Re: [R] tcltk lost after update r-base 2.11.1-2hardy0 to 2.11.1-5hardy0
On 09/06/2010 07:38 AM, Christian Lederer wrote: Dear R-Users, after the last upgrade from r-base 2.11.1-2hardy0 to 2.11.1-5hardy0 i lost the tlctk package. I this a general problem under Ubuntu Hardy or should i search for a configuration error on my system? Assuming that you didn't just msispell it (as you did above), I'd suspect a build configuration error on the Ubuntu side. If you file a bug report with them, you'll likely get a quicker resonse than from people on this list (yea, let's all go see if we can break our productions systems like Christian did...) In any case, people do need to know what the symptoms are, and what you did to upgrade? -- Peter Dalgaard Center for Statistics, Copenhagen Business School 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] How to use lm() output for systemfit() 'Seemingly unrelated regression'
Thanks, it works fine now. -- View this message in context: http://r.789695.n4.nabble.com/How-to-use-lm-output-for-systemfit-Seemingly-unrelated-regression-tp2525418p2527946.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] representing NULL values in a vector
Hi, I have a vector who contents should look like this, c d NULL e f etc or 4 5 6 NULL 7 8 9 how can I represent the null value? [[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] representing NULL values in a vector
Perhaps you mean NA rather than NULL. If NA is not what you want, then I think you'll need to explain your application. On 06/09/2010 06:00, raje...@cse.iitm.ac.in wrote: Hi, I have a vector who contents should look like this, c d NULL e f etc or 4 5 6 NULL 7 8 9 how can I represent the null value? [[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. -- Patrick Burns pbu...@pburns.seanet.com http://www.burns-stat.com (home of 'Some hints for the R beginner' and 'The R Inferno') __ 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] how do I transform this to a for loop
Hi Bill, I didn't make the original post, but its pretty similar to some thing i would have queried the list about. But, as an R dilatante i find more curious your question- ...but why would you want to do so? Is this because you'd typically use the given nine lines of explicit code to carve up a single dataset into nine symmetrical variants ? Or that some contextual information may affect how you would write the for() loop? As i lack the experience to know any better, i perceive your for() loop as de rigour in efficient use of R, and the preferance of all experienced R user's. But not having any formal education in R or role models as such, its only an assumption (compeletely ignoring for the moment processing efficiency/speed, rounding error and such). But which i now question! Explicit, simple crude looking code; or, something which demands a little more proficiency with the language? cheers, Karl On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote: sseq- c(1, seq(5, 40, by = 5)) for(i in 1:length(sseq)) assign(paste(arima, i, sep=), arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1))) ...but why would you want to do so? -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of lord12 Sent: Monday, 6 September 2010 10:57 AM To: r-help@r-project.org Subject: [R] how do I transform this to a for loop arima1- arima(data.ts[1:201], order = c(1,1,1)) arima2- arima(data.ts[5:205], order = c(1,1,1)) arima3- arima(data.ts[10:210], order = c(1,1,1)) arima4- arima(data.ts[15:215], order = c(1,1,1)) arima5- arima(data.ts[20:220], order = c(1,1,1)) arima6- arima(data.ts[25:225], order = c(1,1,1)) arima7- arima(data.ts[30:230], order = c(1,1,1)) arima8- arima(data.ts[35:235], order = c(1,1,1)) arima9- arima(data.ts[40:240], order = c(1,1,1)) -- Karl Brand k.br...@erasmusmc.nl Department of Genetics Erasmus MC Dr Molewaterplein 50 3015 GE Rotterdam P +31 (0)10 704 3409 | F +31 (0)10 704 4743 | M +31 (0)642 777 268 __ 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] representing NULL values in a vector
On 09/06/2010 03:00 PM, raje...@cse.iitm.ac.in wrote: Hi, I have a vector who contents should look like this, c d NULL e f etc or 4 5 6 NULL 7 8 9 how can I represent the null value? Hi rajesh, For character vectors, will probably suffice, but for numbers, you are probably stuck with NA. Jim __ 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] replacing functions
Dear All, Is it possible to replace function with my own? I want to apply pca clustering, but to use some strange correlation function. I'm asking about replacing, say, mean() with new content of mean() and use standard other functions, which might use mean() as part. karsar __ 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] how do I transform this to a for loop
Hi Karl,
The why do it like this is probably direct towards creating 9 new
objects for the arima results (Is this right Bill?). A better option
would be to create a list with nine entries. This is much easier for any
subsequent analyses. An example that uses lapply (an efficient syntax
for loops):
sseq - c(1, seq(5, 40, by = 5))
result_list = lapply(sseq, function(num) {
arima(data.ts[num:(num+200)], order=c(1,1,1))
})
cheers,
Paul
On 09/06/2010 10:46 AM, Karl Brand wrote:
Hi Bill,
I didn't make the original post, but its pretty similar to some thing
i would have queried the list about. But, as an R dilatante i find
more curious your question-
...but why would you want to do so?
Is this because you'd typically use the given nine lines of explicit
code to carve up a single dataset into nine symmetrical variants ? Or
that some contextual information may affect how you would write the
for() loop?
As i lack the experience to know any better, i perceive your for()
loop as de rigour in efficient use of R, and the preferance of all
experienced R user's. But not having any formal education in R or role
models as such, its only an assumption (compeletely ignoring for the
moment processing efficiency/speed, rounding error and such).
But which i now question! Explicit, simple crude looking code; or,
something which demands a little more proficiency with the language?
cheers,
Karl
On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:
sseq- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste(arima, i, sep=),
arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))
...but why would you want to do so?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of lord12
Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop
arima1- arima(data.ts[1:201], order = c(1,1,1))
arima2- arima(data.ts[5:205], order = c(1,1,1))
arima3- arima(data.ts[10:210], order = c(1,1,1))
arima4- arima(data.ts[15:215], order = c(1,1,1))
arima5- arima(data.ts[20:220], order = c(1,1,1))
arima6- arima(data.ts[25:225], order = c(1,1,1))
arima7- arima(data.ts[30:230], order = c(1,1,1))
arima8- arima(data.ts[35:235], order = c(1,1,1))
arima9- arima(data.ts[40:240], order = c(1,1,1))
--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone: +3130 253 5773
http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] replacing functions
Hi Karsar, To replace mean you can make a new function with the same name: l = runif(10) mean(l) mean = function(x) return(1) mean(l) But there must be a better way... cheers, Paul On 09/06/2010 11:52 AM, Karen Sargsyan wrote: Dear All, Is it possible to replace function with my own? I want to apply pca clustering, but to use some strange correlation function. I'm asking about replacing, say, mean() with new content of mean() and use standard other functions, which might use mean() as part. karsar __ 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. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 253 5773 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 __ 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] how do I transform this to a for loop
Hi Karl,
I think the question here is why would you want to create different
objects in the loop using assign().
Usually, using lists is better (more efficient?), although I sometimes
use assign() too in this context. I do it when I want to export the
object as separate files (xls, Rbin, svg, etc). If I don't need separate
files, I use lists.
I'm no expert so I'm not even sure I use the correct approach, but it
might help you get a better understanding.
It would be something like:
sseq- c(1, seq(5, 40, by = 5))
arima - vector(mode=list, length=length(sseq)) ##not sure it is
necessary, but it might be
for(i in 1:length(sseq)) {
arima[[i]] - arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1))
}
HTH,
Ivan
Le 9/6/2010 10:46, Karl Brand a écrit :
Hi Bill,
I didn't make the original post, but its pretty similar to some thing
i would have queried the list about. But, as an R dilatante i find
more curious your question-
...but why would you want to do so?
Is this because you'd typically use the given nine lines of explicit
code to carve up a single dataset into nine symmetrical variants ? Or
that some contextual information may affect how you would write the
for() loop?
As i lack the experience to know any better, i perceive your for()
loop as de rigour in efficient use of R, and the preferance of all
experienced R user's. But not having any formal education in R or role
models as such, its only an assumption (compeletely ignoring for the
moment processing efficiency/speed, rounding error and such).
But which i now question! Explicit, simple crude looking code; or,
something which demands a little more proficiency with the language?
cheers,
Karl
On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:
sseq- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste(arima, i, sep=),
arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))
...but why would you want to do so?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of lord12
Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop
arima1- arima(data.ts[1:201], order = c(1,1,1))
arima2- arima(data.ts[5:205], order = c(1,1,1))
arima3- arima(data.ts[10:210], order = c(1,1,1))
arima4- arima(data.ts[15:215], order = c(1,1,1))
arima5- arima(data.ts[20:220], order = c(1,1,1))
arima6- arima(data.ts[25:225], order = c(1,1,1))
arima7- arima(data.ts[30:230], order = c(1,1,1))
arima8- arima(data.ts[35:235], order = c(1,1,1))
arima9- arima(data.ts[40:240], order = c(1,1,1))
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
[[alternative HTML version deleted]]
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[R] Finding the two most recent dates
Dear R help, I have the following data frame: structure(list(prochi = c(ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1), date_1st_event = structure(c(14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784 ), class = Date), bp_date = structure(c(12660, 14571, 13392, 13080, 12012, 13080, 13894, 14622, 12654, 13894), class = Date), SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L, 151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L, 91L)), .Names = c(prochi, date_1st_event, bp_date, SBP, DBP), row.names = 108:117, class = data.frame) It consists of repeated measures for the same individual. What I want to do is find the two most recent blood pressure readings (SBP and DBP) using date_1st_event and bp_date. What I would do to find the most recent date is to subtract date_1st_event-bp_date and then aggregate by min. I'm not sure how to find the two most recent dates. Are there some functions that can help me or will I have to write a function from scratch. Any help just to point me in the right direction. Thanks, Natalie -- View this message in context: http://r.789695.n4.nabble.com/Finding-the-two-most-recent-dates-tp2528185p2528185.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] R time series analysis
lord12 wrote: I have a data file with a given time series of price data and I would like to split the time series into a test set and training set. I would then like to build an ARIMA model on the training set and apply this model on test set. I had recently the same problem and, after checking documentation and mailing list archives, I discovered that it is not possible to apply the same model on a different data set. Of course you can create the model on a part of the dataset and then check the prediction with the remaining part, as a testing set. But, if you have new data you and you want to apply the same model on them...nothing! I checked the source code of ARIMA functions but it was too complex and I hadn't enough time to learn all that stuff. However I found a little workaround: 1. I calibrate the model on the training part 2. I create a new model with the same parameters, using fixed (check arima documentation) on the new data 3. go to step 2. every time you have new data It worked for me. -- View this message in context: http://r.789695.n4.nabble.com/R-time-series-analysis-tp2527513p2528200.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] representing NULL values in a vector
On 06/09/2010 1:00 AM, raje...@cse.iitm.ac.in wrote: Hi, I have a vector who contents should look like this, c d NULL e f etc or 4 5 6 NULL 7 8 9 how can I represent the null value? As others have said, you probably want NA rather than NULL. If you really want NULL, then use a list (a generic vector). So x - list(c, d, NULL, e, f) y - list(4,5,6,NULL, 7,8,9) You need to be careful when setting values, because y[[1]] - NULL will *remove* element 1, not set it to NULL. To set it to NULL, use y[1] - list(NULL) Duncan Murdoch __ 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] size limit of string/parse a string and convert to vector
Hi,
I have a loop as follows,
dataStr - character(0)
repeat{
fstr-read.socket(sockfd)
if(fstr==)
break
dataStr-paste(dataStr,fstr)
}
at what point does dataStr stop accepting(gets full)? I'm sending millions of
records over the socket and need to know if all of it can go into dataStr.
Also, Incase all of it cannot go into dataStr, I need to parse each
read.socket. In such a case,
I have a string as follows,
|1,ab,2.34|2,cd,3.44| how can I parse this to become a list of 2 string
vectors, namely,
list(c(1,ab,2.34),c(2,cd,3.44))
Any help is appreciated
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[R] Strange behavior of interval values in optimize()
Hi all,
I'm using optimize() to find the minimum of the following function f, and
minimize it (without
f-function(delta,P,U){
minimiz-P+delta*U
x-minimiz[1]
y-minimiz[2]
z-100*(y-x^2)^2+(1-x)^2
return(z)
}
result-optimize(f, interval=c(-1, 1), P=c(0.99,1.01), U=c(1,0))
result
Nothing unexpected in the output so far:
result-optimize(f, interval=c(-1, 1), P=c(0.99,1.01), U=c(1,0))
result
$minimum
[1] 0.01498144
$objective
[1] 2.482991e-05
But, when I choose a larger Interval in the optimization method:
result-optimize(f, interval=c(-100, 10), P=c(0.99,1.01), U=c(1,0))
result
result-optimize(f, interval=c(-100, 10), P=c(0.99,1.01), U=c(1,0))
result
$minimum
[1] -1.989997
$objective
[1] 4.01
The result gets worse (even though the old interval is included in the new one).
In fact I don't want any restrictons for the interval values (something like
interval=(-Inf, Inf) or at least the interval should be as large as possible.
Does anyone know this strange behavior? Where does it come from?
Thank you,
Fabian
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Re: [R] representing NULL values in a vector
NA is good.thanks - Original Message - From: Patrick Burns pbu...@pburns.seanet.com To: r-help@r-project.org, raje...@cse.iitm.ac.in Sent: Mon, 06 Sep 2010 13:55:34 +0530 (IST) Subject: Re: [R] representing NULL values in a vector Perhaps you mean NA rather than NULL. If NA is not what you want, then I think you'll need to explain your application. On 06/09/2010 06:00, raje...@cse.iitm.ac.in wrote: Hi, I have a vector who contents should look like this, c d NULL e f etc or 4 5 6 NULL 7 8 9 how can I represent the null value? [[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. -- Patrick Burns pbu...@pburns.seanet.com http://www.burns-stat.com (home of 'Some hints for the R beginner' and 'The R Inferno') [[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] how to cluster vectors of factors
If I understand you correctly and each factor consists of binary data, you may want to check out monothethic analysis, available in the package 'cluster'. For a simple example and short description of the method to get you started, just type in: require(cluster) ?mona As far as i know there's nothing strictly theoretically invalid in using hierarchical clustering on binary data, and you may want to see how results may differ by trying the example data (and your own data) with other clustering methods. For example, compare: require(cluster) data(animals) plot(mona(animals)) plot(agnes(animals)) 2010/9/2 tueken hannat_tue...@hotmail.com Hello all I wonder what can i use to cluster vectors which composed of several factors. lets say around 30 different factors compose a vector, and if the factor is present then it encoded as 1, if not presented then it will be encoded as 0. I was thinking of using hierarchical clustering, as i know the distance between two vector were calculated through euclidean distance function, but i dont think this distance would be correct to separate the data, cause two vector with different composition, could end up having similar distance to another vector. hope someone could give me some clue! -- View this message in context: http://r.789695.n4.nabble.com/how-to-cluster-vectors-of-factors-tp2514654p2514654.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. [[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] Creating named.list from two matrix columns
Hi Friends, I am new to R. On R utility class pages, creating named.list is described with this command : new(named.list,a=1,b=2) For large matrix having two columns, such as : row1 2334 row2 347 row3 379 ... I want to create a named.list like : $row1 [1] 2334 $row2 [1] 347 ... Can anyone explain how named.list variable can be created by using two specified columns of a dataframe or matrix object, where one of the two columns is assigned as a name (string) and other as its corresponding value ? Thanks [[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] Finding the two most recent dates
Hi Natalie, By far the easiest thing to do is to convert the date to a special date class. See as.POSIXct for example. I'm not sure that 14784 means, nor what the data says in the bp_date column. Probably the two combine into a specific date? Once you've converted the columns into a POSIXct object, you can use the min() function to find the minimum. cheers, Paul On 09/06/2010 12:45 PM, Newbie19_02 wrote: Dear R help, I have the following data frame: structure(list(prochi = c(ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1, ind_1), date_1st_event = structure(c(14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784 ), class = Date), bp_date = structure(c(12660, 14571, 13392, 13080, 12012, 13080, 13894, 14622, 12654, 13894), class = Date), SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L, 151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L, 91L)), .Names = c(prochi, date_1st_event, bp_date, SBP, DBP), row.names = 108:117, class = data.frame) It consists of repeated measures for the same individual. What I want to do is find the two most recent blood pressure readings (SBP and DBP) using date_1st_event and bp_date. What I would do to find the most recent date is to subtract date_1st_event-bp_date and then aggregate by min. I'm not sure how to find the two most recent dates. Are there some functions that can help me or will I have to write a function from scratch. Any help just to point me in the right direction. Thanks, Natalie -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 253 5773 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 __ 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] Two images functions
Hello everyone. I would like to ask you what happens when two functions with the same name exist. I discovered this today when I wrote ?images (I was trying to understand how it works) ?images gave me the following output: Help on topic 'image' was found in the following packages: Image (in package raster in library /home/apa/R/x86_64-unknown-linux-gnu-library/2.11) Display a Color Image (in package graphics in library /usr/lib64/R/library) How can I be sure which function is called when I write Image(x,y,f) or Image(f) I would like to thank you in advance for your help. Best Regards Alex [[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] anova of glm output
francogrex wrote: out - glm(response~Var1+Var2+Var3..,family=binomial,data=mydata) summary(out) stepAIC(out) anova(out, test='Chisq') I understand that stepAIC is used to select the model with the lowest AIC (the best model) but can someone explain what is the purpose of doing the anova: anova(out, test='Chisq')? What extra information does it bring? Thanks out is of class glm (and lm, if that matters). So we have anova working on a glm class, which is documented in anova.glm. You should be aware that this anova depends on the order of you Varx, so be cautious. Dieter -- View this message in context: http://r.789695.n4.nabble.com/anova-of-glm-output-tp2528336p2528354.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] Creating named.list from two matrix columns
Is this what you want: x V1 V2 1 row1 2334 2 row2 347 3 row3 379 x.list - as.list(x$V2) names(x.list) - x$V1 x.list $row1 [1] 2334 $row2 [1] 347 $row3 [1] 379 On Mon, Sep 6, 2010 at 7:55 AM, Viki S is...@live.com wrote: Hi Friends, I am new to R. On R utility class pages, creating named.list is described with this command : new(named.list,a=1,b=2) For large matrix having two columns, such as : row1 2334 row2 347 row3 379 ... I want to create a named.list like : $row1 [1] 2334 $row2 [1] 347 ... Can anyone explain how named.list variable can be created by using two specified columns of a dataframe or matrix object, where one of the two columns is assigned as a name (string) and other as its corresponding value ? Thanks [[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 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] dataframe row names from list
Hi, I have a list which looks like this... str(y) List of 10 $ : chr [1:4] ABCD 5 0 1 $ : chr [1:4] DEF 15 1 16 $ : chr [1:4] AAA 2 17 8 $ : chr [1:4] SSS 15 25 1 $ : chr [1:4] III 15 26 4 $ : chr [1:4] OPQ 7 30 4 $ : chr [1:4] TYR 14 34 8 $ : chr [1:4] IRTS 15 42 1 $ : chr [1:4] LLL 15 43 2 $ : chr [1:4] AQW 3 45 4 I need to create a dataframe whose row names are chr[1] of each vector..ie ABCD,DEF,AAA ETC. how can I do this? [[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] replacing functions
Karen Sargsyan wrote: Is it possible to replace function with my own? I want to apply pca clustering, but to use some strange correlation function. I'm asking about replacing, say, mean() with new content of mean() and use standard other functions, which might use mean() as part. The usual way would be to get the source code for the function you are trying to change (or, fast way, with getAnywhere(pca)), copy it to an editor, and make the changes. However, better always rename it before using, e.g. pcaKaren - function If the code you are trying to change is not directly called, but indirectly, things are more complicated. In most cases, it's easiest to get the whole package, put it into a different namespace, and make the changes. Easiest does not mean not easy, though. Dieter -- View this message in context: http://r.789695.n4.nabble.com/replacing-functions-tp2528143p2528361.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] inserting a vector as a row in a data.frame
Hi, is it possible to insert a vector as a row in a data.frame? [[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] Strange behavior of interval values in optimize()
Michael Bernsteiner wrote: I'm using optimize() to find the minimum of the following function f, and minimize it (without . But, when I choose a larger Interval in the optimization method: The result gets worse (even though the old interval is included in the new one). In fact I don't want any restrictons for the interval values (something like interval=(-Inf, Inf) or at least the interval should be as large as possible. The most likely cause is a second minimum. Dieter -- View this message in context: http://r.789695.n4.nabble.com/Strange-behavior-of-interval-values-in-optimize-tp2528208p2528366.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] anova of glm output
Hi, this is more related to understanding some statistics while using R; I've see such output in a paper: out - glm(response~Var1+Var2+Var3..,family=binomial,data=mydata) summary(out) stepAIC(out) anova(out, test='Chisq') I understand that stepAIC is used to select the model with the lowest AIC (the best model) but can someone explain what is the purpose of doing the anova: anova(out, test='Chisq')? What extra information does it bring? Thanks -- View this message in context: http://r.789695.n4.nabble.com/anova-of-glm-output-tp2528336p2528336.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] How to get mypkg-manual.pdf
I am building a package say mypkg. Five months ago, when I built the package I got the mypkg manual in pdf format. Today, after making updates, I build the same package, same name, and steps; unfortunately I do not get the manual in pdf format. Rather I get the following message: cd: can't cd to /cygdrive/c/Documents saving output to 'mypkg-manual.pdf' ...Done Could any one help me on how to get the mypkg-manual.pdf. Thank you in advance for your help. JN [[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] size limit of string/parse a string and convert to vector
try this:
x - |1,ab,2.34|2,cd,3.44|
# split by the | and remove vectors of zero characters
x.sp - strsplit(x, '|', fixed = TRUE)[[1]]
x.sp - x.sp[nchar(x.sp) 0]
# now split by comma
x.comma - strsplit(x.sp, ',')
# you can now access you data
x.comma
[[1]]
[1] 1ab 2.34
[[2]]
[1] 2cd 3.44
On Mon, Sep 6, 2010 at 6:06 AM, raje...@cse.iitm.ac.in
raje...@cse.iitm.ac.in wrote:
Hi,
I have a loop as follows,
dataStr - character(0)
repeat{
fstr-read.socket(sockfd)
if(fstr==)
break
dataStr-paste(dataStr,fstr)
}
at what point does dataStr stop accepting(gets full)? I'm sending millions of
records over the socket and need to know if all of it can go into dataStr.
Also, Incase all of it cannot go into dataStr, I need to parse each
read.socket. In such a case,
I have a string as follows,
|1,ab,2.34|2,cd,3.44| how can I parse this to become a list of 2 string
vectors, namely,
list(c(1,ab,2.34),c(2,cd,3.44))
Any help is appreciated
[[alternative HTML version deleted]]
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+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] dataframe row names from list
Hi! I'm sure there's an easier way, but that works for me: test_list - list(c(ABC,5,0), c(DEF,10,1)) ##just a part of your example, think about using dput() to create a copy/pastable example test_df - t(as.data.frame(test_list)[-1,]) rownames(test_df) - t(as.data.frame(test_list)[1,]) HTH, Ivan Le 9/6/2010 13:41, raje...@cse.iitm.ac.in a écrit : Hi, I have a list which looks like this... str(y) List of 10 $ : chr [1:4] ABCD 5 0 1 $ : chr [1:4] DEF 15 1 16 $ : chr [1:4] AAA 2 17 8 $ : chr [1:4] SSS 15 25 1 $ : chr [1:4] III 15 26 4 $ : chr [1:4] OPQ 7 30 4 $ : chr [1:4] TYR 14 34 8 $ : chr [1:4] IRTS 15 42 1 $ : chr [1:4] LLL 15 43 2 $ : chr [1:4] AQW 3 45 4 I need to create a dataframe whose row names are chr[1] of each vector..ie ABCD,DEF,AAA ETC. how can I do this? [[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. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[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] max limit of list size and vector size?
It is easy to store a list of that size: x - list(1:1e6, 1:1e6, 1:1e6) object.size(x) 12000112 bytes str(x) List of 3 $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ... $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ... $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ... Now it really depends on how you are processing the data. If you are appending to a vector each time: myData - c(myData, newData) This is not efficient since it may be making copies of the vector each time to extend it. It is best to preallocate the vector and then store the result: myData - numeric(100) # hold 1 million values for (i in 1:1e6) myData[i] - newData My rule of thumb is that a single object should take up no more than 25% of the available RAM so that copies can be made during the processing. HTH On Mon, Sep 6, 2010 at 2:54 AM, raje...@cse.iitm.ac.in raje...@cse.iitm.ac.in wrote: Hi, Is it possible for me to store a list of vectors of 1 million entries? like, cc-list(c(1,2,1million),c(1,2,1million)) also what is the length of the longest string in R? I keep getting info from a socket and keep pasting on a string...when will this start becoming a problem? [[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 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.
Re: [R] inserting a vector as a row in a data.frame
Hi again, see ?rbind Ivan Le 9/6/2010 14:11, raje...@cse.iitm.ac.in a écrit : Hi, is it possible to insert a vector as a row in a data.frame? [[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. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[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] Correct coefficients from treatment contrasts?
On Sep 6, 2010, at 4:03 AM, B W wrote: Snipped out formatting detritus and added back many missing speces. -Hello,I am trying to take the information from the summary of my best fit logisticregression model for the occurrence of a high elevation plant spp. and create the appropriate equation that will calculate probability of occurrence, given the data. My predictors include both continuous variables (slope and a second orderpolynomial of elevation) and a discrete variable for aspect (warm and cool). I have left unchanged the default contrasts option, so I believe that thefollowing coefficients were created using treatment contrasts. My question how can I take this summary output and create the logistic equation that will allow me to calculate probability of occurrence. My interests are touse this to spatially display this info in a GIS environment. I think you should: -- Read the Posting Guide where you should learn that this is a plain text mailing list and that you need to change the configuration of your mail client. -- Read the help page and read other documentation regarding the use of the predict function. I have made adraft equation (shown below) that uses the coefficients from this summaryoutput, but this appears to be incorrect values always return zeroprobabilities. Presumably I need to adjust the values in some way but I am unclear as to how to proceed. Anyguidance would be appreciated! summary ( Call:glm(formula= Po ~ Slope + poly(Elevation, 2) + Aspect_2, family = quasibinomial) DevianceResiduals: Min 1Q Median 3Q Max -1.0532 -0.4167 -0.2760 -0.1823 3.3376 Coefficients: Estimate Std. Error t valuePr(| t|)(Intercept) -4.577707 0.222406 -20.583 2e-16 *** Slope 0.039959 0.003593 11.121 2e-16 *** poly(Elevation,2)1 8.050898 5.601956 1.437 0.1508 poly(Elevation,2)2 -37.694521 6.297806 -5.985 2.39e-09 *** Aspect_2w 0.429229 0.174760 2.456 0.0141 * --- You may get predictions at the original data points with: pred predict(model.Slope.Elevation.Aspect) (1/ (1 + exp(-1 * (-4.577707 + 0.039959*Slope + 8.050898 * poly(Elevation, 2)1 + -37.694521 * poly(Elevation, 2)2 + 0.429229* Aspect_2w) Brendan Wilson 2530 Alexis Road Shoreacres BC Canada V1N 4P6 Ph: 1.250.359.5905 [[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] extracting x,y coordinates from a contour plot
Thank you, David: I obviously didn't look hard enough. This is exactly what I need. Charles Annis, P.E. charles.an...@statisticalengineering.com 561-352-9699 http://www.StatisticalEngineering.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius Sent: Monday, September 06, 2010 12:43 AM To: charles.an...@statisticalengineering.com Cc: r-help@r-project.org Subject: Re: [R] extracting x,y coordinates from a contour plot On Sep 5, 2010, at 11:48 PM, Charles Annis, P.E. wrote: Requisite info: R version 2.11.1 (2010-05-31) running on a 64 bit HP Windows 7 machine. Doubt that makes much of a difference here. I have used contour() for several years. Now I would like to extract from a contour plot the x, y coordinates of a contour z=constant. This seems as though it would be straight-forward but I've been unsuccessful in my searches of CRAN. Suggest you read the help page for contour and the pages to which it links as well as working the examples. The answer is illustrated in the examples on that page. Can anyone provide a hint? x - 10*1:nrow(volcano) y - 10*1:ncol(volcano) xy160 - contourLines(x, y, volcano, nlevels=1, levels=160) str(xy160) List of 2 $ :List of 3 ..$ level: num 160 ..$ x: num [1:165] 110 108 105 102 103 ... ..$ y: num [1:165] 295 300 310 320 330 ... $ :List of 3 ..$ level: num 160 ..$ x: num [1:31] 270 263 262 260 260 ... ..$ y: num [1:31] 310 320 330 340 350 ... -- David. 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-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] nlme Output
Everyone - What do the NaN's mean here? Is this analysis a problem? Linear mixed-effects model fit by maximum likelihood Data: tmp.dat AIC BIClogLik 1611.251 1638.363 -797.6253 Random effects: Formula: ~1 | group_id (Intercept) Residual StdDev: 0.0003077668 9.236715 Fixed effects: AvgTrials ~ time + factor(group_id) + time * factor(group_id) Value Std.Error DF t-value p-value (Intercept)18.159722 3.576664 213 5.077279 0. time4.192708 1.655674 213 2.532327 0.0121 factor(group_id)2 -6.929563 5.235700 0 -1.323522 NaN factor(group_id)3 -1.654554 4.189575 0 -0.394922 NaN time:factor(group_id)2 1.729911 2.423658 213 0.713760 0.4762 time:factor(group_id)3 -2.555111 1.939396 213 -1.317478 0.1891 Correlation: (Intr) time fc(_)2 fc(_)3 t:(_)2 time -0.926 factor(group_id)2 -0.683 0.632 factor(group_id)3 -0.854 0.790 0.583 time:factor(group_id)2 0.632 -0.683 -0.926 -0.540 time:factor(group_id)3 0.790 -0.854 -0.540 -0.926 0.583 Standardized Within-Group Residuals: Min Q1Med Q3Max -1.8842754 -0.6979785 -0.3370998 0.5666704 3.0943948 Number of Observations: 219 Number of Groups: 3 Warning message: In pt(q, df, lower.tail, log.p) : NaNs produced [[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 on write.xlsx library(xlsx)
Hi Adrian,
dat=data.frame(matrix(0,3,3))
write.xlsx(dat,z:/dat.xlsx,sheetName=sheet1,append=F)
write.xlsx(dat,z:/dat.xlsx,sheetName=sheet2,append=F)
The above code works and creates new worksheets. But if I want to append
to an existing worksheet I seem to get an error.
write.xlsx(dat,z:/dat.xlsx,sheetName=sheet2,append=T) - This gives
an error saying The workbook already contains a sheet of this name
Could you please let me know how to append data to an existing sheet
and not create a new sheet each time?
Thank you
Ravi
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
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Re: [R] how do I transform this to a for loop
Hi Paul, Ivan,
Hartstikke bedankt and thanks alot for sharing these thoughts. I can see
'listing up' multiple symmetrical data sets makes a lot of sense. As
does using lapply() on them which i understand to be more
efficient/faster than for().
Goodo- with your concensus (and helpful examples) i can tell myself
investing the extra time to use lapply on lists /will/ pay off in the
long run vs. copying and pasting (nearly) the same line of code 10 times
for every data manipulation...
thanks again,
Karl
On 9/6/2010 12:09 PM, Paul Hiemstra wrote:
Hi Karl,
The why do it like this is probably direct towards creating 9 new
objects for the arima results (Is this right Bill?). A better option
would be to create a list with nine entries. This is much easier for any
subsequent analyses. An example that uses lapply (an efficient syntax
for loops):
sseq - c(1, seq(5, 40, by = 5))
result_list = lapply(sseq, function(num) {
arima(data.ts[num:(num+200)], order=c(1,1,1))
})
cheers,
Paul
On 09/06/2010 10:46 AM, Karl Brand wrote:
Hi Bill,
I didn't make the original post, but its pretty similar to some thing
i would have queried the list about. But, as an R dilatante i find
more curious your question-
...but why would you want to do so?
Is this because you'd typically use the given nine lines of explicit
code to carve up a single dataset into nine symmetrical variants ? Or
that some contextual information may affect how you would write the
for() loop?
As i lack the experience to know any better, i perceive your for()
loop as de rigour in efficient use of R, and the preferance of all
experienced R user's. But not having any formal education in R or role
models as such, its only an assumption (compeletely ignoring for the
moment processing efficiency/speed, rounding error and such).
But which i now question! Explicit, simple crude looking code; or,
something which demands a little more proficiency with the language?
cheers,
Karl
On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:
sseq- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste(arima, i, sep=),
arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))
...but why would you want to do so?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of lord12
Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop
arima1- arima(data.ts[1:201], order = c(1,1,1))
arima2- arima(data.ts[5:205], order = c(1,1,1))
arima3- arima(data.ts[10:210], order = c(1,1,1))
arima4- arima(data.ts[15:215], order = c(1,1,1))
arima5- arima(data.ts[20:220], order = c(1,1,1))
arima6- arima(data.ts[25:225], order = c(1,1,1))
arima7- arima(data.ts[30:230], order = c(1,1,1))
arima8- arima(data.ts[35:235], order = c(1,1,1))
arima9- arima(data.ts[40:240], order = c(1,1,1))
--
Karl Brand k.br...@erasmusmc.nl
Department of Genetics
Erasmus MC
Dr Molewaterplein 50
3015 GE Rotterdam
P +31 (0)10 704 3409 | F +31 (0)10 704 4743 | M +31 (0)642 777 268
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Re: [R] nlme Output
Dear Edward, You have no degrees of freedom left to estimate those p-values. Your design does not allows for the model your implemented. We need a brief summary of your design in order to help you further. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek team Biometrie Kwaliteitszorg Gaverstraat 4 9500 Geraardsbergen Belgium Research Institute for Nature and Forest team Biometrics Quality Assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Edward Patzelt Verzonden: maandag 6 september 2010 15:43 Aan: r-help@r-project.org Onderwerp: [R] nlme Output Everyone - What do the NaN's mean here? Is this analysis a problem? Linear mixed-effects model fit by maximum likelihood Data: tmp.dat AIC BIClogLik 1611.251 1638.363 -797.6253 Random effects: Formula: ~1 | group_id (Intercept) Residual StdDev: 0.0003077668 9.236715 Fixed effects: AvgTrials ~ time + factor(group_id) + time * factor(group_id) Value Std.Error DF t-value p-value (Intercept)18.159722 3.576664 213 5.077279 0. time4.192708 1.655674 213 2.532327 0.0121 factor(group_id)2 -6.929563 5.235700 0 -1.323522 NaN factor(group_id)3 -1.654554 4.189575 0 -0.394922 NaN time:factor(group_id)2 1.729911 2.423658 213 0.713760 0.4762 time:factor(group_id)3 -2.555111 1.939396 213 -1.317478 0.1891 Correlation: (Intr) time fc(_)2 fc(_)3 t:(_)2 time -0.926 factor(group_id)2 -0.683 0.632 factor(group_id)3 -0.854 0.790 0.583 time:factor(group_id)2 0.632 -0.683 -0.926 -0.540 time:factor(group_id)3 0.790 -0.854 -0.540 -0.926 0.583 Standardized Within-Group Residuals: Min Q1Med Q3Max -1.8842754 -0.6979785 -0.3370998 0.5666704 3.0943948 Number of Observations: 219 Number of Groups: 3 Warning message: In pt(q, df, lower.tail, log.p) : NaNs produced [[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. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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] Aggregating the matrices
Hello everyone. Say we have the following: a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010, c(ES, PT, Z , CF, GX, ST, EO))) b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT, CF, AT, EM, ST))) d - cbind(a, b) I want to calculate sums of the columns that have similar column names and then output this summary What I want to have is an array that looks like: ES PT Z CF... -75 -2 5 11... I tried the following, but it did not work: aggregate(d, list(colnames(d)), sum) How can I achieve my objective? Thank you in advance. Sergey __ 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] Finding the two most recent dates
Dear all, Thanks very much for the replies and for the help. This whole data set consists of about 7000 individuals who have had multiple blood pressure measures taken over time so I just used one individual as an example. I'm sorry if it looked like homework...it isn't. Jim your solution worked really well thanks. I was thinking of things in terms of diff time and not order, so thanks. Natalie -- View this message in context: http://r.789695.n4.nabble.com/Finding-the-two-most-recent-dates-tp2528185p2528417.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] Aggregating the matrices
On Mon, Sep 6, 2010 at 9:56 AM, Sergey Goriatchev serg...@gmail.com wrote: Hello everyone. Say we have the following: a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010, c(ES, PT, Z , CF, GX, ST, EO))) b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT, CF, AT, EM, ST))) d - cbind(a, b) I want to calculate sums of the columns that have similar column names and then output this summary What I want to have is an array that looks like: ES PT Z CF... -75 -2 5 11... I tried the following, but it did not work: aggregate(d, list(colnames(d)), sum) How can I achieve my objective? Thank you in advance. Try this: library(plyr) colSums(rbind.fill(as.data.frame(a), as.data.frame(b)), na.rm = TRUE) ES PT Z CF GX ST EO AT EM -75 -2 5 11 2 8 5 4 12 -- 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] Aggregating the matrices
On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote: Hello everyone. Say we have the following: a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010, c(ES, PT, Z , CF, GX, ST, EO))) b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT, CF, AT, EM, ST))) d - cbind(a, b) I want to calculate sums of the columns that have similar column names and then output this summary What I want to have is an array that looks like: ES PT Z CF... -75 -2 5 11... I tried the following, but it did not work: aggregate(d, list(colnames(d)), sum) ES is not in the duplicated column names so perhaps your English specification is not what you meant: d ES PT Z CF GX ST EO PT CF AT EM ST 06092010 -75 3 5 9 2 3 5 -5 2 4 12 5 dupled - colnames(d)[duplicated(colnames(d))] sapply(dupled, function(x) sum( d[, x])) PT CF ST 3 9 3 If you wanted simple a sum over unique column names then it would have been somewhat simpler (no need to construct a duplicated set): sapply(unique(colnames(d)), function(x) sum( d[, x])) ES PT Z CF GX ST EO AT EM -75 3 5 9 2 3 5 4 12 How can I achieve my objective? Thank you in advance. Sergey __ 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] Time Series
Hi How would I analyse time series with - different lengths (i.e. one has 9 entries and the other has 14 entries) - different frequency (i.e. dates are random - no repeated length) - multiple values for the same time entry (e.g. 2009-10-23 below) i.e. my data takes the form: 1st time series 2009-10-07 0.009378 2009-10-19 0.014790 2009-10-23 -0.005946 2009-10-23 0.009096 2009-11-08 0.004189 2009-11-10 -0.004592 2009-11-17 0.009397 2009-11-24 0.003411 2009-12-02 0.003300 2010-01-15 0.010873 2010-01-20 0.010712 2010-01-20 0.022237 2nd time series 2009-09-23 0.076253 2009-10-07 0.039255 2010-02-17 0.039045 2010-03-09 0.024201 2010-03-25 -0.039810 2010-04-13 -0.012428 I am unable to get any functions to work. A simple plot would be nice! Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Time-Series-tp2528444p2528444.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] Aggregating the matrices
Gabor, David, thank you. David, your last suggestion is what I need. Regards, Sergey On Mon, Sep 6, 2010 at 16:12, David Winsemius dwinsem...@comcast.net wrote: On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote: Hello everyone. Say we have the following: a - matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list(06092010, c(ES, PT, Z , CF, GX, ST, EO))) b - matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list(06092010, c(PT, CF, AT, EM, ST))) d - cbind(a, b) I want to calculate sums of the columns that have similar column names and then output this summary What I want to have is an array that looks like: ES PT Z CF... -75 -2 5 11... I tried the following, but it did not work: aggregate(d, list(colnames(d)), sum) ES is not in the duplicated column names so perhaps your English specification is not what you meant: d ES PT Z CF GX ST EO PT CF AT EM ST 06092010 -75 3 5 9 2 3 5 -5 2 4 12 5 dupled - colnames(d)[duplicated(colnames(d))] sapply(dupled, function(x) sum( d[, x])) PT CF ST 3 9 3 If you wanted simple a sum over unique column names then it would have been somewhat simpler (no need to construct a duplicated set): sapply(unique(colnames(d)), function(x) sum( d[, x])) ES PT Z CF GX ST EO AT EM -75 3 5 9 2 3 5 4 12 How can I achieve my objective? Thank you in advance. Sergey __ 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 -- Kniven skärpes bara mot stenen. __ 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] mac: lib/gtk.pkg
On Sep 5, 2010, at 10:32 AM, Daniele Sluijters wrote: Hello, I'm sorry to just pop-up on the mailing list like this and ask a relatively non-R related question but I had no idea whom else to contact on this matter. I'm working on a completely different port of an application to OS X which requires GTK and through Google'ing stumbled on a rather recent GTK installer for Mac at: http://r.research.att.com/ I was wondering if anyone here knows how GTK and its dependencies were packaged into that pkg? It'd be a lifesaver if someone could point me in the right direction. Again, sorry for the non-R related question but this place seemed like the only option. The is an R-SIG-Mac mailing list. It's webpage is at the top of the SIG entries on: http://www.r-project.org/mail.html I suspect that Simon Urbanek, who maintains the ATT webpages and very probably created that package, sometimes reads rhelp but I'm not sure on what schedule. You might see if he makes his email address available on those pages. -- 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] Aggregate certain rows in a matrix
Hi, I have a matrix that looks like this a - c(1,1,1,1,2,2,3,3,3,3) b - c(2,2,2,3,4,4,4,5,5,6) c - c(1,2,3,4,5,6,7,8,9,10) M - matrix(nr=10,nc=3) M[,1] - a M[,2] - b M[,3] - c M [,1] [,2] [,3] [1,]121 [2,]122 [3,]123 [4,]134 [5,]245 [6,]246 [7,]347 [8,]358 [9,]359 [10,]36 10 I want to reduce the matrix according to the following: If the values of the two first columns are the same in two or more rows the values in the third column of the corresponding rows should be added and only one of the rows should be keept. Hence the matrix M above should look like this 1 2 6 1 3 4 2 4 11 3 4 7 3 5 17 3 6 10 Thank you Henrik -- View this message in context: http://r.789695.n4.nabble.com/Aggregate-certain-rows-in-a-matrix-tp2528454p2528454.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] calculating area between plot lines
Hi everyone. I have these data: probClass-seq(0,0.9,0.1) prob1-c(0.0070,0.0911,0.1973,0.2949,0.3936,0.5030,0.5985,0.6869,0.7820,0.8822) prob2-c(0.0066,0.0791,0.2358,0.3478,0.3714,0.3860,0.6667,0.6400,0.7000,1.) # which I'm plotting as follows: plot(probClass,prob1,xlim=c(0,1),ylim=c(0,1),xaxs='i',yaxs='i',type=n) lines(probClass,prob1) lines(probClass,prob2) polygon(c(probClass,rev(probClass)),c(prob2,rev(prob1)),col=red,border=NA) Given that the total area of the plot is 1, how can I calculate the area between the plotted lines (red polygon)? I have only found the areapl function in the splancs package, but it doesn't work for self-intersecting polygons.. Any help will be gratefully received. Cheers, Márcia __ 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] Aggregate certain rows in a matrix
one way is the following: M - cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6), c(1,2,3,4,5,6,7,8,9,10)) ind - do.call(paste, c(as.data.frame(M[, 1:2], sep = \r))) M[, 3] - ave(M[, 3], ind, FUN = sum) unique(M) I hope it helps. Best, Dimitris On 9/6/2010 4:29 PM, Kennedy wrote: Hi, I have a matrix that looks like this a- c(1,1,1,1,2,2,3,3,3,3) b- c(2,2,2,3,4,4,4,5,5,6) c- c(1,2,3,4,5,6,7,8,9,10) M- matrix(nr=10,nc=3) M[,1]- a M[,2]- b M[,3]- c M [,1] [,2] [,3] [1,]121 [2,]122 [3,]123 [4,]134 [5,]245 [6,]246 [7,]347 [8,]358 [9,]359 [10,]36 10 I want to reduce the matrix according to the following: If the values of the two first columns are the same in two or more rows the values in the third column of the corresponding rows should be added and only one of the rows should be keept. Hence the matrix M above should look like this 1 2 6 1 3 4 2 4 11 3 4 7 3 5 17 3 6 10 Thank you Henrik -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ 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] Aggregate certain rows in a matrix
On Mon, Sep 6, 2010 at 3:29 PM, Kennedy henrik.aldb...@gmail.com wrote:
I want to reduce the matrix according to the following: If the values of the
two first columns are the same in two or more rows the values in the third
column of the corresponding rows should be added and only one of the rows
should be keept. Hence the matrix M above should look like this
1 2 6
1 3 4
2 4 11
3 4 7
3 5 17
3 6 10
Use library(plyr), convert to data frame, do ddply, convert back to
matrix if you want. I'm surprised mmply doesn't do it, but I dont
think it does:
Md=data.frame(M)
ddply(Md,c(1,2),function(r){sum(r[,3])})
X1 X2 V1
1 1 2 6
2 1 3 4
3 2 4 11
4 3 4 7
5 3 5 17
6 3 6 10
plyr is on CRAN and that's the third time today I've told someone to use it.
Barry
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[R] How R converts data between objects
Hello everyone. I would kindly request your help concerning how R converts data between different structrures. In the following example please keep attention on the following two 1) I create f - GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget, scale, alpha)) with image(x,y,f) and image(f) I get exactly the same image. then 2)I try to store f inside a raster layer using: r - setValues(r,as.matrix(f)) then comes the tricky part. I am trying to use image again and get the same output as the ouput I got by image(f). I tried image(as.matrix(getValues(r))) which give a completely different output. image function expects a matrix... but for some reason as.matrix(getValues(r)) returns a huge vector :( also tried as.matrix(getValues(r),ncol=ncol(r),nrow=nrow(r)) to force as.matrix to return the appropriate matrix but this also failed and I only got back a vector again Could you please let me understand why this might happening? Best Regards Alex P.S Below you will find part of my code. ...(lines omitted, declarations) x - seq(1, dimx, step) y - seq(1, dimy, step) f - GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget, scale, alpha)) # image(x, y, f) Displays the matrix # f is a 2x2 matrix of dimension (x,y) r - raster(nrow=dimx, ncol=dimy) r - setValues(r,as.matrix(f)) # getValues(r) [[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] How to get mypkg-manual.pdf
On 06/09/2010 9:19 AM, Juliet Ndukum wrote: I am building a package say mypkg. Five months ago, when I built the package I got the mypkg manual in pdf format. Today, after making updates, I build the same package, same name, and steps; unfortunately I do not get the manual in pdf format. Rather I get the following message: cd: can't cd to /cygdrive/c/Documents saving output to 'mypkg-manual.pdf' ...Done Could any one help me on how to get the mypkg-manual.pdf. It looks as though you are trying to put it into a file path with spaces in it, and the version of R you're using doesn't like the spaces. Try saving it somewhere else. I believe the current release doesn't care about spaces, but I'm not sure of that. Building a pdf depends on external tools that are sometimes out of our control. Duncan Murdoch Thank you in advance for your help. JN [[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] Time Series
On Mon, Sep 6, 2010 at 10:24 AM, trb1 thomasrbol...@yahoo.co.uk wrote: Hi How would I analyse time series with - different lengths (i.e. one has 9 entries and the other has 14 entries) - different frequency (i.e. dates are random - no repeated length) - multiple values for the same time entry (e.g. 2009-10-23 below) i.e. my data takes the form: 1st time series 2009-10-07 0.009378 2009-10-19 0.014790 2009-10-23 -0.005946 2009-10-23 0.009096 2009-11-08 0.004189 2009-11-10 -0.004592 2009-11-17 0.009397 2009-11-24 0.003411 2009-12-02 0.003300 2010-01-15 0.010873 2010-01-20 0.010712 2010-01-20 0.022237 2nd time series 2009-09-23 0.076253 2009-10-07 0.039255 2010-02-17 0.039045 2010-03-09 0.024201 2010-03-25 -0.039810 2010-04-13 -0.012428 I am unable to get any functions to work. A simple plot would be nice! Try this. We read in each series taking the last point in the event that there are multiple points with the same date. Then we plot each. If we wish to plot them on the same plot then we merge each series (for producing points) together with linear interpolations of both series (for producing lines) and plot. Lines1 - 2009-10-07 0.009378 2009-10-19 0.014790 2009-10-23 -0.005946 2009-10-23 0.009096 2009-11-08 0.004189 2009-11-10 -0.004592 2009-11-17 0.009397 2009-11-24 0.003411 2009-12-02 0.003300 2010-01-15 0.010873 2010-01-20 0.010712 2010-01-20 0.022237 Lines2 - 2009-09-23 0.076253 2009-10-07 0.039255 2010-02-17 0.039045 2010-03-09 0.024201 2010-03-25 -0.039810 2010-04-13 -0.012428 library(zoo) z1 - read.zoo(textConnection(Lines1), aggregate = function(x) tail(x, 1)) z2 - read.zoo(textConnection(Lines2), aggregate = function(x) tail(x, 2)) plot(z1, type = o) plot(z2, type = o) # or together on the same plot zz - merge(z1, z2) zz - merge(z1, z2, na.approx(zz, na.rm = FALSE)) plot(zz, type = c(p, p, l, l), screen = 1:2) -- 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] rbind() overwriting data.frame()
Hi, first off, I wanna ask how do I declare a data.frame of 0 rows and n columns? Coming to my problem, I have a data.frame of 22 columns by dynamic rows which I insert using rbind. The total number of rows could go upto 2,00,000. The problem is that after about 800 or 900 get inserted rbind starts overwriting the data.frame and I end up with a total of 800-900 rows. What is up with that? The 22 columns are all strings each having about 10 characters -- Rajesh.J [[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] rbind() overwriting data.frame()
Hi again! I'm trying to follow your general goal from your questions today but it's not easy. First, declaring a data.frame of 0 rows is a bad idea. It is much faster to define the length and number of rows from the beginning and to fill it then. Second, I don't know how to do it! What I know is that, to my knowledge (maybe I overlooked some posts in the archive), there is no easy way to do it, such as lists or vectors. The easiest might be to create a list with the correct length with list(), fill it with whatever data and then convert it to a data.frame with as.data.frame() when it's finished. Third, for your problem, maybe do.call() can help you. I don't know what you did up to now, but it sounds that you tried to do it iteratively (in a loop) instead of vectorizing it (though I don't know if do.call() can be really called vectorized). There was a post yesterday/today on do.call(). You'll surely find it if you look with RSiteSearch(). Last, I don't know if it is relevant for you, but I've read on the list many times that matrices are faster to deal with. If all your columns and rows have the same type, then you can use matrices. There are surely guys that know more about this stuff somewhere on the list, but I hope it can get you started. Ivan Le 9/6/2010 16:56, rajesh j a écrit : Hi, first off, I wanna ask how do I declare a data.frame of 0 rows and n columns? Coming to my problem, I have a data.frame of 22 columns by dynamic rows which I insert using rbind. The total number of rows could go upto 2,00,000. The problem is that after about 800 or 900 get inserted rbind starts overwriting the data.frame and I end up with a total of 800-900 rows. What is up with that? The 22 columns are all strings each having about 10 characters -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[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] Aggregate certain rows in a matrix
On Sep 6, 2010, at 10:47 AM, Dimitris Rizopoulos wrote: one way is the following: M - cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6), c(1,2,3,4,5,6,7,8,9,10)) ind - do.call(paste, c(as.data.frame(M[, 1:2], sep = \r))) M[, 3] - ave(M[, 3], ind, FUN = sum) unique(M) I had been working on a similar approach with ave( ,paste(), sum) inside a datafrmae, but I liked your approach of setting up the results of the paste operation as a vector outside of M. (Skips the dataframe operation I was using.) The above solution is destructive, so I constructed this similar alternative that returns the results without altering M: cbind(M, ave(M[ , 3], list(M[,1], M[,2]), FUN=sum))[ !duplicated(M[,1:2]), c(1,2,4)] [,1] [,2] [,3] [1,]126 [2,]134 [3,]24 11 [4,]347 [5,]35 17 [6,]36 10 I hope it helps. Best, Dimitris On 9/6/2010 4:29 PM, Kennedy wrote: Hi, I have a matrix that looks like this a- c(1,1,1,1,2,2,3,3,3,3) b- c(2,2,2,3,4,4,4,5,5,6) c- c(1,2,3,4,5,6,7,8,9,10) M- matrix(nr=10,nc=3) M[,1]- a M[,2]- b M[,3]- c M [,1] [,2] [,3] [1,]121 [2,]122 [3,]123 [4,]134 [5,]245 [6,]246 [7,]347 [8,]358 [9,]359 [10,]36 10 I want to reduce the matrix according to the following: If the values of the two first columns are the same in two or more rows the values in the third column of the corresponding rows should be added and only one of the rows should be keept. Hence the matrix M above should look like this 1 2 6 1 3 4 2 4 11 3 4 7 3 5 17 3 6 10 Thank you Henrik -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ 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] rbind() overwriting data.frame()
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough dataframe and then overwrite the rows.
And it is faster too...
nrows - 2000
ncols - 22
system.time({
+ tmp - data.frame(matrix(nrow = 0, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp - rbind(tmp, rnorm(ncols))
+ }
+ })
user system elapsed
7.830.027.86
system.time({
+ tmp - data.frame(matrix(nrow = nrows, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp[i, ] - rnorm(ncols)
+ }
+ })
user system elapsed
3.750.003.76
#In this case an apply construction was even faster
system.time({
+ tmp - t(sapply(seq_len(nrows), function(i){
+ rnorm(ncols)
+ }))
+ })
user system elapsed
0.020.000.02
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Namens rajesh j
Verzonden: maandag 6 september 2010 16:57
Aan: r-help@r-project.org
Onderwerp: [R] rbind() overwriting data.frame()
Hi,
first off, I wanna ask how do I declare a data.frame of 0
rows and n columns?
Coming to my problem,
I have a data.frame of 22 columns by dynamic rows which I
insert using rbind. The total number of rows could go upto
2,00,000. The problem is that after about 800 or 900 get
inserted rbind starts overwriting the data.frame and I end up
with a total of 800-900 rows. What is up with that?
The 22 columns are all strings each having about 10 characters
--
Rajesh.J
[[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.
Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
door een geldig ondertekend document. The views expressed in this message
and any annex are purely those of the writer and may not be regarded as stating
an official position of INBO, as long as the message is not confirmed by a duly
signed document.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Strange behavior of interval values in optimize()
that was my first idea as well, but as the result shows, the minimized function
value of the wider interval is greater.
In
addidtion, the problem also exists, if the minimized parameter in the
case of the larger interval also already lies within the smaller
interval:
f-function(delta,P,U){
minimiz-P+delta*U
x-minimiz[1]
y-minimiz[2]
z-100*(y-x^2)^2+(1-x)^2
return(z)
}
result-optimize(f, interval=c(-10, 1), P=c(0.99,1.01), U=c(1,0))
result
$minimum
[1] 0.01496807
$objective
[1] 2.483522e-05
result-optimize(f, interval=c(-100, 1), P=c(0.99,1.01), U=c(1,0))
result
$minimum
[1] -1.989994
$objective
[1] 4.01
This does still make no sense for me
Fabian
Date: Mon, 6 Sep 2010 06:18:16 -0700
From: dieter.me...@menne-biomed.de
To: r-help@r-project.org
Subject: Re: [R] Strange behavior of interval values in optimize()
Michael Bernsteiner wrote:
I'm using optimize() to find the minimum of the following function f, and
minimize it (without
.
But, when I choose a larger Interval in the optimization method:
The result gets worse (even though the old interval is included in the new
one).
In fact I don't want any restrictons for the interval values (something
like interval=(-Inf, Inf) or at least the interval should be as large as
possible.
The most likely cause is a second minimum.
Dieter
--
View this message in context:
http://r.789695.n4.nabble.com/Strange-behavior-of-interval-values-in-optimize-tp2528208p2528366.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] path analysis
Hi. which package i need to install to be able to run Path analysis using r? many thanks, Guy -- Guy Rotem Department of Life Sciences The Spatial Ecology Lab Ben Gurion University of the Negev P.O.B. 653 Beer-Sheva 84105 ISRAEL +972-52-3354485 (mobile) +972-8-6461350 (lab) [[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] boxplot knowing Q1, Q3, median, upper and lower whisker value
Dear list, I am using a external program that outputs Q1, Q3, median, upper and lower whisker values for various datasets simultaneously in a tab delimited format. After importing this text file into R, I would like to plot a boxplot using these given values and not the original series of data points, i.e. not using something like boxplot(mydata). Is there an easy way for doing this? If I am not wrong, boxplot() does not accept these values as parameters. Cheers, Dave [[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] rbind() overwriting data.frame()
But If I do that how will I resize later?
On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry
thierry.onkel...@inbo.bewrote:
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough dataframe and then overwrite the rows.
And it is faster too...
nrows - 2000
ncols - 22
system.time({
+ tmp - data.frame(matrix(nrow = 0, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp - rbind(tmp, rnorm(ncols))
+ }
+ })
user system elapsed
7.830.027.86
system.time({
+ tmp - data.frame(matrix(nrow = nrows, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp[i, ] - rnorm(ncols)
+ }
+ })
user system elapsed
3.750.003.76
#In this case an apply construction was even faster
system.time({
+ tmp - t(sapply(seq_len(nrows), function(i){
+ rnorm(ncols)
+ }))
+ })
user system elapsed
0.020.000.02
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Namens rajesh j
Verzonden: maandag 6 september 2010 16:57
Aan: r-help@r-project.org
Onderwerp: [R] rbind() overwriting data.frame()
Hi,
first off, I wanna ask how do I declare a data.frame of 0
rows and n columns?
Coming to my problem,
I have a data.frame of 22 columns by dynamic rows which I
insert using rbind. The total number of rows could go upto
2,00,000. The problem is that after about 800 or 900 get
inserted rbind starts overwriting the data.frame and I end up
with a total of 800-900 rows. What is up with that?
The 22 columns are all strings each having about 10 characters
--
Rajesh.J
[[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.
Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
weer
en binden het INBO onder geen enkel beding, zolang dit bericht niet
bevestigd is
door een geldig ondertekend document. The views expressed in this message
and any annex are purely those of the writer and may not be regarded as
stating
an official position of INBO, as long as the message is not confirmed by a
duly
signed document.
--
Rajesh.J
[[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] Time Series
Thank you very much for your post. Your answer has been very helpful. Is it possible to merge 2 time series? -- View this message in context: http://r.789695.n4.nabble.com/Time-Series-tp2528444p2528584.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] rbind() overwriting data.frame()
Also, when I create the data.frame with matrix and try to rbind, I get
warnings..
Warning messages:
1: In `[-.factor`(`*tmp*`, ri, value = 4) :
invalid factor level, NAs generated
2: In `[-.factor`(`*tmp*`, ri, value = 5) :
invalid factor level, NAs generated
3: In `[-.factor`(`*tmp*`, ri, value = 6) :
invalid factor level, NAs generated
On Mon, Sep 6, 2010 at 9:17 PM, rajesh j akshay.raj...@gmail.com wrote:
But If I do that how will I resize later?
On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry
thierry.onkel...@inbo.be wrote:
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough dataframe and then overwrite the rows.
And it is faster too...
nrows - 2000
ncols - 22
system.time({
+ tmp - data.frame(matrix(nrow = 0, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp - rbind(tmp, rnorm(ncols))
+ }
+ })
user system elapsed
7.830.027.86
system.time({
+ tmp - data.frame(matrix(nrow = nrows, ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp[i, ] - rnorm(ncols)
+ }
+ })
user system elapsed
3.750.003.76
#In this case an apply construction was even faster
system.time({
+ tmp - t(sapply(seq_len(nrows), function(i){
+ rnorm(ncols)
+ }))
+ })
user system elapsed
0.020.000.02
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research Institute for Nature and Forest
team Biometrics Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Namens rajesh j
Verzonden: maandag 6 september 2010 16:57
Aan: r-help@r-project.org
Onderwerp: [R] rbind() overwriting data.frame()
Hi,
first off, I wanna ask how do I declare a data.frame of 0
rows and n columns?
Coming to my problem,
I have a data.frame of 22 columns by dynamic rows which I
insert using rbind. The total number of rows could go upto
2,00,000. The problem is that after about 800 or 900 get
inserted rbind starts overwriting the data.frame and I end up
with a total of 800-900 rows. What is up with that?
The 22 columns are all strings each having about 10 characters
--
Rajesh.J
[[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.
Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.
Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
weer
en binden het INBO onder geen enkel beding, zolang dit bericht niet
bevestigd is
door een geldig ondertekend document. The views expressed in this message
and any annex are purely those of the writer and may not be regarded as
stating
an official position of INBO, as long as the message is not confirmed by a
duly
signed document.
--
Rajesh.J
--
Rajesh.J
[[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] WriteXLS problem
Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py, line 18, in module import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \__init__.py, line 12, in module from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ 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] WriteXLS problem
Hi, Are you sure you used the correct syntax and object names? It might just be because of that...(reading the error messages) There is another function, xlsReadWrite::write.xls(), that I like a lot: it is really easy to use and does not require Perl or Python. HTH, Ivan Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a écrit : Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py, line 18, inmodule import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \__init__.py, line 12, inmodule from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ 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. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[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] path analysis
There are lots of options for path analysis in R. If you go to http://www.rseek.org and type path analysis into the search box, you will get lots of information on functions/packages, and more general info as well. Beyond that, we'd need more specifics about your task. Sarah On Mon, Sep 6, 2010 at 10:37 AM, Guy rotem rottem...@gmail.com wrote: Hi. which package i need to install to be able to run Path analysis using r? many thanks, Guy -- 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] boxplot knowing Q1, Q3, median, upper and lower whisker value
Hi Dave, You can look at the function ?bxp it might work for you. Alternately, create a meaningless boxplot object, and then just edit that data, in which case I know it will work with bxp(). # Create a boxplot, the data does not matter x - boxplot(1:10) x # view the data for the boxplot x$stats - c() # put the stats here, min Q1, median, Q3, max # or hinges or whatever you like x$n - c() # the number of observations, though you do not need to change this bxp(x) # plot boxplot with updated info HTH, Josh On Mon, Sep 6, 2010 at 8:46 AM, David A. dasol...@hotmail.com wrote: Dear list, I am using a external program that outputs Q1, Q3, median, upper and lower whisker values for various datasets simultaneously in a tab delimited format. After importing this text file into R, I would like to plot a boxplot using these given values and not the original series of data points, i.e. not using something like boxplot(mydata). Is there an easy way for doing this? If I am not wrong, boxplot() does not accept these values as parameters. Cheers, Dave [[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. -- 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.
[R] How to run R on Emacs+ESS
Hi folks, Debian 504 64-bit I found following document; http://www.biostat.wisc.edu/~kbroman/Rintro/ Whether it is the right document for installing Emacs+ESS and R so that R can run on Emacs? TIA B.R. Stephen L __ 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] WriteXLS problem
Thank you Ivan for you answer: El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió: Hi, Are you sure you used the correct syntax and object names? It might just be because of that...(reading the error messages) Im sure, because it works with write.csv or write.table. There is another function, xlsReadWrite::write.xls(), that I like a lot: it is really easy to use and does not require Perl or Python. Unfortunately it works on windows, and I am in a non windows platform (ubuntu). Thank you for you advice and help. Kenneth HTH, Ivan Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a crit : Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py, line 18, inmodule import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \__init__.py, line 12, inmodule from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ 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] How to run R on Emacs+ESS
On 6 September 2010 at 09:18, Stephen Liu wrote: | Hi folks, | | Debian 504 64-bit Good. All you need is sudo apt-get install ess | I found following document; | http://www.biostat.wisc.edu/~kbroman/Rintro/ | | Whether it is the right document for installing Emacs+ESS and R so that R can | run on Emacs? There is nothing else to do. Restart (X)Emacs, whichever variant you use on Debian, and type M-x R. You now run R inside Emacs. After that, see http://ess.r-project.org, esp the Documentation tab. Dirk -- Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.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] WriteXLS problem
On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote: Thank you Ivan for you answer: El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió: Hi, Are you sure you used the correct syntax and object names? It might just be because of that...(reading the error messages) Im sure, because it works with write.csv or write.table. Sure? You are making the incorrect assumption that those write functions have the same syntax. At least for WriteXLS that assumption is false. The help page clearly states that the objects need to be quoted rather than being referred to by their naked names. The error you are getting with your second option suggests to me that you offered an unquoted name of an object. You can offer a vector of quoted names of dataframes to WriteXLS and each named dataframe will be converted to a worksheet within the workbook. -- David. There is another function, xlsReadWrite::write.xls(), that I like a lot: it is really easy to use and does not require Perl or Python. Unfortunately it works on windows, and I am in a non windows platform (ubuntu). Thank you for you advice and help. Kenneth HTH, Ivan Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a crit : Hi R users: I don't know if you have had the following problem trying to export to an xls format file in a non windows platform. I try to use the following packages: 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1) 2. WriteXLS (version 1.9.0) (with perl and testPerl working) Even xlsx package that take too long and do not finish. The data frame I try to export has 269363 row and 116 columns. In the first one (dataframe2xls) I get this message: Traceback (most recent call last): File C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/ csv2xls.py, line 18, inmodule import pyexcelerator File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \__init__.py, line 12, inmodule from Workbook import Workbook File C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python \pyexcelerator \Workbook.py, line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L, sheet.hidden, sheet.name).get()) ^ SyntaxError: invalid syntax Using the second option I get this message: Error en get(as.character(i)),envr=envir) : objeto '089' no encontrado Object '089' not found. Im using this R platform: sessionInfo() R version 2.11.1 Patched (2010-08-30 r52848) Platform: x86_64-unknown-linux-gnu (64-bit) Locale: LC_CTYPE=es_CO.UTF-8 Is the only solution to export to .csv and then to .xls format with other program like openoffice? Thank you for your help and advice. Kenneth __ 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. 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] WriteXLS problem
I use the following sintaxis for the packages: For WriteXLS I use: writeXLS(todo2009,todo2009.xls) And for dataframes2xls I use: dataframe2xls::write.xls(todo2009,todo2009.xls) El lun, 06-09-2010 a las 12:34 -0400, David Winsemius escribió: On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote: Thank you Ivan for you answer: El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió: Hi, Are you sure you used the correct syntax and object names? It might just be because of that...(reading the error messages) Im sure, because it works with write.csv or write.table. Sure? You are making the incorrect assumption that those write functions have the same syntax. At least for WriteXLS that assumption is false. The help page clearly states that the objects need to be quoted rather than being referred to by their naked names. The error you are getting with your second option suggests to me that you offered an unquoted name of an object. You can offer a vector of quoted names of dataframes to WriteXLS and each named dataframe will be converted to a worksheet within the workbook. __ 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] How can I fixe convergence=1 in optim
[forwarding back to r-help for archiving/further discussion] On 10-09-05 08:48 PM, Sally Luo wrote: Prof. Bolker, Thanks for your reply and the helpful info. I still have a few questions. 1. I also tried to use different methods other than BFGS without changing their default maxit values, and got very different results. Since I am not that experienced with the optim funciton, I am not sure which one is correct or trustworthy? *If I use method=SANN, that is,* p-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031, -0.00245, 3.366, 0.5885, -0.8, +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49, 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63, +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56), f, method =SANN, y=y,X=X,W=W) I get: There were 50 or more warnings (use warnings() to see the first 50) p $par [1] -0.2392925 0.4653128 -0.8332286 0.0657000 -0.0031000 -0.0024500 3.366 0.5885000 -0.800 0.0786000 -0.0029200 -0.0008100 [13] 3.266 -0.3632000 -0.490 0.1856000 0.0039400 -0.0019300 -0.889 0.5379000 -0.630 0.213 0.0033800 -0.0002600 [25] -0.8912000 -0.3023000 -0.560 $value [1] -772.3262 $counts function gradient 1 NA $convergence [1] 0 $message NULL warnings() Warning messages: 1: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced 2: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced . . 49: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced 50: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced for SANN, the convergence criterion is not meaningful, because SANN does not use a tolerance-based stopping criterion. As ?optim says, 0 indicates successful completion (which is always the case for SANN). *If I change method to Nelder-Mead, that is:* p-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031, -0.00245, 3.366, 0.5885, -0.8, +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49, 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63, +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56), f, method =Nelder-Mead, y=y,X=X,W=W) Then I get: There were 21 warnings (use warnings() to see them) p $par [1] -0.2392925 0.4653128 -0.8332286 0.0657000 -0.0031000 -0.0024500 3.366 0.5885000 -0.800 0.0786000 -0.0029200 -0.0008100 [13] 3.5184500 -0.3632000 -0.490 0.1856000 0.0039400 -0.0019300 -0.889 0.5379000 -0.630 0.213 0.0033800 -0.0002600 [25] -0.8912000 -0.3023000 -0.560 $value [1] -772.3568 $counts function gradient 192 NA $convergence [1] 10 ACCORDING TO the R manual, convergence=10 indicates degeneracy of the NelderMead simplex. Could you explain to me what this degeneracy means? Does it mean the optimization gets stuck with a local minimal? This means that the Nelder-Mead simplex has shrunk in such a way that in a least one dimension the extent of the simplex has shrunk to a point. It doesn't have anything to do with local minima (there's not really any way that a single run of a local optimizer can detect a local minimum). You could try re-starting the optimization from the point at which the previous run stopped. 2. I also tried to change the maxit value of BFGS to 1, and got the following results. It seems this time the algorithm coverges, but the estimation results are quite different from what I got by using the method SANN. In this case, which method should I use? p-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031, -0.00245, 3.366, 0.5885, -0.8, +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49, 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63, +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56), f, method =BFGS, hessian =TRUE, control=list(maxit=1),y=y,X=X,W=W) There were 50 or more warnings (use warnings() to see the first 50) p $par [1] 1.113491e-01 6.347504e-02 -1.570647e-01 7.793766e-02 7.011026e-02 -3.075866e-03 3.365178e+00 8.123945e-02 -2.670111e-04 [10] 7.941502e-02 -2.249492e-04 -1.388776e-03 3.266022e+00 -4.023881e-01 -6.195116e-03 1.829491e-01 -1.116388e-02 -3.088426e-03 [19] -8.888543e-01 6.394912e-01 3.425666e-03 2.193541e-01 3.743851e-02 8.376799e-05 -8.915029e-01 -5.596738e-01 -1.845092e-03 $value [1] -950.553 $counts function gradient 31321 1741 $convergence [1] 0 The BFGS result (-950) is much better than the SANN or Nelder-Mead results (-722). 3. In Peng's email, he pointed out the importance of choosing good initial values in order to get sensible estimates by using optim. Since I am not confident that my initial values are that good and I got different estimation results under different methods, in such cases, would you recommend any
[R] poisson distribution
Hello! I need some help. How I know it to draw the formula of the poisson distribution? expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda) --- not good on the screen the k! not the Poisson Formula, but factorial(k) Thanx! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Failure to aggregate
I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. n.h1 = sqldf(select distinct h, count(*) from x group by h) Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double' n.h2 = aggregate(x$price, by = x$h, FUN = nrow) Error in names(y) - c(names(by), names(x)) : 'names' attribute [10] must be the same length as the vector [2] Arrgh... -- View this message in context: http://r.789695.n4.nabble.com/Failure-to-aggregate-tp2528613p2528613.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] sample a matrix with one element to be 1 from wishart distribution
Hi, I am not sure if this make sense at all. I'd like to sample a matrix, which follows a wishart / inverted wishart distribution. However, the (1,1) element of this matrix should always be equal to 1. How can I handle it in R? Any suggestion is greatly appreciated. Thanks a lot. Sonia [[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] Creating named.list from two matrix columns
Hi Jim, Thanks, That´s right. But the problem is that it introduces unnecessary quotes, perhaps due to the format of first column data in this case : x-cbind(c(row:1, row:2, row:3), c(4889, 9987, 494)) x1-as.list(x[,2]) names(x1)-x[,1] x1 $`row:1` [1] 4889 $`row:2` [1] 9987 $`row:3` [1] 494 How can I avoid unnecessary ` `quotes around the names ? V Date: Mon, 6 Sep 2010 09:14:23 -0400 Subject: Re: [R] Creating named.list from two matrix columns From: jholt...@gmail.com To: is...@live.com CC: r-help@r-project.org Is this what you want: x V1 V2 1 row1 2334 2 row2 347 3 row3 379 x.list - as.list(x$V2) names(x.list) - x$V1 x.list $row1 [1] 2334 $row2 [1] 347 $row3 [1] 379 On Mon, Sep 6, 2010 at 7:55 AM, Viki S is...@live.com wrote: Hi Friends, I am new to R. On R utility class pages, creating named.list is described with this command : new(named.list,a=1,b=2) For large matrix having two columns, such as : row1 2334 row2 347 row3 379 ... I want to create a named.list like : $row1 [1] 2334 $row2 [1] 347 ... Can anyone explain how named.list variable can be created by using two specified columns of a dataframe or matrix object, where one of the two columns is assigned as a name (string) and other as its corresponding value ? Thanks [[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 Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] Over lay 2 scale in same plot
Hi Everyone, I have two different data set in 2 different scale. I want to plot these two data in the same plot in their respective scale. So the plot will have 2 different scale. I have added an image below to show how it should look. does any bode has any idea how this can be done. 2 different y scale in same plot..?? http://r.789695.n4.nabble.com/file/n2528661/2scale_ovelay.jpg Thanks in advance. Mamun -- View this message in context: http://r.789695.n4.nabble.com/Over-lay-2-scale-in-same-plot-tp2528661p2528661.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] sample a matrix with one element to be 1 from wishart distribution
Hi, I am not sure if this make sense at all. I'd like to sample a matrix, which follows a wishart / inverted wishart distribution. However, the (1,1) element of this matrix should always be equal to 1. How can I handle it in R? Any suggestion is greatly appreciated. Thanks a lot. Sonia [[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] poisson distribution
On Sep 6, 2010, at 1:13 PM, tamas barjak wrote: Hello! I need some help. How I know it to draw the formula of the poisson distribution? expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^- lambda) --- not good ?plotmath (Do not see factorial as a plotmath function Try: expr-expression(P(xi == k) == frac(lambda^k, k*!)*e^-lambda) on the screen the k! not the Poisson Formula, but factorial(k) Thanx! -- David. __ 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] poisson distribution
Successful! Thank you! 2010/9/6 David Winsemius dwinsem...@comcast.net On Sep 6, 2010, at 1:13 PM, tamas barjak wrote: Hello! I need some help. How I know it to draw the formula of the poisson distribution? expr-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda) --- not good ?plotmath (Do not see factorial as a plotmath function Try: expr-expression(P(xi == k) == frac(lambda^k, k*!)*e^-lambda) on the screen the k! not the Poisson Formula, but factorial(k) Thanx! -- David. [[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] ERCIM'10: Submission of abstracts
Dear useRs, the deadline for submission of abstracts is approaching for ERCIM'10. Please upload your abstract until 2010-09-08 if you would like to give a presentation at our track on Statistical Algorithms and Software at the 3rd International Conference of the ERCIM WG on COMPUTING STATISTICS (ERCIM'10) 10-12 December 2010, Senate House, University of London, UK http://www.cfe-csda.org/ercim10 Important Dates are: Submission of abstracts: 2010-09-08 Standard registration:2010-09-14 Tutorial: 2010-12-09 Conference: 2010-12-10 to 2010-12-12 Hope to see you there, Achim, Bettina and Uwe __ 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] Failure to aggregate
On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote: I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. n.h1 = sqldf(select distinct h, count(*) from x group by h) Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double' n.h2 = aggregate(x$price, by = x$h, FUN = nrow) A vector argument (x$price) would only have one row (at most). nrow(c(1,2) NULL Error in names(y) - c(names(by), names(x)) : 'names' attribute [10] must be the same length as the vector [2] Try: tapply(x$price, by = x$h, FUN = length) -- David. __ 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] Over lay 2 scale in same plot
Hi, Looking at the picture, I think you are just talking about plotting two datasets. Here is an example I made up, that looks sort of like your picture: # make a barplot barplot(-50:50) # add points into the existing plot at the coordinates set by x and y # and use a line to connect them points(x = 1:101, y = seq(from = 30, to = -20, length.out = 101), type = l) Do you have some sample data you could send us of what you are trying to plot? We can give more specific feedback if we have some actual data to work with. Hope that helps, Josh On Mon, Sep 6, 2010 at 9:57 AM, mamunbabu2001 mrashi...@hotmail.com wrote: Hi Everyone, I have two different data set in 2 different scale. I want to plot these two data in the same plot in their respective scale. So the plot will have 2 different scale. I have added an image below to show how it should look. does any bode has any idea how this can be done. 2 different y scale in same plot..?? http://r.789695.n4.nabble.com/file/n2528661/2scale_ovelay.jpg Thanks in advance. Mamun -- View this message in context: http://r.789695.n4.nabble.com/Over-lay-2-scale-in-same-plot-tp2528661p2528661.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] Time Series
On Mon, Sep 6, 2010 at 11:56 AM, trb1 thomasrbol...@yahoo.co.uk wrote: Thank you very much for your post. Your answer has been very helpful. Is it possible to merge 2 time series? zz is my posted code was formed by merging two univariate and one multivariate series. -- 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] combining collumns for data.frames
Hi This question is far less simple than the title suggests, please read carefully, thanks. I have 2 sets of data, both read into R data1-read.table (1.txt, header=T, sep=\t) data2-read.table (2.txt, header=T, sep=\t) data1 Taxon stage1 stage2 stage3 stage4 T1 0 0 1 1 T2 0 1 1 0 T3 0 0 0 1 T4 1 0 0 0 data2 # this is a library file, it contains all possible values of stage (Col_1) that may be contained in the data1 file (headers of each column), and what they correspond to # in the Col_2 ie stages 1:2 == Group1 Col_1Col_2 Stage1 Group1 Stage2 Group1 Stage3 Group2 Stage4 Group2 I want to get R to combine the columns in data1 based on the information in data2 (Col_2), eg in this instance reduce the columns in data1 from 4 to 2, summing up the values within each column of data1 to get the result below Taxon group1 group2 T1 0 1 T2 1 1 T3 0 1 T4 1 0 i have many datasets which have different numbers of stage eg one dataset will have stage1-10, another will have stage15-35 (data2, Col_2 has all possilbe stage values so will say what group they correspond to) so far i can isolate the rows of data2 which contains the stages in data1 with this: data1.names-names(data1[,-1])#take the header names from data1 minus the 1st column (this is not found in the data2 library file) row.numbers-match(data1.names, data2[,1]) #match the vector containing the data1 column header names to those found in the library file of data2 data2.small-data2[row.numbers] #reduce the data2 to only include the same stages as found in the data1 file from here on i dont know what to, really i wanted to just be able to change the header names of data1 to their corresponding name that is found in Col_2 and then use some statement that could merge columns in data1 which were the same (and also sum the values at each row and dividing by their value if they were greater than 1 (so i only have 0 or 1 again) but i dont know how to do that. Can someone help me to get the desired result (as in the example above) that doe not require me to manually merge columns? ie get the example output in an automated way that could take any version of the data1 file (ie with different stage values) and using the data2 file (library file - same in each instance) get the output similar as in the example above? Thanks Martin [[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] how to change the xlab name?
I simply put, plot(density(), main=, + xlab = XXX), it says that I have an unexpected = in it. -- View this message in context: http://r.789695.n4.nabble.com/how-to-change-the-xlab-name-tp2528733p2528733.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] Failure to aggregate
On Mon, Sep 6, 2010 at 12:15 PM, Dimitri Shvorob dimitri.shvo...@gmail.com wrote: I have a (very big - 1.5 rows) dataframe with a (POSIXt POSIXlt) column h (hour). Surprisingly, I cannot calculate a simple aggregate over the dataframe. n.h1 = sqldf(select distinct h, count(*) from x group by h) Error in sqliteExecStatement(con, statement, bind.data) : RS-DBI driver: (error in statement: no such table: x) In addition: Warning message: In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double' n.h2 = aggregate(x$price, by = x$h, FUN = nrow) Error in names(y) - c(names(by), names(x)) : 'names' attribute [10] must be the same length as the vector [2] Since you are using group by you don't want distinct. In aggregate use x[price] and x[h] rather than x$price and x$h or use a formula. Also use nrow or length in place of NROW. library(sqldf) x - data.frame(price = 1:4, h = c(1, 1, 2, 3)) sqldf(select h, count(*) from x group by h) aggregate(x[price], by = x[h], FUN = NROW) aggregate(x[price], by = x[h], FUN = length) aggregate(price ~ h, x, FUN = length) -- 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] how to change the xlab name?
On Mon, Sep 6, 2010 at 11:07 AM, tooblue kai...@berkeley.edu wrote: I simply put, plot(density(), main=, + xlab = XXX), it says that I have an unexpected = in it. You just have an extra ' + ' before the xlab argument: plot(density(rnorm(100)), main = , xlab = XXX) ought to do it. Cheers, Josh -- View this message in context: http://r.789695.n4.nabble.com/how-to-change-the-xlab-name-tp2528733p2528733.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] how to change the xlab name?
On Sep 6, 2010, at 2:07 PM, tooblue wrote: I simply put, plot(density(), main=, + xlab = XXX), it says that I have an unexpected = in it. It may be a case of a confused parser. You have an extraneous + in there: = rnorm(100) plot(density(), main=, xlab = XXX) # works If on the other hand you wanted to construct a more complex title then you will probably need to read the expression and bquote help pages and submit a more descriptive problem statement. -- David. __ 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] likelyhood maximization problem with polr
Dear community, I am currently trying to fit an ordinal logistic regression model with the polr function. I often get the same error message : attempt to find suitable starting values failed, for example with : require(MASS) data(iris) polr(Species~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width,iris) (I know the response variable Species should be nominal but I do as levels were ordered for the example). I think this is a likelyhood maximization problem ; I tried to solve this by setting the start option of polr to a null or a random vector by it doesn't garantee to find a good solution at the end. Does anyone have a clue ? Thanks a lot ! -- View this message in context: http://r.789695.n4.nabble.com/likelyhood-maximization-problem-with-polr-tp2528818p2528818.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] two questions
Dear friends, two questions (1) does anyone know if there are any non-parametric equivalents of the two-way ANOVA in R? I have an ordinal non-normally distributed dependent variable and two factors (gender and city of birth). Normally, one would try a two-way anova, but if R has any non-parametric equivalents, that might be great. (2) Also, if the interaction of gender and city of birth is statistically significant, which post-hoc tests should I run? Thanks Jason Dr. Iasonas Lamprianou Assistant Professor (Educational Research and Evaluation) Department of Education Sciences European University-Cyprus P.O. Box 22006 1516 Nicosia Cyprus Tel.: +357-22-713178 Fax: +357-22-590539 Honorary Research Fellow Department of Education The University of Manchester Oxford Road, Manchester M13 9PL, UK Tel. 0044 161 275 3485 iasonas.lampria...@manchester.ac.uk __ 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 with unexpected symbol errors
Hi I have got a long script which will not run for me as i keep getting errors : source(clusterfixV1_4.r) Error in source(clusterfixV1_4.r) : clusterfixV1_4.r: unexpected symbol at 158: eck[k,2] - as.numeric(1) 159: #ClusterInfo[k,2] - Clustered I have sorted all the ones i can but i am having a problem here Can anyone tell me the cause of these problems. Its not a very short or straightforward script so i dont expect you to go through the whole thing but it would be great if you can give me an indication as to what I may be doing wrong. I havent attached the data that the script uses because I'm pretty sure the principles i have used are right when running the script i get the above error on line 158 and 159 I am more than happy to provide further information(e.g the dataset) if it helps Many thanks in advance Amit Patel __ 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] two questions
The usual least-squares methods are fairly robust to departures from normality. Furthermore, it is the residuals that are assumed to be normally distributed (not the marginal distributions that you are probably looking at) , so it does not sound as though you have yet examined the data properly. Tell us what the descriptive stats (say the means, variance, 10th and 90th percentiles) are on the residuals within cells cross-classified by the gender and city-of-birth variables (say the means, variance, 10th and 90th percentiles). On Sep 6, 2010, at 4:34 PM, Iasonas Lamprianou wrote: Dear friends, two questions (1) does anyone know if there are any non-parametric equivalents of the two-way ANOVA in R? I have an ordinal non-normally distributed dependent variable and two factors (gender and city of birth). Normally, one would try a two-way anova, but if R has any non- parametric equivalents, that might be great. There is an entire task view page on robust methods if you decide to press on with this quest. (2) Also, if the interaction of gender and city of birth is statistically significant, which post-hoc tests should I run? How many cities are we talking about? Thanks Jason Dr. Iasonas Lamprianou -- 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] PostScript/PDF graphics with another font
I am using the standard phonetic font Doulos SIL in a graph
(http://scripts.sil.org/cms/scripts/page.php?site_id=nrsiid=DoulosSILfont)
This is an example:
windowsFonts(IPA=TT Doulos SIL)
barplot(c(1,2,3,4,5),names=c(\u{0251},\u{0252},\u{0253},\u{0254},\u{0255}),family=IPA)
However, I am unable to make an eps or pdf file from this graph. I get the
following error message when trying to save as .eps:
Error: family 'IPA' not included in PostScript device
In addition: Warning messages:
1: font family not found in PostScript font database
[...]
And this when I try to save as .pdf:
Error: Invalid font type
In addition: Warning messages:
1: font family not found in PostScript font database
[...]
I have tried various applications of the functions postscript(),
postscriptFonts(), pdf(), but I only get error messages when I try to tell
it to make use of the Doulos SIL font. I've searched for answers in Nabble
and google as well, but haven't found a solution (at least not one I
understand).
I'm using R 2.11 for Windows.
Thank you
--
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Re: [R] dataframe row names from list
On 09/06/2010 09:41 PM, raje...@cse.iitm.ac.in wrote: Hi, I have a list which looks like this... str(y) List of 10 $ : chr [1:4] ABCD 5 0 1 $ : chr [1:4] DEF 15 1 16 $ : chr [1:4] AAA 2 17 8 $ : chr [1:4] SSS 15 25 1 $ : chr [1:4] III 15 26 4 $ : chr [1:4] OPQ 7 30 4 $ : chr [1:4] TYR 14 34 8 $ : chr [1:4] IRTS 15 42 1 $ : chr [1:4] LLL 15 43 2 $ : chr [1:4] AQW 3 45 4 I need to create a dataframe whose row names are chr[1] of each vector..ie ABCD,DEF,AAA ETC. how can I do this? Hi rajesh, Try this: rownames(mydf)-unlist(lapply(y,[,1)) Jim __ 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] limit on read.socket?
On Sat, 4 Sep 2010, raje...@cse.iitm.ac.in wrote:
Hi,
I have the following piece of code,
repeat{
ss-read.socket(sockfd);
if(ss==) break
output-paste(output,ss)
}
but somehow, output is not receiving all the data that is coming through the
socket.My suspicion is on the if statement. what happens if a white space
occurs in between the string arriving over the socket?
That's not the problem. The problem occurs when R reads faster than the data
are provided -- R will read and there will not be any new data available, so ss
will be the empty string and the program will end.
In general you can't rely on the sender transmitting data fast enough to keep
ahead of R. The example in make.socket() is reading a very small amount of
data, so it worked reasonably well back in the days when the finger daemon was
more widely active. You need some more precise way of knowing when the data
stream is over. This could be some sort of 'end of transmission' marker or a
count of the number of bytes or lines to be expected.
-thomas
Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle
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Re: [R] Help with unexpected symbol errors
We would need to at least see 10 lines either side of the error to understand its context. Also take a look with you editor (hex editor would be handy) to see if there is some unprintable character around it. Can you isolate just that portion of the code? Try putting it inside a function to see if it will at least be read in without a syntax error. This is where you start learning to debug your program by executing pieces at a time, or in your case, at least compiling the section of code by trying to source it in with a function to see what happens. You might have an unbalanced quote in your script. On Mon, Sep 6, 2010 at 4:42 PM, Amit Patel amitrh...@yahoo.co.uk wrote: Hi I have got a long script which will not run for me as i keep getting errors : source(clusterfixV1_4.r) Error in source(clusterfixV1_4.r) : clusterfixV1_4.r: unexpected symbol at 158: eck[k,2] - as.numeric(1) 159: #ClusterInfo[k,2] - Clustered I have sorted all the ones i can but i am having a problem here Can anyone tell me the cause of these problems. Its not a very short or straightforward script so i dont expect you to go through the whole thing but it would be great if you can give me an indication as to what I may be doing wrong. I havent attached the data that the script uses because I'm pretty sure the principles i have used are right when running the script i get the above error on line 158 and 159 I am more than happy to provide further information(e.g the dataset) if it helps Many thanks in advance Amit Patel __ 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.
Re: [R] combining collumns for data.frames
Try this (after making sure that Col_1 in data2 matches your column
names in data1
data1 - read.table(textConnection(Taxon stage1 stage2 stage3 stage4
+ T1 0 0 1 1
+ T2 0 1 1 0
+ T3 0 0 0 1
+ T4 1 0 0 0), header=TRUE)
data2 - read.table(textConnection(Col_1Col_2
+ stage1 Group1
+ stage2 Group1
+ stage3 Group2
+ stage4 Group2), header=TRUE, as.is=TRUE)
closeAllConnections()
# get the columns to summarize by
colSumz - split(data2$Col_1, data2$Col_2)
# create the output matrix
result - matrix(0, nrow=nrow(data1), ncol=length(colSumz))
colnames(result) - names(colSumz)
rownames(result) - data1$Taxon
for (i in names(colSumz)){
+ result[, i] - rowSums(data1[, colSumz[[i]]])
+ }
result
Group1 Group2
T1 0 2
T2 1 1
T3 0 1
T4 1 0
On Mon, Sep 6, 2010 at 1:49 PM, Martin Hughes sensei2...@hotmail.com wrote:
Hi
This question is far less simple than the title suggests, please read
carefully, thanks.
I have 2 sets of data, both read into R
data1-read.table (1.txt, header=T, sep=\t)
data2-read.table (2.txt, header=T, sep=\t)
data1
Taxon stage1 stage2 stage3 stage4
T1 0 0 1 1
T2 0 1 1 0
T3 0 0 0 1
T4 1 0 0 0
data2 # this is a library file, it contains all possible values of stage
(Col_1) that may be contained in the data1 file (headers of each column), and
what they correspond to
# in the Col_2 ie stages 1:2 == Group1
Col_1 Col_2
Stage1 Group1
Stage2 Group1
Stage3 Group2
Stage4 Group2
I want to get R to combine the columns in data1 based on the information in
data2 (Col_2), eg in this instance reduce the columns in data1 from 4 to 2,
summing up the
values within each column of data1 to get the result below
Taxon group1 group2
T1 0 1
T2 1 1
T3 0 1
T4 1 0
i have many datasets which have different numbers of stage eg one dataset
will have stage1-10, another will have stage15-35 (data2, Col_2 has all
possilbe stage values so will say what group they correspond to)
so far i can isolate the rows of data2 which contains the stages in data1
with this:
data1.names-names(data1[,-1]) #take the header names
from data1 minus the 1st column (this is not found in the data2 library file)
row.numbers-match(data1.names, data2[,1]) #match the vector containing
the data1 column header names to those found in the library file of data2
data2.small-data2[row.numbers] #reduce the data2 to
only include the same stages as found in the data1 file
from here on i dont know what to, really i wanted to just be able to change
the header names of data1 to their corresponding name that is found in Col_2
and then use some statement that could merge columns in data1 which were the
same (and also sum the values at each row and dividing by their value if they
were greater than 1 (so i only have 0 or 1 again) but i dont know how to do
that.
Can someone help me to get the desired result (as in the example above) that
doe not require me to manually merge columns? ie get the example output in an
automated way that could take any version of the data1 file (ie with
different stage values) and using the data2 file (library file - same in each
instance) get the output similar as in the example above?
Thanks
Martin
[[alternative HTML version deleted]]
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] nlme Output
The design is a repeated measures with 3 instances. There are 3 groups: Controls, Heavy Cocaine Users, Light Cocaine Users. I reshaped the data so that there was one variable for the 3 instances called AvgTrials. Time is the indicator of each instance. Here is the model call: mod5 - lme(AvgTrials ~ time + factor(group_id) + time*factor(group_id), random = ~ 1 | group_id, data = tmp.dat, method = ML, na.action = na.omit) What else would you need here? -Edward On Mon, Sep 6, 2010 at 8:52 AM, ONKELINX, Thierry thierry.onkel...@inbo.bewrote: Dear Edward, You have no degrees of freedom left to estimate those p-values. Your design does not allows for the model your implemented. We need a brief summary of your design in order to help you further. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek team Biometrie Kwaliteitszorg Gaverstraat 4 9500 Geraardsbergen Belgium Research Institute for Nature and Forest team Biometrics Quality Assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Edward Patzelt Verzonden: maandag 6 september 2010 15:43 Aan: r-help@r-project.org Onderwerp: [R] nlme Output Everyone - What do the NaN's mean here? Is this analysis a problem? Linear mixed-effects model fit by maximum likelihood Data: tmp.dat AIC BIClogLik 1611.251 1638.363 -797.6253 Random effects: Formula: ~1 | group_id (Intercept) Residual StdDev: 0.0003077668 9.236715 Fixed effects: AvgTrials ~ time + factor(group_id) + time * factor(group_id) Value Std.Error DF t-value p-value (Intercept)18.159722 3.576664 213 5.077279 0. time4.192708 1.655674 213 2.532327 0.0121 factor(group_id)2 -6.929563 5.235700 0 -1.323522 NaN factor(group_id)3 -1.654554 4.189575 0 -0.394922 NaN time:factor(group_id)2 1.729911 2.423658 213 0.713760 0.4762 time:factor(group_id)3 -2.555111 1.939396 213 -1.317478 0.1891 Correlation: (Intr) time fc(_)2 fc(_)3 t:(_)2 time -0.926 factor(group_id)2 -0.683 0.632 factor(group_id)3 -0.854 0.790 0.583 time:factor(group_id)2 0.632 -0.683 -0.926 -0.540 time:factor(group_id)3 0.790 -0.854 -0.540 -0.926 0.583 Standardized Within-Group Residuals: Min Q1Med Q3Max -1.8842754 -0.6979785 -0.3370998 0.5666704 3.0943948 Number of Observations: 219 Number of Groups: 3 Warning message: In pt(q, df, lower.tail, log.p) : NaNs produced [[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. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. -- Edward H. Patzelt Research Assistant TRiCAM Lab University of Minnesota Psychology/Psychiatry VA Medical Center Office: N437 Elliot Hall - Twin Cities Campus Phone: 612-624-3892 Email: patze...@umn.edu Please consider the environment before printing this email www.psych.umn.edu/research/tricam [[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] Way OT: Does PSA testing reduce prostate cancer mortality?
Hi Folks: I found the following paper to be an interesting example of how even well designed and conducted studies -- randomized trials, even -- can be rendered problematic by systematic effects beyond the control of the investigators: http://www.nature.com/nrurol/journal/v7/n9/full/nrurol.2010.120.html It's about whether PSA testing helps reduce mortality due to prostate cancer -- an important issue with significant healh, economic, and QOL impact. After decades of careful work, the answer still seems to be: we don't know. Perhaps a good lesson in humility for us statisticians ... For those who are interested, enjoy. Cheers, Bert Bert Gunter Genentech Nonclinical Statistics __ 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.
