice for this sort of thing is to use substitute() or maybe the dreaded eval(parse(...)) construction -- but as I said, I don't really understand what you're after. -- Bert On Thu, Jul 14, 2011 at 12:32 PM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics 467-7374 http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
Dmitri: as.matrix makes a matrix out of the dataframe that is passed to it. As a further note I attempted and failed for reasons that are unclear to me to construct a formula that would (I hoped) preserve the column names which are being mangle in the posted effort: form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) myvars-c(value1,value2) example.agg1-aggregate(formula=form,data=example, FUN=sum) Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable traceback() 2: aggregate.formula(formula = form, data = example, FUN = sum) 1: aggregate(formula = form, data = example, FUN = sum) form cbind(value1, value2) ~ group + mydate parse(text=form) expression(~ cbind(value1, value2), group + mydate) So it seems to be correctly dispatched to aggregate.formula but not passing some check or another. Also tried with formula() rather than as.formula with identical error message. Also tried including without naming the argument. -- David On Jul 14, 2011, at 3:32 PM, Dimitri Liakhovitski wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group +mydate,sum,data=example) example.agg1 group mydate example[myvars]NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2 =c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1, 3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group +mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
Thanks a lot! actually, what I tried to do is very simple - just passing tons of variable names into the formula. Maybe that get thing suggested by Bert would work... Dimitri On Thu, Jul 14, 2011 at 4:01 PM, David Winsemius dwinsem...@comcast.net wrote: Dmitri: as.matrix makes a matrix out of the dataframe that is passed to it. As a further note I attempted and failed for reasons that are unclear to me to construct a formula that would (I hoped) preserve the column names which are being mangle in the posted effort: form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) myvars-c(value1,value2) example.agg1-aggregate(formula=form,data=example, FUN=sum) Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable traceback() 2: aggregate.formula(formula = form, data = example, FUN = sum) 1: aggregate(formula = form, data = example, FUN = sum) form cbind(value1, value2) ~ group + mydate parse(text=form) expression(~ cbind(value1, value2), group + mydate) So it seems to be correctly dispatched to aggregate.formula but not passing some check or another. Also tried with formula() rather than as.formula with identical error message. Also tried including without naming the argument. -- David On Jul 14, 2011, at 3:32 PM, Dimitri Liakhovitski wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
David - I tried exactly the thing you did (and after that asked my question to the forum): form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) And it did not work - although it looks clean. At the end I ended up writing a loop across individual variables with this code in the body: myformula-as.formula(paste(i,~myfactor,sep=)) temp-aggregate(myformula,sum,data=mydata) ... I then it worked. Really don't understand why pasting the cbind(...) text does not work. Dimitri On Thu, Jul 14, 2011 at 4:01 PM, David Winsemius dwinsem...@comcast.net wrote: Dmitri: as.matrix makes a matrix out of the dataframe that is passed to it. As a further note I attempted and failed for reasons that are unclear to me to construct a formula that would (I hoped) preserve the column names which are being mangle in the posted effort: form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) myvars-c(value1,value2) example.agg1-aggregate(formula=form,data=example, FUN=sum) Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable traceback() 2: aggregate.formula(formula = form, data = example, FUN = sum) 1: aggregate(formula = form, data = example, FUN = sum) form cbind(value1, value2) ~ group + mydate parse(text=form) expression(~ cbind(value1, value2), group + mydate) So it seems to be correctly dispatched to aggregate.formula but not passing some check or another. Also tried with formula() rather than as.formula with identical error message. Also tried including without naming the argument. -- David On Jul 14, 2011, at 3:32 PM, Dimitri Liakhovitski wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in aggregate formula - based on an existing object (vector)
You may find it easier to use the data.frame method for aggregate instead of the formula method when you are using vectors of column names. E.g., responseVars - c(mpg, wt) byVars - c(cyl, gear) aggregate(mtcars[responseVars], by=mtcars[byVars], FUN=median) gives the same result as aggregate(cbind(mpg, wt) ~ cyl + gear, FUN=median, data=mtcars) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Liakhovitski Sent: Thursday, July 14, 2011 1:45 PM To: David Winsemius Cc: r-help Subject: Re: [R] cbind in aggregate formula - based on an existing object (vector) Thanks a lot! actually, what I tried to do is very simple - just passing tons of variable names into the formula. Maybe that get thing suggested by Bert would work... Dimitri On Thu, Jul 14, 2011 at 4:01 PM, David Winsemius dwinsem...@comcast.net wrote: Dmitri: as.matrix makes a matrix out of the dataframe that is passed to it. As a further note I attempted and failed for reasons that are unclear to me to construct a formula that would (I hoped) preserve the column names which are being mangle in the posted effort: form - as.formula(paste( cbind(, paste( myvars, collapse=,), ) ~ group+mydate, sep= ) ) myvars-c(value1,value2) example.agg1-aggregate(formula=form,data=example, FUN=sum) Error in m[[2L]][[2L]] : object of type 'symbol' is not subsettable traceback() 2: aggregate.formula(formula = form, data = example, FUN = sum) 1: aggregate(formula = form, data = example, FUN = sum) form cbind(value1, value2) ~ group + mydate parse(text=form) expression(~ cbind(value1, value2), group + mydate) So it seems to be correctly dispatched to aggregate.formula but not passing some check or another. Also tried with formula() rather than as.formula with identical error message. Also tried including without naming the argument. -- David On Jul 14, 2011, at 3:32 PM, Dimitri Liakhovitski wrote: Thank you, David, it does work. Could you please explain why? What exactly does changing it to as matrix do? Thank you! Dimitri On Thu, Jul 14, 2011 at 3:25 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 14, 2011, at 3:05 PM, Dimitri Liakhovitski wrote: Hello! I am aggregating using a formula in aggregate - of the type: aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata) However, I actually have an object (vector of my variables to be aggregated): myvars-c(var1,var2,var3) I'd like my aggregate formula (its cbind part) to be able to use my myvars object. Is it possible? Thanks for your help! Not sure I have gotten all the way there, but this does work: example.agg1-aggregate(as.matrix(example[myvars])~group+mydate,sum,data=example) example.agg1 group mydate example[myvars] NA 1 group1 2008-12-01 4 4.2 2 group2 2008-12-01 6 6.2 3 group1 2009-01-01 40 40.2 4 group2 2009-01-01 60 60.2 5 group1 2009-02-01 400 400.2 6 group2 2009-02-01 600 600.2 Dimitri Reproducible example: mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4) value1=c(1,10,100,2,20,200,3,30,300,4,40,400) value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1) example-data.frame(mydate=mydate,value1=value1,value2=value2) example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3)) example$group-as.factor(example$group) (example);str(example) example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example) # this works (example.agg1) ### Building my object (vector of 2 names - in reality, many more): myvars-c(value1,value2) example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example) ### does not work -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com David Winsemius, MD West Hartford, CT -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented
Re: [R] cbind in aggregate formula - based on an existing object (vector)
Hi:
I think Bill's got the right idea for your problem, but for the fun of
it, here's how Bert's suggestion would play out:
# Kind of works, but only for the first variable in myvars...
aggregate(get(myvars) ~ group + mydate, FUN = sum, data = example)
group mydate get(myvars)
1 group1 2008-12-01 4
2 group2 2008-12-01 6
3 group1 2009-01-01 40
4 group2 2009-01-01 60
5 group1 2009-02-01 400
6 group2 2009-02-01 600
# Maybe sapply() with get as the function will work...
aggregate(sapply(myvars, get) ~ group + mydate, FUN = sum, data = example)
group mydate myvars get
1 group1 2008-12-01 4 4.2
2 group2 2008-12-01 6 6.2
3 group1 2009-01-01 40 40.2
4 group2 2009-01-01 60 60.2
5 group1 2009-02-01400 400.2
6 group2 2009-02-01600 600.2
Apart from the variable names, it matches example.agg1. OTOH, Bill's
suggestion matches example.agg1 exactly and has an advantage in terms
of code clarity:
byVars - c('group', 'mydate')
aggregate(example[myvars], by = example[byVars], FUN = sum)
group mydate value1 value2
1 group1 2008-12-01 44.2
2 group2 2008-12-01 66.2
3 group1 2009-01-01 40 40.2
4 group2 2009-01-01 60 60.2
5 group1 2009-02-01400 400.2
6 group2 2009-02-01600 600.2
FWIW,
Dennis
On Thu, Jul 14, 2011 at 12:05 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
I am aggregating using a formula in aggregate - of the type:
aggregate(cbind(var1,var2,var3)~factor1+factor2,sum,data=mydata)
However, I actually have an object (vector of my variables to be aggregated):
myvars-c(var1,var2,var3)
I'd like my aggregate formula (its cbind part) to be able to use my
myvars object. Is it possible?
Thanks for your help!
Dimitri
Reproducible example:
mydate = rep(seq(as.Date(2008-12-01), length = 3, by = month),4)
value1=c(1,10,100,2,20,200,3,30,300,4,40,400)
value2=c(1.1,10.1,100.1,2.1,20.1,200.1,3.1,30.1,300.1,4.1,40.1,400.1)
example-data.frame(mydate=mydate,value1=value1,value2=value2)
example$group-c(rep(group1,3),rep(group2,3),rep(group1,3),rep(group2,3))
example$group-as.factor(example$group)
(example);str(example)
example.agg1-aggregate(cbind(value1,value2)~group+mydate,sum,data=example)
# this works
(example.agg1)
### Building my object (vector of 2 names - in reality, many more):
myvars-c(value1,value2)
example.agg1-aggregate(cbind(myvars)~group+mydate,sum,data=example)
### does not work
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind 3 or more matrices
How can I cbind three or more matrices like A,B and C. This does not work: cbind(A,B,C) -- Thanks, Jim. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind 3 or more matrices
do.call(cbind, list(A, B, C)) On Sat, Jun 4, 2011 at 7:14 PM, Jim Silverton jim.silver...@gmail.com wrote: How can I cbind three or more matrices like A,B and C. This does not work: cbind(A,B,C) -- Thanks, Jim. -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind 3 or more matrices
A, B, C should have the same number of rows. mlist = replicate(3, matrix(rnorm(6), 2), simplify=FALSE) names(mlist) = LETTERS[seq_along(mlist)] with(mlist, cbind(A,B,C)) or, do.call(cbind, mlist) HTH, baptiste On 5 June 2011 11:14, Jim Silverton jim.silver...@gmail.com wrote: How can I cbind three or more matrices like A,B and C. This does not work: cbind(A,B,C) -- Thanks, Jim. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind 3 or more matrices
Jim - In what sense does cbind(A,B,C) not work? A = matrix(rnorm(10),5,2) B = matrix(rnorm(15),5,3) C = matrix(rnorm(20),5,4) cbind(A,B,C) [,1] [,2] [,3] [,4] [,5][,6] [1,] -0.54194873 -1.1105170 -0.479010 0.619911163 0.1610162 0.49028633 [2,] -0.39289246 0.0752089 1.427386 -0.921868090 -0.7637016 -0.34905125 [3,] -0.07082828 -0.1060497 -1.007713 -0.003673573 -0.8384406 -0.88816295 [4,] 0.22733701 -1.6134894 -1.993654 2.277865363 -2.3599239 -0.21704046 [5,] -0.13809337 0.3443488 -1.384425 0.132130433 0.1345938 -0.04170648 [,7] [,8][,9] [1,] -1.7481451 0.4467964 -0.41358420 [2,] -0.2882922 1.0243662 -0.48263684 [3,] 0.9402479 0.5467952 -0.01922035 [4,] 0.6795783 1.4560765 -0.23013826 [5,] 0.9800312 -1.3462175 -0.77064872 - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Sat, 4 Jun 2011, Jim Silverton wrote: How can I cbind three or more matrices like A,B and C. This does not work: cbind(A,B,C) -- Thanks, Jim. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cbind query
Hello jholtman, Thanks very much for the suggestion. I tried using as.is=TRUE and it worked as the way I expected. Sorry for not being clear about the problem in my mail. The characters are very much needed, cos I am trying to create a signaling network using Rgraphviz. Thanks again. Regards, kaarz -- View this message in context: http://r.789695.n4.nabble.com/Cbind-query-tp3006988p3008190.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Cbind query
I am new to R and request your kind help. I have a table like the one below, one two 1 apple fruit 2 ball game 3 chair wood 4 wood plain 5 fruitbanana 6 cloth silk Note: duplicate entries are there the task is to create relations to each each row entries, like apple - fruit . when I tried to combine column1 with column 2 (one, two), using cbind the string is changed to numerical value...something like this [,1] [,2] [1,] 10 53 [2,] 25 562 [3,] 25 462 [4,] 25 1045 [5,] 25 488 [6,] 26 1062 [7,] 27 951 [8,] 27 144 [9,] 27 676 [10,] 27 486 Please suggest me how to get the string names back like the first table in the out put, using cbind. Thanks in advance regards kaarz -- View this message in context: http://r.789695.n4.nabble.com/Cbind-query-tp3006988p3006988.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cbind query
This comes about since when using read.table (which I assume you did, but you did not show us what commands you were using), characters are converted to factors. If you are not using factors, then you probably want the data read in as characters. You should understand the use of 'str' to look at the structure of your objects to understand what might be happening. See example below: tab - read.table(textConnection( one two + 1 apple fruit + 2 ball game + 3 chair wood + 4 wood plain + 5 fruitbanana + 6 cloth silk)) closeAllConnections() str(tab) # note the you have 'factors' 'data.frame': 6 obs. of 2 variables: $ one: Factor w/ 6 levels apple,ball,..: 1 2 3 6 5 4 $ two: Factor w/ 6 levels banana,fruit,..: 2 3 6 4 1 5 cbind(tab$one, tab$two) # this gives numeric values of the factors [,1] [,2] [1,]12 [2,]23 [3,]36 [4,]64 [5,]51 [6,]45 # now read in data and not convert to factors (note: as.is=TRUE) tab - read.table(textConnection( one two + 1 apple fruit + 2 ball game + 3 chair wood + 4 wood plain + 5 fruitbanana + 6 cloth silk), as.is = TRUE) closeAllConnections() str(tab) # now you have characters 'data.frame': 6 obs. of 2 variables: $ one: chr apple ball chair wood ... $ two: chr fruit game wood plain ... cbind(tab$one, tab$two) # this gives character values [,1][,2] [1,] apple fruit [2,] ball game [3,] chair wood [4,] wood plain [5,] fruit banana [6,] cloth silk On Fri, Oct 22, 2010 at 7:06 AM, karthicklakshman karthick.laksh...@gmail.com wrote: I am new to R and request your kind help. I have a table like the one below, one two 1 apple fruit 2 ball game 3 chair wood 4 wood plain 5 fruit banana 6 cloth silk Note: duplicate entries are there the task is to create relations to each each row entries, like apple - fruit . when I tried to combine column1 with column 2 (one, two), using cbind the string is changed to numerical value...something like this [,1] [,2] [1,] 10 53 [2,] 25 562 [3,] 25 462 [4,] 25 1045 [5,] 25 488 [6,] 26 1062 [7,] 27 951 [8,] 27 144 [9,] 27 676 [10,] 27 486 Please suggest me how to get the string names back like the first table in the out put, using cbind. Thanks in advance regards kaarz -- View this message in context: http://r.789695.n4.nabble.com/Cbind-query-tp3006988p3006988.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind
Is there a way to rename the columns to something like A and B in the cbind function? x - rnorm(n = 10, mean = 0, sd = 1) y - rnorm(n = 10, mean = 0, sd = 1) cbind(x,y) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind
cbind(A=x, B=y) On Sat, Aug 21, 2010 at 6:53 PM, r.ookie r.oo...@live.com wrote: Is there a way to rename the columns to something like A and B in the cbind function? x - rnorm(n = 10, mean = 0, sd = 1) y - rnorm(n = 10, mean = 0, sd = 1) cbind(x,y) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind
On 2010-08-21 16:53, r.ookie wrote: Is there a way to rename the columns to something like A and B in the cbind function? x- rnorm(n = 10, mean = 0, sd = 1) y- rnorm(n = 10, mean = 0, sd = 1) cbind(x,y) Unless I completely misunderstand your query, ?cbind tells you: ... vectors or matrices. These can be given as named arguments. Surely, you checked? -Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind
Thanks. On Aug 21, 2010, at 4:01 PM, RICHARD M. HEIBERGER wrote: cbind(A=x, B=y) On Sat, Aug 21, 2010 at 6:53 PM, r.ookie r.oo...@live.com wrote: Is there a way to rename the columns to something like A and B in the cbind function? x - rnorm(n = 10, mean = 0, sd = 1) y - rnorm(n = 10, mean = 0, sd = 1) cbind(x,y) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] CBIND / MERGE two time series objects along time (overlapping indices, redundant data)
Hi there, I need to merge/bind two time series objects (from RPackage: timeSeries) by column. The theory is laid out nicely, even for overlapping indices. In my example, I have overlapping indices (01.01.2001), where in one time series I have one data point and in the other redundant data. Default usage of merge and cbind would lead to the result that the one data point would be replicated (see example below). However, I do not like the replicating feature and would like to have the NA's filled instead. Desired output, would be: TS.1 SS.2 2001-01-01 23.1 23.1 2001-01-01 NA 23.4 Is there a slick way to produce above result? Thank you for your answer - help is really appreciated. Kind regards, Bernhard. --- # R Console example: require(timeSeries) date -timeDate(01.01.2001, format = %d.%m.%Y) test1 - timeSeries(c(23.1), charvec = date) test2 - timeSeries(c(23.1, 23.4), charvec = c(date, date)) colnames(test2) - SS.2 merge(test1, test2) GMT TS.1 SS.2 2001-01-01 23.1 23.1 2001-01-01 23.1 23.4 cbind(test1, test2) Fehler in cbind(deparse.level, ...) : number of rows must match cbind(test2, test1) GMT SS.2 TS.1 2001-01-01 23.1 23.1 2001-01-01 23.4 23.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind in for loops
I have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-in-for-loops-tp2285690p2285690.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in for loops
Hi,
Assuming that you have read the files into R,
and that their names (in R) are held in some object
(e.g., 'file2'), then this works
do.call(what = cbind, args = mget(x = file2, envir = .GlobalEnv)
Here is a reproducible example:
x1 - data.frame(x = 1:10)
x2 - data.frame(y = 1:10)
file.names - c(x1, x2)
do.call(cbind, mget(file.names, envir=.GlobalEnv))
Best regards,
Josh
On Sun, Jul 11, 2010 at 9:08 PM, jd6688 jdsignat...@gmail.com wrote:
I have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-in-for-loops-tp2285690p2285690.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in for loops
On Jul 12, 2010, at 2:32 AM, Joshua Wiley wrote:
Hi,
Assuming that you have read the files into R,
and that their names (in R) are held in some object
(e.g., 'file2'), then this works
do.call(what = cbind, args = mget(x = file2, envir = .GlobalEnv)
Here is a reproducible example:
x1 - data.frame(x = 1:10)
x2 - data.frame(y = 1:10)
file.names - c(x1, x2)
do.call(cbind, mget(file.names, envir=.GlobalEnv))
If you want to automatically retrieve an unknown or large number of
files from the current environment you can use grep with ls
ls()[grep(^v, ls()) ]
[1] v1 v2 v3 v4 varname
I don't have any objects named ...out3 but you might try the above
form with your pattern.
--
DAvid.
Best regards,
Josh
On Sun, Jul 11, 2010 at 9:08 PM, jd6688 jdsignat...@gmail.com wrote:
I have 30 files in the current directories, i would like to perform
the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-in-for-loops-tp2285690p2285690.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind in for loops
Try this:
do.call(cbind, lapply(dir(pattern = '.out3$'), read.table))
On Mon, Jul 12, 2010 at 1:08 AM, jd6688 jdsignat...@gmail.com wrote:
I have 30 files in the current directories, i would like to perform the
cbind(fil1,file2,file3,file4file30)
how could i do this in a for loop:
such as:
file2 - list.files(pattern=.out3$)
for (j in file2) {
cbind(j)...how to implement cbind here
}
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-in-for-loops-tp2285690p2285690.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind with vectors of different lengths?
I wrote a function that cbinds vectors of different lengths. Basically, the
function adds NAs to shorter vectors and cbinds in the end. I'm attaching
the code for guidance.
# This function takes in a list() of vectors and cbinds them into a
data.frame.
timerMelt - function(x, write.down = FALSE, filename = output) {
stopifnot(is.list(x))
filename - paste(filename, .txt, sep = )
len1 - length(x[[1]])
len2 - length(x[[2]])
len3 - length(x[[3]])
max.len - max(c(len1, len2, len3))
x.cat1 - data.frame(rep(NA, max.len))
x.cat2 - data.frame(rep(NA, max.len))
x.cat3 - data.frame(rep(NA, max.len))
if (len1 max.len) {
x.cat1[1:len1,] - data.frame(x[[1]])
} else {
x.cat1 - data.frame(x[[1]])
}
if (len2 max.len) {
x.cat2[1:len2,] - data.frame(x[[2]])
} else {
x.cat2 - data.frame(x[[2]])
}
if (len3 max.len) {
x.cat3[1:len3,] - data.frame(x[[3]])
} else {
x.cat3 - data.frame(x[[3]])
}
result - cbind(x.cat1, x.cat2, x.cat3)
names(result) - c(s, p, r)
if (write.down == TRUE) {
write.table(result, filename, row.names = FALSE)
}
return(result)
}
Cheers,
Roman
--
View this message in context:
http://r.789695.n4.nabble.com/cbind-with-vectors-of-different-lengths-tp2249680p2250025.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] cbind with vectors of different lengths?
Hello R help I have a dataframe, with 71 samples (rows) and 30 variables. I got linear models for some of the variables, and I want to join fitted and residuals of these models to the data frame. Sometimes, these vectors have the same length of the dependant variable, but in a few cases, NA values can be found on my data, and therefore, both fitted and residuals have a few rows less than the original data frame. As I try cbind, R answers with error, because both vectors have different lenghts. I have tried with merge but... suddenly I had a lot of rows of repeated values. I think (with my small idea of R and the manuals and helps I did read) that first, I have to force residuals and fitted of my model to be a data frame. as.data.frame(fitted(lm)) this gives, for example [1] 1.1 [3] 3.8 [4] 1.3 [5] 0.9 instead of the original fitted(lm) 1 3 4 5 1.1 3.8 1.3 0.9 that I want to join to my data frame [1] 17.0 [2] 15.2 [3] 17.3 [4] 15.0 [5] 17.4 as you can see, row 2 does not exist in fitted... how can I tell R to leave this row as NA, as below? [1] 17.0 1.1 [2] 15.2 NA [3] 17.3 3.8 [4] 15.0 1.3 [5] 17.4 0.9 thanks and greets. Arantzazu Blanco Bernardeau Dpto de Química Agrícola, Geología y Edafología Universidad de Murcia-Campus de Espinardo _ ¿Un navegador seguro buscando estás? ¡Protegete ya en www.ayudartepodria.com! www.ayudartepodria.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind and automatic type conversion
On 01.05.2010 21:09, Giovanni Azua wrote: Hello, I have three method types and 100 generalization errors for each, all in the range [0.65,0.81]. I would like to make a stacked histogram plot using ggplot2 with this data ... Therefore I need a data frame of the form e.g. Method GE -- -- Classic0.76 Classic0.79 Own Bootstrap 0.81 Own Bootstrap 0.79 R Bootstrap0.71 R Bootstrap0.75 So I combine the data in the following way: normalerrors- rbind(cbind(rep(Classic,S),classicge[,1]), cbind(rep(Own Bootstrap,S),ownge[,1]),cbind(rep(R Bootstrap,S),rbootge[,1])) normalerrors- data.frame(method=factor(normalerrors[,1]),ge=normalerrors[,2]) But doing it in this way my GE coefficients get automatically converted to string type ... how can I avoid this conversion when doing the cbind? Not at all, since cbind() constructs a matrix which is of exactly one type. You probably want to construct the data.frame directly as in labels - c(Classic, Own Bootstrap, R Bootstrap) normalerrors - data.frame( method = gl(length(labels), S, labels = labels), ge = c(classicge[,1], ownge[,1], rbootge[,1])) Best, Uwe Ligges TIA, Best regards, Giovanni [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind and automatic type conversion
Hello, I have three method types and 100 generalization errors for each, all in the range [0.65,0.81]. I would like to make a stacked histogram plot using ggplot2 with this data ... Therefore I need a data frame of the form e.g. Method GE -- -- Classic0.76 Classic0.79 Own Bootstrap 0.81 Own Bootstrap 0.79 R Bootstrap0.71 R Bootstrap0.75 So I combine the data in the following way: normalerrors - rbind(cbind(rep(Classic,S),classicge[,1]), cbind(rep(Own Bootstrap,S),ownge[,1]),cbind(rep(R Bootstrap,S),rbootge[,1])) normalerrors - data.frame(method=factor(normalerrors[,1]),ge=normalerrors[,2]) But doing it in this way my GE coefficients get automatically converted to string type ... how can I avoid this conversion when doing the cbind? TIA, Best regards, Giovanni [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] cbind, row names
Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, x and y: x - 1:20 y - 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Could you please help? Thank you Sören __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, row names
This gives what you want: rbind.data.frame( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) On Fri, Jan 29, 2010 at 8:49 AM, soeren.vo...@eawag.ch wrote: Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, x and y: x - 1:20 y - 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Could you please help? Thank you Sören __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cbind, row names
Hi! 29.01.2010 12:49, soeren.vo...@eawag.ch wrote: Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, x and y: x - 1:20 y - 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Maybe because the cbinds in your code produce matrices: is.matrix(cbind(N=length(x), M=mean(x), SD=sd(x))) [1] TRUE Quote ?rbind: For cbind (rbind) the column (row) names are taken from the colnames (rownames) of the arguments if these are matrix-like. HTH, Kimmo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
