[R] Simulation
I tried looking for help but I couldn't locate the exact solution. I have data that has several variables. I want to do several sample simulations using only two of the variables (eg: say you have data between people and properties owned. You only want to check how many in the samples will come up with bicycles) to estimate probabilities and that sort of thing. Now, I can only do a simulation in terms of this code: sample(1:10, size = 15, replace = TRUE). I do not know how select specific variables only. I'll appreciate the help -- View this message in context: http://r.789695.n4.nabble.com/Simulation-tp3329173p3329173.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] Help
Hi, I've been working with R the past few days in trying to get a proper table set up but have not had any luck at all and the manuals I've read have not worked for me either. I was wondering if anyone here would be able to help me in setting this data up in R both as a table and as a bargraph as I have not been able to do it myself. If somebody could help me with doing this and send me an overview in a document on how it should look, I would really appreciate it. Thank you. Laura Diet: Binger-yes: Binger-No: Total: None 24 134 158 Healthy 9 52 61 Unhealthy 23 72 95 Dangerous 12 15 27 [[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] Explained variance for ICA
Hello, I think to use FastICA package for microarray data clusterization, but one question stops me: can I know how much variance explain each component (or all components together) ? I will be very thankful for the help. Thanks, Pavel [[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] Help
Hi Laura, Have you read this documentation http://cran.r-project.org/doc/manuals/R-data.pdf ? If not, you should. Specifically, see read.table() and read.csv(). When you'll have this, you can look for functions that can import xls or xlsx spreadsheets directly, but you're not that far yet. There are also plenty of documentation for beginners that you should read, see for example http://www.burns-stat.com/pages/Tutor/hints_R_begin.html This one is always good too http://www.burns-stat.com/pages/Tutor/hints_R_begin.html Concerning your data, it already looks good (except for the :). R can work with Excel spreadsheets without problems. I think that if you read that, you will know how to do barplots too. If you then have problems using some specific functions or doing some specific task, you can ask the list again, with a reproducible example, with the code you tried, with the error/warning you got, and with the desired output if relevant. For this, read the posting guide if you still haven't. Now good luck with the readings, Ivan Le 3/1/2011 05:05, Laura Clasemann a écrit : Hi, I've been working with R the past few days in trying to get a proper table set up but have not had any luck at all and the manuals I've read have not worked for me either. I was wondering if anyone here would be able to help me in setting this data up in R both as a table and as a bargraph as I have not been able to do it myself. If somebody could help me with doing this and send me an overview in a document on how it should look, I would really appreciate it. Thank you. Laura Diet: Binger-yes: Binger-No: Total: None 24 134 158 Healthy 9 52 61 Unhealthy 23 72 95 Dangerous 12 15 27 [[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/1525_8_1.php __ 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] Simulation
Well, knowing how your data looks like would definitely help! Say your data object is called mydata, just paste the output from dput(mydata) into the email you want to send to the list. Ivan Le 3/1/2011 04:18, bwaxxlo a écrit : I tried looking for help but I couldn't locate the exact solution. I have data that has several variables. I want to do several sample simulations using only two of the variables (eg: say you have data between people and properties owned. You only want to check how many in the samples will come up with bicycles) to estimate probabilities and that sort of thing. Now, I can only do a simulation in terms of this code: sample(1:10, size = 15, replace = TRUE). I do not know how select specific variables only. I'll appreciate the help -- 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/1525_8_1.php __ 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] stuk at another point: simple question
Hi:
This is *really* ugly, but given the number of variables you have in mind,
it seems rather necessary, at least to me.
# Given an original data frame dataf, find the max ID number:
Mvars - names(dataf)[grep('^M', names(dataf))]
# Pick out the number of variables that start with M:
n - max(as.numeric(sapply(strsplit(Mvars, ''), function(x) unlist(x)[2])))
# Generate all the new variables, similar to those in the transform()
statement to get df2
# In your example, you showed the new variables for one pair - this extends
to all pairs,
# repeated up to the largest parent number
newvars - c(paste('m', rep(1:max(c(Parent1, Parent2), na.rm = TRUE), each =
n),
rep(c('a', 'b')), rep(c('p1', 'p2'), each = 2), sep
= ''))
# Generate a two column matrix of the M*a and M*b names
v - matrix(names(dataf)[-(1:4)], ncol = 2, byrow = TRUE)
# cbind v with itself, transpose the result and collapse it to a vector
vnames - as.vector(t(cbind(v, v)))
# vnames and newvars should have the same length
# Generate a character string of the generate the set of calls,
# extending the call you used to get the new variables in df2
nvarx - paste(newvars, '=', vnames, '[', rep(c('Parent1', 'Parent2'), each
= 2), ']', sep = '')
nvarx.str - paste(nvarx, collapse = ', ')
# Generate the text string that comprises the call to transform()
txtexp - paste('transform(dataf, ', nvarx.str, ')', sep = '')
# Parse the text string in txtemp and evaluate it to get the extended data
frame df3
df3 - eval(parse(text = txtexp))
# Next, we work on producing the averages. To do this, we first generate a
matrix of names,
# then write a function for each row of the matrix that
# * produces text strings that select the proper names for each of the
calls in hP1, hP2, t1 and t2
# * evaluates the parsed strings
# * produces the average
# Matrix of names for each parent
namesmat - matrix(c(paste('M', 1:n, 'a', sep = ''), paste('M', 1:n, 'b',
sep = ''),
paste('m', 1:n, 'ap1', sep = ''), paste('m', 1:n,
'bp1', sep = ''),
paste('m', 1:n, 'ap2', sep = ''), paste('m', 1:n,
'bp2', sep = '')),
nrow = n)
# Function to apply to each row of the above matrix
findavg - function(x) {
# x is an input string of variable names
# we construct the calls as strings and then evaluate them
# final output is the average
x - as.vector(x)
hP1 - eval(parse(text = 'as.numeric(df3[x[3]] != df3[x[4]])'))
hP2 - eval(parse(text ='as.numeric(df3[x[5]] != df3[x[6]])'))
t1 - eval(parse(text ='as.numeric(df3[x[1]] != df3[x[3]])'))
t2 - eval(parse(text ='as.numeric(df3[x[2]] != df3[x[5]])'))
C - (hP1*(t1-0.25)+ hP2 *(t2-0.25))
yv - df3$y
mean(C * yv, na.rm = TRUE)
}
# Apply it to the matrix of names:
apply(namesmat, 1, findavg)
# For the example data given,
[1] -1.17 6.50 -1.17 2.916667
Please double check this on your example below to make sure it's doing the
right thing - I didn't check whether or not the averages were right.
After all that eval(parse(text = *))ing, I need a shower...I feel dirty :)
If there's a better way, I'd love to see it.
HTH,
Dennis
On Mon, Feb 28, 2011 at 5:00 AM, Umesh Rosyara rosyar...@gmail.com wrote:
Dear R-community members.
I am really appreciate R-help group. Dennis has been extrremely helpful to
solve some of my questions. I am following Dennis recommendation in the
following email, yet I am stuck at another point (hope this will took me to
end of this project.
Ind - c(1:5)
Parent1 - c(NA,NA,1,1,3)
Parent2 - c(NA,NA,2,2,4)
y - c(6,5,8,10,7)
M1a - c(1,2,1,1,1)
M1b - c(1,2,2,2,1)
M2a - c(3,3,1,1,3)
M2b - c(1,1,3,3,3)
M3a - c(4,4,4,4,4)
M3b - c(4,4,1,1,4)
M4a - c(1,4,4,1,4)
M4b - c(4,4,4,4,4)
dataf - data.frame (Ind, Parent1, Parent2, y, M1a, M1b,M2a,M2b,
M3a,M3b,M4a, M4b) # I have more than 1000 variables pair
# pair1 (M1a,M1b) pair2 (M2a, M2b), pair3 (M3a, M3b)...
df2 - transform(dataf,m1ap1 = dataf$M1a[dataf$Parent1],
m1bp1 = dataf$M1b[dataf$Parent1],
m1ap2 = dataf$M1a[dataf$Parent2],
m1bp2 = dataf$M1b[dataf$Parent2])
# downstream calculations
hP1 - ifelse(df2$m1ap1==df2$m1bp1,0,1)
hP2 - ifelse(df2$m1bp2==df2$m1bp2,0,1)
t1 - ifelse(df2$M1a==df2$m1ap1,1,0)
t2 - ifelse(df2$M1b==df2$m1ap2,1,0)
C - (hP1*(t1-0.25)+ hP2 *(t2-0.25))
yv - df2$y
Cy - C*yv
avgCy - mean(Cy, na.rm=T)
avgCy # I want to store this value to new dataframe with first model i.e.
How can I loop the process to output the second pair( here M2a, M2b), third
pair (here M3a, M3b) to all pairs (I have more than 1000)
Mode1 avgCy
1 1.75 # from pair M1a and M1b
2 # from pair M2a and M2b
3 # from pair M3a and M3b
4 # from pair M4a and M4b
to the end of the file
Thank you in advance
Umesh R
--
*From:* Dennis Murphy [mailto:djmu...@gmail.com]
Re: [R] tricky (for me) merging of data...more clarity
Hi Again, Thanks very much for your response. It seems my example got rearranged (transposed?) after I posted it. Hopefully this example will be more clear. I have one file (ex. sheet 1) that will have a column for individuals (ind) and a column for the date (date). I would like to merge this with another file (ex. sheet 2) that has both the 'ind' and date column as well as associated body condition measurements (BC1 and BC2). My problem: The body condition values were measured intermittently throughout an individuals history but for our purposes we would like to treat them as representative of 15 days before and after measurement and days outside of this window should have NAs (there are other data associated with those days). When I merge these to files, is there a way to write these body condition values forward and back 15 days?This would give me something that looks like sheet 3. Thank you! Sheet 1 ind date 1 01/02/87 1 02/02/87 1 03/02/87 1 04/02/87 1 05/02/87 1 06/02/87 1 07/02/87 1 08/02/87 1 09/02/87 1 10/02/87 1 11/02/87 1 12/02/87 1 13/02/87 1 14/02/87 1 15/02/87 1 16/02/87 1 17/02/87 1 18/02/87 1 19/02/87 1 20/02/87 1 21/02/87 1 22/02/87 1 23/02/87 1 24/02/87 1 25/02/87 1 26/02/87 1 27/02/87 1 28/02/87 1 01/03/87 1 02/03/87 1 03/03/87 1 04/03/87 1 05/03/87 1 06/03/87 1 07/03/87 1 08/03/87 1 09/03/87 1 10/03/87 1 11/03/87 1 12/03/87 1 13/03/87 1 14/03/87 1 15/03/87 1 16/03/87 1 17/03/87 1 18/03/87 1 19/03/87 1 20/03/87 1 21/03/87 1 22/03/87 1 23/03/87 1 24/03/87 Sheet 2 ind DateBC1 BC2 1 01/02/87 33 3 1 03/03/87 44 3 Sheet 3 ind date BC1 BC2 1 01/02/87 33 3 1 02/02/87 33 3 1 03/02/87 33 3 1 04/02/87 33 3 1 05/02/87 33 3 1 06/02/87 33 3 1 07/02/87 33 3 1 08/02/87 33 3 1 09/02/87 33 3 1 10/02/87 33 3 1 11/02/87 33 3 1 12/02/87 33 3 1 13/02/87 33 3 1 14/02/87 33 3 1 15/02/87 33 3 1 16/02/87 NA NA 1 17/02/87 NA NA 1 18/02/87 44 3 1 19/02/87 44 3 1 20/02/87 44 3 1 21/02/87 44 3 1 22/02/87 44 3 1 23/02/87 44 3 1 24/02/87 44 3 1 25/02/87 44 3 1 26/02/87 44 3 1 27/02/87 44 3 1 28/02/87 44 3 1 01/03/87 44 3 1 02/03/87 44 3 1 03/03/87 44 3 1 04/03/87 44 3 1 05/03/87 44 3 1 06/03/87 44 3 1 07/03/87 44 3 1 08/03/87 44 3 1 09/03/87 44 3 1 10/03/87 44 3 1 11/03/87 44 3 1 12/03/87 44 3 1 13/03/87 44 3 1 14/03/87 44 3 1 15/03/87 44 3 1 16/03/87 44 3 1 17/03/87 44 3 1 18/03/87 44 3 1 19/03/87 44 3 1 20/03/87 44 3 1 21/03/87 NA NA 1 22/03/87 NA NA 1 23/03/87 NA NA 1 24/03/87 NA NA On 27 February 2011 20:49, Tal Galili tal.gal...@gmail.com wrote: Hi Grant, I don't have a solution, but just to be clearer on your situation: One row from sheet 2 looks like this: BC1 BC2 1 01/02/87 33 3 1 03/03/87 44 3 ? Are you using only the first 6 columns for the data to be replicated, and using the other columns as some sort of indicators on when a sequence ends? If so, I would suggest asking the group how you might be able to turn sheet 2 so that it will have as many rows as you needs (which you will then merge with sheet 1). And for clarity sake, consider using ?dput for your objects. Looking at data the way you pasted them (also, without column names) is not very easy for the reader (which might reduce your chances of getting help). Cheers, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Sun, Feb 27, 2011 at 6:41 PM, Grant Gillis grant.j.gil...@gmail.comwrote: BC1 BC2 1 01/02/87 33 3 1 03/03/87 44 3 [[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 Save R library data into xls or dta format
Thanks in advance. I'm having a trouble with data saving. I want to run the same data which is in Ecdat library at different statistic programs(excel, stata and matlab) The data I want to use is library(Ecdat) data(Housing) and I want to extract this data our of R as *.dta *.xls formats. So, my first try was to open the data in R window and drag and paste to excel or notepad. BUT, it didn't work. Do you have any decent skills to extract this library data? Please, share to me. -- View this message in context: http://r.789695.n4.nabble.com/How-to-Save-R-library-data-into-xls-or-dta-format-tp3329489p3329489.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] How to Save R library data into xls or dta format
for excel .. see library(xlsx) On Tue, Mar 1, 2011 at 2:57 PM, JoonGi joo...@hanmail.net wrote: Thanks in advance. I'm having a trouble with data saving. I want to run the same data which is in Ecdat library at different statistic programs(excel, stata and matlab) The data I want to use is library(Ecdat) data(Housing) and I want to extract this data our of R as *.dta *.xls formats. So, my first try was to open the data in R window and drag and paste to excel or notepad. BUT, it didn't work. Do you have any decent skills to extract this library data? Please, share to me. -- View this message in context: http://r.789695.n4.nabble.com/How-to-Save-R-library-data-into-xls-or-dta-format-tp3329489p3329489.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-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] RWinEdt difficulties
Hello Everyone I have just upgraded my PC to Windows 7 (64 bit) and I have installed R 2.12.2. R seems to be working fine. I am having problems getting RWinEdt working with it though. I have tried installing WinEdt 6.0 and WinEdt 5.5. But both fail with the same error using R as 64 bit or 32 bit. I install the package using Administrator rights. library(RWinEdt) Warning message: In shell(paste(\\, .gW$InstallRoot, \\WinEdt.exe\ -C=\R-WinEdt\ -E=, : 'C:\Program Files (x86)\WinEdt Team\WinEdt 6\WinEdt.exe -C=R-WinEdt -E=C:\Program Files (x86)\WinEdt Team\WinEdt 6\R.ini' execution failed with error code 1 Does it matter if you are using 64 bit R and the 32 bit WinEdt? (I have tried 32 bit R). Does RWinEdt work with WinEdt 6.0? Can anybody suggest a solution? Thanks for any help. Regards John Seers ** ** [[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] Help
Hi,
It is a bit unclear what it is you are trying to do, as mentioned in
replies by a variety of people previously, if you are just trying to get
your data into R and label rows / columns, then
tt = matrix(c(24,134,158,9,52,61,23,72,95,12,15,27),ncol=3,byrow=T)
rownames(tt) = c(None, Healthy,Unhealthy,Dangerous)
colnames(tt) = c(Binger-yes:, Binger-no:, Total)
is sufficient, and bar charts can be created using the barplot command,
for example:
barplot(t(tt[,1:2]),legend.text=colnames(tt[,1:2]),xlab=Diet)
If you are trying to do something other than this then you will need to
supply some additional information
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Laura Clasemann
Sent: 01 March 2011 04:05
To: r-help@r-project.org
Subject: [R] Help
Hi,
I've been working with R the past few days in trying to get a proper
table set up but have not had any luck at all and the manuals I've read
have not worked for me either. I was wondering if anyone here would be
able to help me in setting this data up in R both as a table and as a
bargraph as I have not been able to do it myself. If somebody could help
me with doing this and send me an overview in a document on how it
should look, I would really appreciate it. Thank you.
Laura
Diet: Binger-yes: Binger-No:
Total:
None 24 134 158
Healthy 9 52 61
Unhealthy 23 72 95
Dangerous 12 15 27
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Re: [R] getting attributes of list without the names.
On 11-02-28 11:17 PM, Jeroen Ooms wrote:
I am trying to encode arbitrary S3 objects by recursively looping over the
object and all its attributes. However, there is an unfortunate feature of
the attributes() function that is causing trouble. From the manual for
?attributes:
The names of a pairlist are not stored as attributes, but are reported as if
they were (and can be set by the replacement method for attributes).
Now because of this, my program ends up in infinite recursion, because it
will try to encode
attributes(attributes(attributes(attributes(list(foo=123 etc. I can't
remove the 'names' attribute, because this will actually affect the list
structure. And even when I do:
attributes(attributes(obj)[names(attributes(obj)) != names])
This will keep giving me a named list. Is there any way I can get the
attributes() of a list without it reporting the names of a list as
attributes? I.e it should hold that:
atr1- attributes(list(foo=bar));
atr2- attributes(list());
identical(atr1,atr2);
The names of a list (a generic vector) are attributes, just like the
names of other vectors. The documentation is talking about pairlists, a
mostly internal structure, used for example to store parts of
expressions. So your premise might be wrong about the cause of the
recursion...
But assuming you really want to see all attributes except names. Then
just write your own version:
nonameattributes - function(obj) {
result - attributes(obj)
if (!is.null(result$names))
result$names - NULL
# This removes the empty names of the result if there were no other
# attributes. It's optional, but you said you wanted
# identical(atr1, atr2)
if (!length(result))
names(result) - NULL
result
}
You can make the conditional more complicated, only making the change
for pairlists, etc., using tests on typeof(obj) or other tests.
Duncan Murdoch
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Re: [R] How to Save R library data into xls or dta format
On Tue, 1 Mar 2011, Santosh Srinivas wrote: for excel .. see library(xlsx) (That's for Excel = 2007 only, .xlsx not .xls are requested.) Simply consult the relevant manual, 'R Data Import/Export': all of this is covered there. There is a very new package XLConnect that is only covered in the latest version (post R 2.12.2). On Tue, Mar 1, 2011 at 2:57 PM, JoonGi joo...@hanmail.net wrote: Thanks in advance. I'm having a trouble with data saving. I want to run the same data which is in Ecdat library at different statistic programs(excel, stata and matlab) The data I want to use is library(Ecdat) data(Housing) and I want to extract this data our of R as *.dta *.xls formats. So, my first try was to open the data in R window and drag and paste to excel or notepad. BUT, it didn't work. Do you have any decent skills to extract this library data? Please, share to me. We already did: we wrote a manual for you! -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] df.residual for rlm()
Hello,
for testing coefficients of lm(), I wrote the following function (with
the kind support of this mailing list):
# See Verzani, simpleR (pdf), p. 80
coeff.test - function(lm.result, idx, value) {
# idx = 1 is the intercept, idx1 the other coefficients
# null hypothesis: coeff = value
# alternative hypothesis: coeff != value
coeff - coefficients(lm.result)[idx]
SE - coefficients(summary(lm.result))[idx,Std. Error]
n - df.residual(lm.result)
t - (coeff - value )/SE
2 * pt(-abs(t),n) # times two because problem is two-sided
}
This works fine for lm() objects, but fails for rlm() because
df.residual() is NA.
Can I get the degrees of freedom by calculating
n = length(lm.result) - length(coefficients(lm.result))
Thanks for any help!
Jan
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[R] which does the S.D. returned by {Hmisc} rcorr.cens measure?
Dear R-help,
This is an example in the {Hmisc} manual under rcorr.cens function:
set.seed(1)
x - round(rnorm(200))
y - rnorm(200)
round(rcorr.cens(x, y, outx=F),4)
C IndexDxy S.D. nmissing
uncensored Relevant Pairs Concordant Uncertain
0.4831-0.0338 0.0462 200. 0.
200. 39800. 19228. 0.
That S.D. confuses me!!
It is obviously not the standard deviation of x or y.. but there is only one
realization of the c-index or Dxy for this sample dataset, where does the
variation come from..?? if I use the conventional formula for calculating
the standard deviation of proportions: sqrt((C Index)*(1-C Index)/n), I get
0.0353 instead of 0.0462..
Any advice is appreciated.
Vikki
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[R] Is there any Command showing correlation of all variables in a dataset?
Thanks in advance. I want to derive correlations of variables in a dataset Specifically library(Ecdat) data(Housing) attach(Housing) cor(lotsize, bathrooms) this code results only the correlationship between two variables. But I want to examine all the combinations of variables in this dataset. And I will finally make a table in Latex. How can I test correlations for all combinations of variables? with one simple command? -- View this message in context: http://r.789695.n4.nabble.com/Is-there-any-Command-showing-correlation-of-all-variables-in-a-dataset-tp3329599p3329599.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] bootstrap resampling question
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on the
resampling part of it. I try to sum it up in a simplified way so that I would
not confuse anybody.
I have a small database consisting of 20 observations (basically numbers from 1
to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
I would like to resample this database many times for the bootstrap process
with the following two conditions. The resampled databases should also have 20
observations and you can select each of the previously mentioned 20 numbers
with replacement. I guess it is obvious so far. Now the more difficult second
condition is that one number can be selected only maximum 5 times. In order to
make this clear I try to show you an example. So there can be resampled
databases like the following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
(4 different numbers are chosen, each selected 5 times)
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
(Two numbers - 8 and 6 - selected 5 times, number 1 selected four times, the
others selected less than 4 times)
My very first guess that came to my mind whilst thinking about the problem was
the sample function where there are settings like replace=TRUE and prob=...
where you can create a probability vector i.e. how much should be the
probability of selecting a number. So I tried to calculate probabilities first.
I thought the problem can basically described as a k-combination with
repetitions. Unfortunately the only thing I could calculate so far is the total
number of all possible selections which amounts to 137 846 527 049.
Anybody knows how to implement my second tricky condition into one of the R
functions? Are 'boot' and 'bootstrap' packages capable of managing this? I
guess they are, I just couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
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reprodukálása, másolása, vagy egyéb más úton történŠterjesztése,
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- cÃmzett(ek)hez történÅ - eljuttatásáért, valamint semmilyen
késésért, kapcsolat megszakadásból eredŠhibáért, vagy az információ
felhasználásából vagy annak megbÃzhatatlanságából eredÅ kárért.
Az üzenetek EBH-n kÃvüli küldÅje vagy cÃmzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az
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Re: [R] SetInternet2, RCurl and proxy
Dear all, I am facing a problem. I am trying to install packages using a proxy, but I am not able to call the setInternet2 function, either with the small or capital s. What package do I have to call then? And, could there be a reason why this does not function? Thanks, Marco -- View this message in context: http://r.789695.n4.nabble.com/SetInternet2-RCurl-and-proxy-tp3248576p3329624.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] Robust variance estimation with rq (failure of the bootstrap?)
Matt: Thanks for your prompt reply. The disparity between the bootstrap and sandwich variance estimates derived when modeling the highly skewed outcome suggest that either (A) the empirical robust variance estimator is underestimating the variance or (B) the bootstrap is breaking down. The bootstrap variance estimate of a robust location estimate is not necessarily robust, see Statistics Probability Letters 50 (2000) 49-53. Since submitting my earlier post, I have noticed that the the robust kernel variance estimate is similar to the bootstrap estimate. Under what conditions would one expect Koenker and Machado's sandwich variance estimator, which uses a local estimate of the sparsity, to fail? -- Jim On Mon, Feb 28, 2011 at 8:59 PM, Matt Shotwell m...@biostatmatt.com wrote: Jim, If repeated measurements on patients are correlated, then resampling all measurements independently induces an incorrect sampling distribution (= incorrect variance) on a statistic of these data. One solution, as you mention, is the block or cluster bootstrap, which preserves the correlation among repeated observations in resamples. I don't immediately see why the cluster bootstrap is unsuitable. Beyond this, I would be concerned about *any* variance estimates that are blind to correlated observations. The bootstrap variance estimate may be larger than the asymptotic variance estimate, but that alone isn't evidence to favor one over the other. Also, I can't justify (to myself) why skew would hamper the quality of bootstrap variance estimates. I wonder how it affects the sandwich variance estimate... Best, Matt On Mon, 2011-02-28 at 17:50 -0600, James Shaw wrote: I am fitting quantile regression models using data collected from a sample of 124 patients. When modeling cross-sectional associations, I have noticed that nonparametric bootstrap estimates of the variances of parameter estimates are much greater in magnitude than the empirical Huber estimates derived using summary.rq's nid option. The outcome variable is severely skewed, and I am afraid that this may be affecting the consistency of the bootstrap variance estimates. I have read that the m out of n bootstrap can be used to overcome this problem. However, this procedure requires both the original sample (n) and the subsample (m) sizes to be large. The version implemented in rq.boot does not appear to provide any improvement over the naive bootstrap. Ultimately, I am interested in using median regression to model changes in the outcome variable over time. Summary.rq's robust variance estimator is not applicable to repeated-measures data. I question whether the block (cluster) bootstrap variance estimator, which can accommodate intraclass correlation, would perform well. Can anyone suggest alternatives for variance estimation in this situation? Regards, Jim James W. Shaw, Ph.D., Pharm.D., M.P.H. Assistant Professor Department of Pharmacy Administration College of Pharmacy University of Illinois at Chicago 833 South Wood Street, M/C 871, Room 266 Chicago, IL 60612 Tel.: 312-355-5666 Fax: 312-996-0868 Mobile Tel.: 215-852-3045 __ 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. -- James W. Shaw, Ph.D., Pharm.D., M.P.H. Assistant Professor Department of Pharmacy Administration College of Pharmacy University of Illinois at Chicago 833 South Wood Street, M/C 871, Room 266 Chicago, IL 60612 Tel.: 312-355-5666 Fax: 312-996-0868 Mobile Tel.: 215-852-3045 __ 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] Is there any Command showing correlation of all variables in a dataset?
Date: Tue, 1 Mar 2011 02:41:00 -0800 From: joo...@hanmail.net To: r-help@r-project.org Subject: [R] Is there any Command showing correlation of all variables in a dataset? Thanks in advance. I want to derive correlations of variables in a dataset Specifically library(Ecdat) data(Housing) attach(Housing) cor(lotsize, bathrooms) this code results only the correlationship between two variables. But I want to examine all the combinations of variables in this dataset. And I will finally make a table in Latex. How can I test correlations for all combinations of variables? with one simple command? This appears to work as expected although I don't have enough intuition to know if these numbers are plausible beyond rough guess( diag of course is one LOL), df-data.frame(a=rnorm(100), b=runif(100),c=runif(100)+rnorm(100)) str(df) 'data.frame': 100 obs. of 3 variables: $ a: num 0.1841 0.2296 -1.2251 0.0898 -0.961 ... $ b: num 0.586 0.343 0.821 0.41 0.352 ... $ c: num -0.373 2.225 0.102 1.186 -0.737 ... cor(df) a b c a 1. 0.07710107 0.11088579 b 0.07710107 1. -0.02424471 c 0.11088579 -0.02424471 1. ?cor df-data.frame(a=.1*rnorm(100), b=(1:100)/100,c=(1:100)/100+.1*rnorm(100)) cor(df) a b c a 1.0 -0.01970874 -0.003665239 b -0.019708737 1. 0.950375445 c -0.003665239 0.95037544 1.0 __ 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] Hi
Hi, I am facing problem for classification based on graph kernels. we calculated the kernel between two molecule data set.Then I am confused about classification -- View this message in context: http://r.789695.n4.nabble.com/Hi-tp3329650p3329650.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] Simulation
Date: Mon, 28 Feb 2011 19:18:18 -0800
From: kadodamb...@hotmail.com
To: r-help@r-project.org
Subject: [R] Simulation
I tried looking for help but I couldn't locate the exact solution.
I have data that has several variables. I want to do several sample
simulations using only two of the variables (eg: say you have data between
people and properties owned. You only want to check how many in the samples
will come up with bicycles) to estimate probabilities and that sort of
thing.
Now, I can only do a simulation in terms of this code: sample(1:10, size =
15, replace = TRUE).
I do not know how select specific variables only.
I'll appreciate the help
This is probably not the best R but you can do something like either of these.
Note that this is just the easiest derivative of stuff I already had
and can be fixed to your needs, I usually use runif instead of sample for
example.
The first example probably being much less efficient than the second,
df-data.frame(a=.1*rnorm(100), b=(1:100)/100,c=(1:100)/100+.1*rnorm(100))
res=1:100; for ( i in 1:100) {res[i]=cor(df[which(runif(100).9),])[1,3] }
res
hist(res)
res=1:100; for ( i in 1:100) {wh=which(runif(100).9);
res[i]=cor(df$a[wh],df$c[wh]); }
res
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[R] Components of variance with lme
R 2.10
Windows Vista
Is it possible to run a variance-components analysis using lme? I looked at
Pinheiro and Bates' book and don't see code that will perform these analyses.
If the analyses can not be done using lme, what package might I try?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
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Re: [R] nested case-control study
1. Using offset(logweight) in coxph is the same as using an offset logweight; statement in SAS, and neither is the same as case weights. 2. For a nested case control, which is what you said you have, the strata controls who is in what risk set. No trickery with start,stop times is needed. It does no harm, but is not needed. In a case-cohort design the start= stop-epsilon trick is one way to set risk sets up properly. But that's not the design you gave for your study. Terry T. On Mon, 2011-02-28 at 17:02 -0800, array chip wrote: Terry, thanks very much! Professor Langholz used a SAS software trick to estimate absolute risk by creating a fake variable entry_time that is 0.001 less than the variable exit_time (i.e. time to event), and then use both variables in Phreg. Is this equivalent to your creating a dummy survival with time=1? Another question is, is using offset(logweight) inside the formula of coxph() the same as using weight=logweight argument in coxph(), because my understanding of Professor Langholz's approach for nested case-control study is weighted regression? Thank you very much for the help. John __ 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] Components of variance with lme
VarCorr() and then calculate percentage variance for each component from output
of VarCorr
On Tuesday, March 1, 2011 at 6:55 AM, John Sorkin wrote:
R 2.10
Windows Vista
Is it possible to run a variance-components analysis using lme? I looked at
Pinheiro and Bates' book and don't see code that will perform these analyses.
If the analyses can not be done using lme, what package might I try?
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:12}}
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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] mlogit.data
http://r.789695.n4.nabble.com/file/n3329821/workdata.csv workdata.csv The code I posted is exactly what I am running. What you need is this data. Here is the code again. hbwmode-mlogit.data(worktrips.csv, shape=long, choice=CHOSEN, alt.var=ALTNUM) hbwmode-mlogit.data(hbwtrips, shape=long, choice=CHOSEN, alt.var=ALTNUM) -- View this message in context: http://r.789695.n4.nabble.com/mlogit-data-tp3328739p3329821.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] unicodepdf font problem RESOLVED
Just to add to this (I've been looking through the archive) problem with
display unicode fonts in pdf document in R
If you can use the Cairo package to create pdf on Mac, it seems quite happy
with pushing unicode characters through (probably still font family dependant
whether it will display)
probstring - c(' \u2264 0.2',' \u2268 0.4',' \u00FC 0.6',' \u2264
0.8',' \u2264 1.0')
Cairo(type='pdf', file='outputs/demo.pdf', width=9,height=12,
units='in', bg='transparent')
plot(1:5,1:5, type='n')
text(1:5,1:5,probstring)
dev.off()
?Cairo suggests encoding is ignored if you do try to set it.
cheers
Ben
On 14/01/2011, at 7:00 PM, r-help-requ...@r-project.org wrote:
Date: Thu, 13 Jan 2011 10:47:09 -0500
From: David Winsemius dwinsem...@comcast.net
To: Sascha Vieweg saschav...@gmail.com
Cc: r-help@r-project.org
Subject: Re: [R] unicodepdf font problem RESOLVED
Message-ID: 74fa099f-4ce5-45c7-a05a-4a1de6c87...@comcast.net
Content-Type: text/plain; charset=ISO-8859-1; format=flowed; delsp=yes
On Jan 13, 2011, at 10:41 AM, Sascha Vieweg wrote:
I have many German umlauts in my data sets and code them UTF-8. When
it comes to plotting on pdf, I figured out that CP1257 is a good
choice to output Umlauts. I have no experiences with CP1250, but
maybe this small hint helps:
pdf(file=paste(sharepath, /filename.pdf, sep=), 9, 6, pointsize
= 11, family = Helvetica, encoding = CP1257)
Just an FYI for the archives, that encoding fails with
pdf(encoding=CP1257) on a Mac when printing that target umlaut.
David.
*S*
On 11-01-13 16:17, tde...@cogpsyphy.hu wrote:
Date: Thu, 13 Jan 2011 16:17:04 +0100 (CET)
From: tde...@cogpsyphy.hu
To: David Winsemius dwinsem...@comcast.net
Cc: r-help@r-project.org
Subject: Re: [R] unicodepdf font problem RESOLVED
Dear David,
Thank you for your efforts. Inspired by your remarks, I started a new
google-search and found this:
http://stackoverflow.com/questions/3434349/sweave-not-printing-localized-characters
SO HERE COMES THE SOLUTION (it works on both OSs):
pdf.options(encoding = CP1250)
pdf()
plot(1,type=n)
text(1,1,\U0171)
dev.off()
CP1250 should work for all Central-European languages:
http://en.wikipedia.org/wiki/Windows-1250
Thank you again,
Denes
On Jan 13, 2011, at 7:01 AM, tde...@cogpsyphy.hu wrote:
Hi!
Sorry for the missing specs, here they are:
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 12.1
year 2010
month 12
day16
svn rev53855
language R
version.string R version 2.12.1 (2010-12-16)
OS: Windows 7 (English version, 32 bit)
You are after what Adobe calls: udblacute; 0171. It is recognized
in
the list of adobe glyphs:
str(tools::Adobe_glyphs[371, ])
'data.frame': 1 obs. of 2 variables:
$ adobe : chr udblacute
$ unicode: chr 0171
Consulted the help pages
points {graphics}
postscript {grDevices}
pdf {grDevices}
charsets {tools}
postscriptFonts {grDevices}
I have tried a variety of the pdfFonts installed on my Mac without
success. You can perhaps make a list of fonts on your machines with
names(pdfFonts()). Perhaps the range of fonts and the glyphs they
contain is different on your machines. I get consistently warning
messages saying there is a conversion failure:
pdf(trial.pdf, family=Helvetica)
# also tried with font=Helvetica but I think that is erroneous
plot(1,type=n)
text(1,1,print \U0170\U0171)
Warning messages:
1: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for c5
2: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for b0
3: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for c5
4: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for b1
5: In text.default(1, 1, print ) :
font metrics unknown for Unicode character U+0170
6: In text.default(1, 1, print ) :
font metrics unknown for Unicode character U+0171
7: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for c5
8: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for b0
9: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for c5
10: In text.default(1, 1, print ) :
conversion failure on 'print ' in 'mbcsToSbcs': dot
substituted
for b1
And this is despite my system saying the \U0170 and \U0171 are
present
in the Helvetica font. Also tried family=URWHelvetica and
family=NimbusSanand and a bunch of
Re: [R] Explained variance for ICA
You determine the variance explained by *any* unit vector by taking its inner product with the data points, then finding the variance of the results. In the case of FastICA, the variance explained by the ICs collectively is exactly the same as the variance explained by the principal components (collectively) from which they are derived. HTH Rex -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Pavel Goldstein Sent: Tuesday, March 01, 2011 1:24 AM To: r-help@r-project.org Subject: [R] Explained variance for ICA Hello, I think to use FastICA package for microarray data clusterization, but one question stops me: can I know how much variance explain each component (or all components together) ? I will be very thankful for the help. Thanks, Pavel [[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. message may contain confidential information. If you are not the designated recipient, please notify the sender immediately, and delete the original and any copies. Any use of the message by you is prohibited. __ 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] what does the S.D. returned by {Hmisc} rcorr.cens measure?
Vikki,
The formula you used for std. error of C is not correct. C is not a simple
per-observation proportion.
SD in the output is the standard error of Dxy. Dxy = 2(C - .5). Backsolve
for std err of C.
Variation in Dxy or C comes from the usual source: sampling variability.
You can also see this by sampling from the original dataset (a la
bootstrap).
Frank
vikkiyft wrote:
Dear R-help,
This is an example in the {Hmisc} help manual in the section of
rcorr.cens:
set.seed(1)
x - round(rnorm(200))
y - rnorm(200)
round(rcorr.cens(x, y, outx=F),4)
C IndexDxy S.D. nmissing
uncensored Relevant Pairs Concordant Uncertain
0.4831-0.0338 0.0462 200. 0.
200. 39800. 19228. 0.
That S.D. confuses me!!
It is obviously not the standard deviation of x or y.. but there is only
one realization of the c-index or Dxy for this sample dataset, where does
the variation come from..?? if I use the conventional formula for
calculating the standard deviation of proportions: sqrt((C Index)*(1-C
Index)/n), I get 0.0353 instead of 0.0462..
Any advice is appreciated.
Vikki
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to Save R library data into xls or dta format
On Tue, Mar 1, 2011 at 4:27 AM, JoonGi joo...@hanmail.net wrote: Thanks in advance. I'm having a trouble with data saving. I want to run the same data which is in Ecdat library at different statistic programs(excel, stata and matlab) The data I want to use is library(Ecdat) data(Housing) and I want to extract this data our of R as *.dta *.xls formats. So, my first try was to open the data in R window and drag and paste to excel or notepad. For excel see: http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windowss=excel -- 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] High standard error
Hi to everyone, if the estimate of the parameter results in 0.196 and his standard error is 0.426, can I say that this parameter is not significant for the model? Thank you very much Pippo -- View this message in context: http://r.789695.n4.nabble.com/High-standard-error-tp3329903p3329903.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] High standard error
Sure why not? You do realize, do you not that no one has the slightest idea of what you are doing? PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. --- On Tue, 3/1/11, danielepippo dan...@hotmail.it wrote: From: danielepippo dan...@hotmail.it Subject: [R] High standard error To: r-help@r-project.org Received: Tuesday, March 1, 2011, 9:07 AM Hi to everyone, if the estimate of the parameter results in 0.196 and his standard error is 0.426, can I say that this parameter is not significant for the model? Thank you very much Pippo -- View this message in context: http://r.789695.n4.nabble.com/High-standard-error-tp3329903p3329903.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] Robust variance estimation with rq (failure of the bootstrap?)
Jim, Thanks for pointing me to this article. The authors argue that the bootstrap intervals for a robust estimator may not be as robust as the estimator. In this context, robustness is measured by the breakdown point, which is supposed to measure robustness to outliers. Even so, the authors found that the upper bound of a quantile bootstrap interval for the sample median was nearly as robust as the sample median. That brings some comfort in using quantile bootstrap intervals in quantile regression. Does the sandwich estimator assume that errors are independent? And a related question: Does the rq function allow the user to specify clusters/grouping among the observations? Best, Matt On Tue, 2011-03-01 at 05:35 -0600, James Shaw wrote: Matt: Thanks for your prompt reply. The disparity between the bootstrap and sandwich variance estimates derived when modeling the highly skewed outcome suggest that either (A) the empirical robust variance estimator is underestimating the variance or (B) the bootstrap is breaking down. The bootstrap variance estimate of a robust location estimate is not necessarily robust, see Statistics Probability Letters 50 (2000) 49-53. Since submitting my earlier post, I have noticed that the the robust kernel variance estimate is similar to the bootstrap estimate. Under what conditions would one expect Koenker and Machado's sandwich variance estimator, which uses a local estimate of the sparsity, to fail? -- Jim On Mon, Feb 28, 2011 at 8:59 PM, Matt Shotwell m...@biostatmatt.com wrote: Jim, If repeated measurements on patients are correlated, then resampling all measurements independently induces an incorrect sampling distribution (= incorrect variance) on a statistic of these data. One solution, as you mention, is the block or cluster bootstrap, which preserves the correlation among repeated observations in resamples. I don't immediately see why the cluster bootstrap is unsuitable. Beyond this, I would be concerned about *any* variance estimates that are blind to correlated observations. The bootstrap variance estimate may be larger than the asymptotic variance estimate, but that alone isn't evidence to favor one over the other. Also, I can't justify (to myself) why skew would hamper the quality of bootstrap variance estimates. I wonder how it affects the sandwich variance estimate... Best, Matt On Mon, 2011-02-28 at 17:50 -0600, James Shaw wrote: I am fitting quantile regression models using data collected from a sample of 124 patients. When modeling cross-sectional associations, I have noticed that nonparametric bootstrap estimates of the variances of parameter estimates are much greater in magnitude than the empirical Huber estimates derived using summary.rq's nid option. The outcome variable is severely skewed, and I am afraid that this may be affecting the consistency of the bootstrap variance estimates. I have read that the m out of n bootstrap can be used to overcome this problem. However, this procedure requires both the original sample (n) and the subsample (m) sizes to be large. The version implemented in rq.boot does not appear to provide any improvement over the naive bootstrap. Ultimately, I am interested in using median regression to model changes in the outcome variable over time. Summary.rq's robust variance estimator is not applicable to repeated-measures data. I question whether the block (cluster) bootstrap variance estimator, which can accommodate intraclass correlation, would perform well. Can anyone suggest alternatives for variance estimation in this situation? Regards, Jim James W. Shaw, Ph.D., Pharm.D., M.P.H. Assistant Professor Department of Pharmacy Administration College of Pharmacy University of Illinois at Chicago 833 South Wood Street, M/C 871, Room 266 Chicago, IL 60612 Tel.: 312-355-5666 Fax: 312-996-0868 Mobile Tel.: 215-852-3045 __ 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. -- James W. Shaw, Ph.D., Pharm.D., M.P.H. Assistant Professor Department of Pharmacy Administration College of Pharmacy University of Illinois at Chicago 833 South Wood Street, M/C 871, Room 266 Chicago, IL 60612 Tel.: 312-355-5666 Fax: 312-996-0868 Mobile Tel.: 215-852-3045 __ 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
Re: [R] nls not solving
Here is a reply by Bart: Yes you're right (I should have taken off my glasses and looked closer). However, the argument is essentially the same: Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc] + [bc] + (bc) *exp(kl')exp(-kx) is also a solution, where l' = l - log(c)/k . Cheers, Bert (Feel free to post this correction if you like) This is from me: The problem with dropping the l parameter is that it is supposed to account for the lag component. This equation was published in the literature and has been being solved in SAS. When I put it in excel, it solves, but not very well as it comes to a different solution for each time that I change the starting values. As such, I'm not sure how SAS solves for it and I'm not sure what I should do about the equation. Maybe I should just drop the parameter a. Thanks for the help. -- View this message in context: http://r.789695.n4.nabble.com/nls-not-solving-tp3328647p3329936.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] Quantreg model error and goodness of fit
Hi All, I am using the package quantreg to create a 'model' that I can then use to predict the response variable (volume) from a larger set of explanatory variables (environmental factors): ie- #model- fit - rqss(volume~qss(env.factor1,lambda=1)+ qss(env.factor2,lambda=1), tau = 0.9) summary(fit) predict volume from new environmental factors for a larger area where I do not know the volume predi-predict(fit, new, interval = none, level = 0.9) However I am getting the following error message: Error in predict.qss1(qss, newdata = newd, ...) : no extrapolation allowed in predict.qss Is there a way around this? and also with the initial model, is there a way to test goodness of fit? If so how? because #summary(fit) only tells me if the model is significant not how good it is (like with a linear regression you get and R square value which tells you how good the model is). Thank you All help is appreciated, If you would like anything clarified please contact me, Kitty [[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] Is there any Command showing correlation of all variables in a dataset?
?cor answers that question. If Housing is a dataframe, cor(Housing) should do it. Surprisingly, ??correlation doesn't point you to ?cor. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of JoonGi Sent: Tuesday, March 01, 2011 5:41 AM To: r-help@r-project.org Subject: [R] Is there any Command showing correlation of all variables in a dataset? Thanks in advance. I want to derive correlations of variables in a dataset Specifically library(Ecdat) data(Housing) attach(Housing) cor(lotsize, bathrooms) this code results only the correlationship between two variables. But I want to examine all the combinations of variables in this dataset. And I will finally make a table in Latex. How can I test correlations for all combinations of variables? with one simple command? -- View this message in context: http://r.789695.n4.nabble.com/Is-there-any-Command-showing-correlation-of-all-variables-in-a-dataset-tp3329599p3329599.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. message may contain confidential information. If you are not the designated recipient, please notify the sender immediately, and delete the original and any copies. Any use of the message by you is prohibited. __ 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] bootstrap resampling - simplified
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on the
resampling part of it. I try to sum it up in a simplified way so that I would
not confuse anybody.
I have a small database consisting of 20 observations (basically numbers from 1
to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
I would like to resample this database many times for the bootstrap process
with the following conditions. Firstly, every resampled database should also
include 20 observations. Secondly, when selecting a number from the
above-mentioned 20 numbers, you can do this selection with replacement. The
difficult part comes now: one number can be selected only maximum 5 times. In
order to make this clear I show you a couple of examples. So the resampled
databases might be like the following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
4 different numbers are chosen (1, 2, 3, 4), each selected - for the maximum
possible - 5 times.
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
Two numbers - 8 and 6 - selected 5 times (the maximum possible times), number 1
selected 4 times, the others selected less than 4 times.
(3rd database) 1,1,2,2,3,3,4,4,9,9,9,10,10,13,10,9,3,9,2,1
Number 9 chosen for the maximum possible 5 times, number 10, 3, 2, 1 chosen for
3 times, number 4 selected twice and number 13 selected only once.
...
Anybody knows how to implement my tricky condition into one of the R
functions - that one number can be selected only 5 times at most? Are 'boot'
and 'bootstrap' packages capable of managing this? I guess they are, I just
couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos és/vagy
jogilag, szakmailag vagy más módon védett információt tartalmazhat.
Amennyiben nem Ãn a levél cÃmzettje akkor a levél tartalmának közlése,
reprodukálása, másolása, vagy egyéb más úton történŠterjesztése,
felhasználása szigorúan tilos. Amennyiben tévedésbÅl kapta meg ezt az
üzenetet kérjük azonnal értesÃtse az üzenet küldÅjét. Az Erste Bank
Hungary Zrt. (EBH) nem vállal felelÅsséget az információ teljes és pontos
- cÃmzett(ek)hez történÅ - eljuttatásáért, valamint semmilyen
késésért, kapcsolat megszakadásból eredŠhibáért, vagy az információ
felhasználásából vagy annak megbÃzhatatlanságából eredÅ kárért.
Az üzenetek EBH-n kÃvüli küldÅje vagy cÃmzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az
EBH folytonos munkamenetének biztosÃtása érdekében.
This e-mail and any attached files are confidential and/...{{dropped:19}}
__
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] Problem on flexmix when trying to apply signature developed in one model to a new sample
Problem on flexmix when trying to apply signature developed in one model to a new sample. Dear R Users, R Core Team, I have a problem when trying to know the classification of the tested cases using two variables with the function of flexmix: After importing the database and creating a matrix: BM-cbind(Data$var1,Data$var2) I see that the best model has 2 groups and use: ex2 - flexmix(BM~1, k=2, model=FLXMCmvnorm(diagonal=FALSE)) print(ex2) plotEll(ex2, BM) Then I want to test to which group one subset of patients belongs, so I import a smaller sample of the previous data: BM2-data.frame (Data2$var1,Data2$var2) However when I test the results I get are from the complete training sample I used in ex2 and not from the new sample BM2. ProbMCI-posterior(ex2, BM2) And if I do the following I get double the number of entered cases (I think because I entered 2 variables): BM2-cbind (Data2$var1,Data2$var2) p-posterior(ex2)[BMMCI,] max.col(p) (The same with clusters(ex2)[BM2]) In the future I would like to test the result of this mixture also in new samples. Thank you in advance [[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] bootstrap resampling question
Here are a couple of thoughts.
If you want to use the boot package then the statistic function you give it
just receives the bootstrapped indexes, you could test the indexes for your
condition of not more than 5 of each and if it fails return an NA instead of
computing the statistic. Then in your output just remove the NAs (you should
increase the total number of samples tried so that you have a reasonable number
after deletion).
If you want to do it by hand, just use the rep function to create 5 replicates
of your data, then sample from that without replacement. You can get up to 5
copies of each value (from the hand replication), but no more since it samples
without replacement.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Bodnar Laszlo EB_HU
Sent: Tuesday, March 01, 2011 3:31 AM
To: 'r-help@r-project.org'
Subject: [R] bootstrap resampling question
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on
the resampling part of it. I try to sum it up in a simplified way so
that I would not confuse anybody.
I have a small database consisting of 20 observations (basically
numbers from 1 to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
I would like to resample this database many times for the bootstrap
process with the following two conditions. The resampled databases
should also have 20 observations and you can select each of the
previously mentioned 20 numbers with replacement. I guess it is obvious
so far. Now the more difficult second condition is that one number can
be selected only maximum 5 times. In order to make this clear I try to
show you an example. So there can be resampled databases like the
following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
(4 different numbers are chosen, each selected 5 times)
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
(Two numbers - 8 and 6 - selected 5 times, number 1 selected four
times, the others selected less than 4 times)
My very first guess that came to my mind whilst thinking about the
problem was the sample function where there are settings like
replace=TRUE and prob=... where you can create a probability vector
i.e. how much should be the probability of selecting a number. So I
tried to calculate probabilities first. I thought the problem can
basically described as a k-combination with repetitions. Unfortunately
the only thing I could calculate so far is the total number of all
possible selections which amounts to 137 846 527 049.
Anybody knows how to implement my second tricky condition into one of
the R functions? Are 'boot' and 'bootstrap' packages capable of
managing this? I guess they are, I just couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
___
_
Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos
és/vagy jogilag, szakmailag vagy más módon védett információt
tartalmazhat. Amennyiben nem Ön a levél címzettje akkor a levél
tartalmának közlése, reprodukálása, másolása, vagy egyéb más úton
történő terjesztése, felhasználása szigorúan tilos. Amennyiben
tévedésből kapta meg ezt az üzenetet kérjük azonnal értesítse az üzenet
küldőjét. Az Erste Bank Hungary Zrt. (EBH) nem vállal felelősséget az
információ teljes és pontos - címzett(ek)hez történő - eljuttatásáért,
valamint semmilyen késésért, kapcsolat megszakadásból eredő hibáért,
vagy az információ felhasználásából vagy annak megbízhatatlanságából
eredő kárért.
Az üzenetek EBH-n kívüli küldője vagy címzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az
EBH folytonos munkamenetének biztosítása érdekében.
This e-mail and any attached files are confidential
and/...{{dropped:19}}
__
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] bootstrap resampling question
A simple way of sampling with replacement from 1:20, with the additional
constraint that each number can be selected at most five times is
sample(rep(1:20, 5), 20)
HTH,
Giovanni
On Tue, 2011-03-01 at 11:30 +0100, Bodnar Laszlo EB_HU wrote:
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on the
resampling part of it. I try to sum it up in a simplified way so that I would
not confuse anybody.
I have a small database consisting of 20 observations (basically numbers from
1 to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
I would like to resample this database many times for the bootstrap process
with the following two conditions. The resampled databases should also have
20 observations and you can select each of the previously mentioned 20
numbers with replacement. I guess it is obvious so far. Now the more
difficult second condition is that one number can be selected only maximum 5
times. In order to make this clear I try to show you an example. So there can
be resampled databases like the following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
(4 different numbers are chosen, each selected 5 times)
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
(Two numbers - 8 and 6 - selected 5 times, number 1 selected four times,
the others selected less than 4 times)
My very first guess that came to my mind whilst thinking about the problem
was the sample function where there are settings like replace=TRUE and
prob=... where you can create a probability vector i.e. how much should be
the probability of selecting a number. So I tried to calculate probabilities
first. I thought the problem can basically described as a k-combination with
repetitions. Unfortunately the only thing I could calculate so far is the
total number of all possible selections which amounts to 137 846 527 049.
Anybody knows how to implement my second tricky condition into one of the R
functions? Are 'boot' and 'bootstrap' packages capable of managing this? I
guess they are, I just couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos és/vagy
jogilag, szakmailag vagy más módon védett információt tartalmazhat.
Amennyiben nem Ön a levél címzettje akkor a levél tartalmának közlése,
reprodukálása, másolása, vagy egyéb más úton történő terjesztése,
felhasználása szigorúan tilos. Amennyiben tévedésből kapta meg ezt az
üzenetet kérjük azonnal értesítse az üzenet küldőjét. Az Erste Bank Hungary
Zrt. (EBH) nem vállal felelősséget az információ teljes és pontos -
címzett(ek)hez történő - eljuttatásáért, valamint semmilyen késésért,
kapcsolat megszakadásból eredő hibáért, vagy az információ felhasználásából
vagy annak megbízhatatlanságából eredő kárért.
Az üzenetek EBH-n kívüli küldője vagy címzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az EBH
folytonos munkamenetének biztosítása érdekében.
This e-mail and any attached files are confidential and/...{{dropped:19}}
__
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.
--
Giovanni Petris gpet...@uark.edu
Associate Professor
Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)
http://definetti.uark.edu/~gpetris/
__
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.
[R] problems with playwith
hello, i tried to run playwith but : library(playwith) Loading required package: lattice Loading required package: cairoDevice Loading required package: gWidgetsRGtk2 Loading required package: gWidgets Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared object 'H:/R/cran/RGtk2/libs/i386/RGtk2.dll': LoadLibrary failure: The specified procedure could not be found. Failed to load RGtk2 dynamic library, attempting to install it. Learn more about GTK+ at http://www.gtk.org If the package still does not load, please ensure that GTK+ is installed and that it is on your PATH environment variable IN ANY CASE, RESTART R BEFORE TRYING TO LOAD THE PACKAGE AGAIN Error : .onAttach failed in attachNamespace() for 'gWidgetsRGtk2', details: call: .Call(name, ..., PACKAGE = PACKAGE) error: C symbol name S_gtk_icon_factory_new not in DLL for package RGtk2 Error: package 'gWidgetsRGtk2' could not be loaded Sys.getenv(PATH) PATH H:\\R/GTK/bin;H:\\R/GTK/lib;H:\\R/ImageMagick;C:\\windows\\system32;C:\\windows;C:\\windows\\System32\\Wbem;C:\\windows\\System32\\WindowsPowerShell\\v1.0\\;C:\\Program Files\\Common Files\\Ulead Systems\\MPEG;C:\\Program Files\\QuickTime\\QTSystem\\;H:\\R\\GTK\\GTK2-Runtime\\bin;H:\\PortableUSB/PortableApps/MikeTex/miktex/bin packages(lattice, cairoDevice, gWidgetsRGtk2, gWidgets, RGtk2, playwith) were reinstalled program GTK was reinstalled. using R-2-12-2 on Windows 7 Can anybody suggest a solution? thanks R.Heberto Ghezzo Ph.D. Montreal - Canada __ 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] Does POSIXlt extract date components properly?
I would like to use POSIX classes to store dates and extract components of
dates. Following the example in Spector (Data Manipulation in R), I
create a date
mydate = as. POSIXlt('2005-4-19 7:01:00')
I then successfully extract the day with the command
mydate$day
[1] 19
But when I try to extract the month
mydate$mon
[1] 3
it returns the wrong month. And mydate$year is off by about 2,000 years.
Am I doing something wrong?
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
1731 Research Park Drive, Davis California
sbige...@fs.fed.us / ph. 530 759 1718
[[alternative HTML version deleted]]
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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.
[R] Speed up sum of outer products?
Hi, I'm new to R and stats, and I'm trying to speed up the following sum,
for (i in 1:n){
C = C + (X[i,] %o% X[i,]) # the sum of outer products - this is very
slow
according to Rprof()
}
where X is a data matrix (nrows=1000 X ncols=50), and n=1000. The sum has to
be calculated over 10,000 times for different X.
I think it is similar to estimating a co-variance matrix for demeaned data
X. I tried using cov, but got different answers, and it was'nt much quicker?
Any help gratefully appreciated,
--
View this message in context:
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Re: [R] mlogit.data
My previous posting seems to have got mangled. This reposts it. On Mar 01, 2011; 03:32pm gmacfarlane wrote: workdata.csv The code I posted is exactly what I am running. What you need is this data. Here is the code again. hbwmode-mlogit.data(worktrips.csv, shape=long, choice=CHOSEN, alt.var=ALTNUM) hbwmode-mlogit.data(hbwtrips, shape=long, choice=CHOSEN, alt.var=ALTNUM) You still have not done what the posting guide asks for but have expected me (or someone else) to scrutinize a large unknown data set (22003 rows). Fortunately there are other routes. Had you studied Yves Croissant's examples (?mlogit.data), which do work, you would have seen that your input or raw data have to have a particular format for mlogit.data to work. In particular, the alt.var (mode in the TravelMode data set and ALTNUM in your data set) has to go through all its levels in sequence. Yours don't (your variable has 6 levels but sometimes runs from 1 to 5, sometimes from 2 to 6, and so on). Within each run there must be only one choice. ## library(mlogit) data(TravelMode, package = AER) head(TravelMode, n= 20) individual mode choice wait vcost travel gcost income size 1 1 air no 695910070 351 2 1 train no 343137271 351 3 1 bus no 352541770 351 4 1 caryes01018030 351 5 2 air no 6458 6868 302 6 2 train no 443135484 302 7 2 bus no 532539985 302 8 2 caryes01125550 302 9 3 air no 69 115125 129 401 10 3 train no 3498892 195 401 11 3 bus no 3553882 149 401 12 3 caryes023720 101 401 13 4 air no 6449 6859 703 14 4 train no 442635479 703 15 4 bus no 532139981 703 16 4 caryes0 518032 703 17 5 air no 646014482 452 18 5 train no 443240493 452 19 5 bus no 532644994 452 20 5 caryes0 860099 452 When we look at just the relevant part of your data we have the following: hbwtrips-read.csv(E:/Downloads/workdata.csv, header=TRUE, sep=,, dec=., row.names=NULL) head(hbwtrips[, c(2:11)], n=25) HHID PERID CASE ALTNUM NUMALTS CHOSEN IVTT OVTT TVTT COST 1 2 11 1 5 1 13.38 2.00 15.38 70.63 2 2 11 2 5 0 18.38 2.00 20.38 35.32 3 2 11 3 5 0 20.38 2.00 22.38 20.18 4 2 11 4 5 0 25.90 15.20 41.10 115.64 5 2 11 5 5 0 40.50 2.00 42.50 0.00 6 3 12 1 5 0 29.92 10.00 39.92 390.81 7 3 12 2 5 0 34.92 10.00 44.92 195.40 8 3 12 3 5 0 21.92 10.00 31.92 97.97 9 3 12 4 5 1 22.96 14.20 37.16 185.00 103 12 5 5 0 58.95 10.00 68.95 0.00 115 13 1 4 1 8.60 6.00 14.60 37.76 125 13 2 4 0 13.60 6.00 19.60 18.88 135 13 3 4 0 15.60 6.00 21.60 10.79 145 13 4 4 0 16.87 21.40 38.27 105.00 156 14 1 4 0 30.60 8.50 39.10 417.32 166 14 2 4 0 35.70 8.50 44.20 208.66 176 14 3 4 0 22.70 8.50 31.20 105.54 186 14 4 4 1 24.27 9.00 33.27 193.49 198 25 2 4 1 23.04 3.00 26.04 29.95 208 25 3 4 0 25.04 3.00 28.04 17.12 218 25 4 4 0 25.04 23.50 48.54 100.00 228 25 5 4 0 34.35 3.00 37.35 0.00 238 36 2 5 0 11.14 3.50 14.64 14.00 248 36 3 5 0 13.14 3.50 16.64 8.00 258 36 4 5 1 3.95 16.24 20.19 100.00 To show you that this is so we will mock up two variables that have the characteristics described above and use them to execute the function. ## hbwtrips$CHOICEN - rep(c(rep(0,10),1), 2003) hbwtrips$ALTNUMTest - gl(11,1,22033, labels=LETTERS[1:11]) hbwtrips[1:30, c(1:11,44,45)] hbwmode - mlogit.data(hbwtrips, varying=c(8:11), shape=long, choice=CHOICEN, alt.var=ALTNUMTest) Hope that helps, Regards, Mark. -- View this message in context: http://r.789695.n4.nabble.com/mlogit-data-tp3328739p3330148.html Sent from the R help mailing list archive at Nabble.com. __
Re: [R] Is there any Command showing correlation of all variables in a dataset?
cor.prob function gives matrix of correlation coefficients and p-values together
### Function for calculating correlation matrix, corrs below diagonal,
### and P-values above diagonal cor.prob - function(X, dfr = nrow(X) - 2)
{ R - cor(X) above - row(R) col(R) r2 - R[above]^2 Fstat - r2 * dfr / (1
- r2) R[above] - 1 - pf(Fstat, 1, dfr) R }
Scott
On Tuesday, March 1, 2011 at 9:54 AM, rex.dw...@syngenta.com wrote:
?cor answers that question. If Housing is a dataframe, cor(Housing) should do
it. Surprisingly, ??correlation doesn't point you to ?cor.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of JoonGi
Sent: Tuesday, March 01, 2011 5:41 AM
To: r-help@r-project.org
Subject: [R] Is there any Command showing correlation of all variables in a
dataset?
Thanks in advance.
I want to derive correlations of variables in a dataset
Specifically
library(Ecdat)
data(Housing)
attach(Housing)
cor(lotsize, bathrooms)
this code results only the correlationship between two variables.
But I want to examine all the combinations of variables in this dataset.
And I will finally make a table in Latex.
How can I test correlations for all combinations of variables?
with one simple command?
--
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Re: [R] problems with playwith
On Tue, Mar 1, 2011 at 5:58 PM, R Heberto Ghezzo, Dr heberto.ghe...@mcgill.ca wrote: hello, i tried to run playwith but : library(playwith) Loading required package: lattice Loading required package: cairoDevice Loading required package: gWidgetsRGtk2 Loading required package: gWidgets Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared object 'H:/R/cran/RGtk2/libs/i386/RGtk2.dll': LoadLibrary failure: The specified procedure could not be found. Failed to load RGtk2 dynamic library, attempting to install it. Did you install RGtk2? [1] Liviu [1] https://code.google.com/p/playwith/ Learn more about GTK+ at http://www.gtk.org If the package still does not load, please ensure that GTK+ is installed and that it is on your PATH environment variable IN ANY CASE, RESTART R BEFORE TRYING TO LOAD THE PACKAGE AGAIN Error : .onAttach failed in attachNamespace() for 'gWidgetsRGtk2', details: call: .Call(name, ..., PACKAGE = PACKAGE) error: C symbol name S_gtk_icon_factory_new not in DLL for package RGtk2 Error: package 'gWidgetsRGtk2' could not be loaded Sys.getenv(PATH) PATH H:\\R/GTK/bin;H:\\R/GTK/lib;H:\\R/ImageMagick;C:\\windows\\system32;C:\\windows;C:\\windows\\System32\\Wbem;C:\\windows\\System32\\WindowsPowerShell\\v1.0\\;C:\\Program Files\\Common Files\\Ulead Systems\\MPEG;C:\\Program Files\\QuickTime\\QTSystem\\;H:\\R\\GTK\\GTK2-Runtime\\bin;H:\\PortableUSB/PortableApps/MikeTex/miktex/bin packages(lattice, cairoDevice, gWidgetsRGtk2, gWidgets, RGtk2, playwith) were reinstalled program GTK was reinstalled. using R-2-12-2 on Windows 7 Can anybody suggest a solution? thanks R.Heberto Ghezzo Ph.D. Montreal - Canada __ 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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ 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] Is there any Command showing correlation of all variables in a dataset?
On Tue, Mar 1, 2011 at 11:41 AM, JoonGi joo...@hanmail.net wrote: Thanks in advance. I want to derive correlations of variables in a dataset Specifically library(Ecdat) data(Housing) attach(Housing) cor(lotsize, bathrooms) this code results only the correlationship between two variables. But I want to examine all the combinations of variables in this dataset. And I will finally make a table in Latex. How can I test correlations for all combinations of variables? with one simple command? See Rcmdr for an example. Liviu -- View this message in context: http://r.789695.n4.nabble.com/Is-there-any-Command-showing-correlation-of-all-variables-in-a-dataset-tp3329599p3329599.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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ 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] Does POSIXlt extract date components properly?
Month counts from 0 in POSIXlt objects, so that April is month 3 in your
example, January being month 0.
Year counts from 1900 in POSIXlt objects, so that 2005 should return as 105
in your example.
All of the other fields in POSIXlt should return values that you might
expect them to a priori. Keep an eye on daylight savings time adjustments if
they apply in your time zone.
On Tue, Mar 1, 2011 at 12:01 PM, Seth W Bigelow sbige...@fs.fed.us wrote:
I would like to use POSIX classes to store dates and extract components of
dates. Following the example in Spector (Data Manipulation in R), I
create a date
mydate = as. POSIXlt('2005-4-19 7:01:00')
I then successfully extract the day with the command
mydate$day
[1] 19
But when I try to extract the month
mydate$mon
[1] 3
it returns the wrong month. And mydate$year is off by about 2,000 years.
Am I doing something wrong?
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
1731 Research Park Drive, Davis California
sbige...@fs.fed.us / ph. 530 759 1718
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Northeastern University
Nahant, MA
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Re: [R] SetInternet2, RCurl and proxy
On 01.03.2011 12:00, Manta wrote: Dear all, I am facing a problem. I am trying to install packages using a proxy, but I am not able to call the setInternet2 function, either with the small or capital s. What package do I have to call then? And, could there be a reason why this does not function? What does the error message say? Is this a recent version of R? Uwe Ligges Thanks, Marco __ 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] Speed up sum of outer products?
What you're doing is breaking up the calculation of X'X
into n steps. I'm not sure what you mean by very slow:
X = matrix(rnorm(1000*50),1000,50)
n = 1000
system.time({C=matrix(0,50,50);for(i in 1:n)C = C + (X[i,] %o% X[i,])})
user system elapsed
0.096 0.008 0.104
Of course, you could just do the calculation directly:
system.time({C1 = t(X) %*% X})
user system elapsed
0.008 0.000 0.007
all.equal(C,C1)
[1] TRUE
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 1 Mar 2011, AjayT wrote:
Hi, I'm new to R and stats, and I'm trying to speed up the following sum,
for (i in 1:n){
C = C + (X[i,] %o% X[i,]) # the sum of outer products - this is very
slow
according to Rprof()
}
where X is a data matrix (nrows=1000 X ncols=50), and n=1000. The sum has to
be calculated over 10,000 times for different X.
I think it is similar to estimating a co-variance matrix for demeaned data
X. I tried using cov, but got different answers, and it was'nt much quicker?
Any help gratefully appreciated,
--
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http://r.789695.n4.nabble.com/Speed-up-sum-of-outer-products-tp3330160p3330160.html
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Re: [R] Speed up sum of outer products?
Isn't the following the canonical (R-ish) way of doing this:
X = matrix(rnorm(1000*50),1000,50)
system.time({C1 = t(X) %*% X}) # Phil's example
C2 - crossprod(X) # use crossprod instead
all.equal(C1,C2)
[1] TRUE
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Phil Spector
Sent: Tuesday, March 01, 2011 12:31 PM
To: AjayT
Cc: r-help@r-project.org
Subject: Re: [R] Speed up sum of outer products?
What you're doing is breaking up the calculation of X'X
into n steps. I'm not sure what you mean by very slow:
X = matrix(rnorm(1000*50),1000,50)
n = 1000
system.time({C=matrix(0,50,50);for(i in 1:n)C = C + (X[i,] %o% X[i,])})
user system elapsed
0.096 0.008 0.104
Of course, you could just do the calculation directly:
system.time({C1 = t(X) %*% X})
user system elapsed
0.008 0.000 0.007
all.equal(C,C1)
[1] TRUE
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 1 Mar 2011, AjayT wrote:
Hi, I'm new to R and stats, and I'm trying to speed up the following sum,
for (i in 1:n){
C = C + (X[i,] %o% X[i,]) # the sum of outer products - this is very
slow
according to Rprof()
}
where X is a data matrix (nrows=1000 X ncols=50), and n=1000. The sum has to
be calculated over 10,000 times for different X.
I think it is similar to estimating a co-variance matrix for demeaned data
X. I tried using cov, but got different answers, and it was'nt much quicker?
Any help gratefully appreciated,
--
View this message in context: http://r.789695.n4.nabble.com/Speed-up-sum-of-
outer-products-tp3330160p3330160.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] nls not solving
On 2011-03-01 06:38, Schatzi wrote: Here is a reply by Bart: Yes you're right (I should have taken off my glasses and looked closer). However, the argument is essentially the same: Suppose you have a solution with a,b,k,l. Then for any positive c, [a+b-bc] + [bc] + (bc) *exp(kl')exp(-kx) is also a solution, where l' = l - log(c)/k . Cheers, Bert (Feel free to post this correction if you like) This is from me: The problem with dropping the l parameter is that it is supposed to account for the lag component. This equation was published in the literature and has been being solved in SAS. When I put it in excel, it solves, but not very well as it comes to a different solution for each time that I change the starting values. As such, I'm not sure how SAS solves for it and I'm not sure what I should do about the equation. Maybe I should just drop the parameter a. Thanks for the help. When you say 'published in the literature' you should provide a reference; you may be misinterpreting what's published. If SAS provides a 'solution', then there's an added assumption being made (perhaps 'l' is being fixed?). What Excel does is of little interest. 'Dropping' the parameter 'a' is equivalent to setting a=0. You could also set, say, a = -10 or l = 50, or ... The point is that, as Bert says, the model is nonidentifiable. Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SetInternet2, RCurl and proxy
It says the function does not exist. The version is around 2.8, cant check right now. Is it because it's an older version? If so, is there any way to do it in a different way then? -- View this message in context: http://r.789695.n4.nabble.com/SetInternet2-RCurl-and-proxy-tp3248576p3330244.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] Data type problem when extract data from SQLite to R by using RSQLite
Hi Seth, Thanks so much for identifying the problem and explaining everything. I think the first solution that you suggest--make sure the schema has well defined types--would work the best for me. But, I have one question about how to implement it, which is more about sqlite itself. First, I found out that the columns that don't have the expected data types in the table annual_data3 are created by aggregate functions in a separate table. These columns are later combined with other columns that do. I read the link that you provide, http://www.sqlite.org/datatype3.html. One paragraph says When grouping values with the GROUP BY clause values with different storage classes are considered distinct, except for INTEGER and REAL values which are considered equal if they are numerically equal. No affinities are applied to any values as the result of a GROUP by clause. If I understand it correctly, the columns created by aggregate functions with a GROUP by clause do not have any expected data types. My solution is to use CREATE TABLE clause to declare the expected datatype and then insert the values of columns created by the aggregate functions with the GROUP by clause. However, this solution requires a CREATE TABLE cause every time the aggregate function and the GROUP by clause is used. My question is: Is this the best way to make sure that the columns as a result of a GROUP by clause have the expected data types? Thanks. Best, Jia On Tue, Mar 1, 2011 at 1:16 AM, Seth Falcon s...@userprimary.net wrote: Hi Jia, On Mon, Feb 28, 2011 at 6:57 PM, chen jia chen_1...@fisher.osu.edu wrote: The .schema of table annual_data3 is sqlite .schema annual_data3 CREATE TABLE annual_data3( PERMNO INT, DATE INT, CUSIP TEXT, EXCHCD INT, SICCD INT, SHROUT INT, PRC REAL, RET REAL, ... pret_var, pRET_sd, nmret, pya_var, [snip] Is there a reason that you've told SQLite the expected data type for only some of the columns? Interestingly, I find that the problem I reported does not for columns labeled real in the schema info. For example, the type of column RET never changes no matter what the first observation is. Yes, that is expected and I think it is the solution to your problem: setup your schema so that all columns have a declared type. For some details on SQLite's type system see http://www.sqlite.org/datatype3.html. RSQLite currently maps NA values to NULL in the database. Pulling data out of a SELECT query, RSQLite uses the sqlite3_column_type SQLite API to determine the data type and map it to an R type. If NULL is encountered, then the schema is inspected using sqlite3_column_decltype to attempt to obtain a type. If that fails, the data is mapped to a character vector at the R level. The type selection is done once after the first row has been fetched. To work around this you can: - make sure your schema has well defined types (which will help SQLite perform its operations); - check whether the returned column has the expected type and convert if needed at the R level. - remove NA/NULL values from the db or decide on a different way of encoding them (e.g you might be able to use -1 in the db in some situation to indicate missing). Your R code would then need to map these to proper NA. Hope that helps. + seth -- Seth Falcon | @sfalcon | http://userprimary.net/ -- 700 Fisher Hall 2100 Neil Ave. Columbus, Ohio 43210 http://www.fisher.osu.edu/~chen_1002/ __ 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] Adjusting values via vectorization
I'm adjusting values in a list based on a couple of matrixes. One matrix
specifies the row to be taken from the adjustment matrix, while using the
aligned column values. I have an approach which works, but I might find an
approach with vectorization.
Here is code with my solution:
--
nids - 10
npredictors - 2
ncol - 4
values - sample(c(-1,0,1),nids,replace=TRUE)
input - matrix(sample(1:ncol,nids*npredictors,replace=TRUE),nrow=nids)
values.adjust - matrix(rnorm(ncol*npredictors),ncol=npredictors)
for(i in 1:nids){
for(j in 1:npredictors){
values[i] - values[i] + values.adjust[input[i,j],j]
}
}
--
I'm using this as an example to hopefully better understand R syntax w.r.t.
vectorization. Is there such a way to replace my for loops?
Thanks,
-albert
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Re: [R] Does POSIXlt extract date components properly?
On Tue, 1 Mar 2011, Seth W Bigelow wrote:
I would like to use POSIX classes to store dates and extract components of
dates. Following the example in Spector (Data Manipulation in R), I
create a date
mydate = as. POSIXlt('2005-4-19 7:01:00')
I then successfully extract the day with the command
mydate$day
[1] 19
But when I try to extract the month
mydate$mon
[1] 3
it returns the wrong month. And mydate$year is off by about 2,000 years.
Am I doing something wrong?
Not reading the documentation (nor the posting guide).
?DateTimeClasses says
‘mon’ 0-11: months after the first of the year.
‘year’ years since 1900.
That is the POSIX standard ... you could also have looked there.
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
1731 Research Park Drive, Davis California
sbige...@fs.fed.us / ph. 530 759 1718
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Please note what the posting guide said about that!
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--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] boa library and plots
Many thanks for your response, and I am sorry I did not post correctly. I have found dev.copy2eps() useful. Emma -- View this message in context: http://r.789695.n4.nabble.com/boa-library-and-plots-tp3322508p3330299.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] bootstrap resampling question
I'm not sure that is equivalent to sampling with replacement, since if the
first draw is 1, then the probability that the next draw will be one is
4/100 instead of the 1/20 it would be in sampling with replacement. I
think the way to do this would be what Greg suggested - something like:
bigsamp - sample(1:20, 100, T)
idx - sort(unlist(sapply(1:20, function(x) which(bigsamp ==
x)[1:5])))[1:20]
samp - bigsamp[idx]
--
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Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the word... imbue it.
- Jubal Early, Firefly
r-help-boun...@r-project.org wrote on 03/01/2011 09:37:31 AM:
[image removed]
Re: [R] bootstrap resampling question
Giovanni Petris
to:
Bodnar Laszlo EB_HU
03/01/2011 11:58 AM
Sent by:
r-help-boun...@r-project.org
Cc:
'r-help@r-project.org'
A simple way of sampling with replacement from 1:20, with the additional
constraint that each number can be selected at most five times is
sample(rep(1:20, 5), 20)
HTH,
Giovanni
On Tue, 2011-03-01 at 11:30 +0100, Bodnar Laszlo EB_HU wrote:
Hello there,
I have a problem concerning bootstrapping in R - especially
focusing on the resampling part of it. I try to sum it up in a
simplified way so that I would not confuse anybody.
I have a small database consisting of 20 observations (basically
numbers from 1 to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
I would like to resample this database many times for the
bootstrap process with the following two conditions. The resampled
databases should also have 20 observations and you can select each
of the previously mentioned 20 numbers with replacement. I guess it
is obvious so far. Now the more difficult second condition is that
one number can be selected only maximum 5 times. In order to make
this clear I try to show you an example. So there can be resampled
databases like the following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
(4 different numbers are chosen, each selected 5 times)
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
(Two numbers - 8 and 6 - selected 5 times, number 1 selected
four times, the others selected less than 4 times)
My very first guess that came to my mind whilst thinking about the
problem was the sample function where there are settings like
replace=TRUE and prob=... where you can create a probability vector
i.e. how much should be the probability of selecting a number. So I
tried to calculate probabilities first. I thought the problem can
basically described as a k-combination with repetitions.
Unfortunately the only thing I could calculate so far is the total
number of all possible selections which amounts to 137 846 527 049.
Anybody knows how to implement my second tricky condition into
one of the R functions? Are 'boot' and 'bootstrap' packages capable
of managing this? I guess they are, I just couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos
és/vagy jogilag, szakmailag vagy más módon védett információt
tartalmazhat. Amennyiben nem Ön a levél címzettje akkor a levél
tartalmának közlése, reprodukálása, másolása, vagy egyéb más úton
történő terjesztése, felhasználása szigorúan tilos. Amennyiben
tévedésből kapta meg ezt az üzenetet kérjük azonnal értesítse az
üzenet küldőjét. Az Erste Bank Hungary Zrt. (EBH) nem vállal
felelősséget az információ teljes és pontos - címzett(ek)hez történő
- eljuttatásáért, valamint semmilyen késésért, kapcsolat
megszakadásból eredő hibáért, vagy az információ felhasználásából
vagy annak megbízhatatlanságából eredő kárért.
Az üzenetek EBH-n kívüli küldője vagy címzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet
az EBH folytonos munkamenetének biztosítása érdekében.
This e-mail and any attached files are confidential
and/...{{dropped:19}}
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--
Giovanni Petris gpet...@uark.edu
Associate Professor
Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)
http://definetti.uark.edu/~gpetris/
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[R] more boa plots questions
I have MCMC output chains A and B for example, I want to produce trace plots
for them using the boa command line...
#loads boa
boa.init()
#reads in chains
boa.chain.add(boa.importMatrix('A'), 'A')
boa.chain.add(boa.importMatrix('B'), 'B')
#plot trace plot
problems arise here!
I know I can get trace plots using boa.plot('trace') but this plots the
parameter chains on the same plot- I want separate plots using
boa.plot.trace()
#from the manual..
boa.plot.trace(lnames, pname, annotate = boa.par(legend))
lnames: Character vector giving the name of the desired MCMC sequence
in the working session list of sequences.
pname: Character string giving the name of the parameters to be
plotted.
annotate: Logical value indicating that a legend be included in the
plot.
I tried boa.plot.trace(B) and boa.plot.trace(B,B) but both do not give me
a trace plot for chain B and print FALSE. I am obviously misinterpreting
lnames and pnames.
Can anyone help please?
Thanks in advance Emma
--
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[R] Lattice: useOuterStrips and axes
Consider the following:
library(lattice)
library(latticeExtra)
temp - expand.grid(
subject = factor(paste('Subject', 1:3)),
var = factor(paste('Variable', 1:3)),
time = 1:10
)
temp$resp - rnorm(nrow(temp), 10 * as.numeric(temp$var), 1)
ylimits - by(temp$resp, temp$var, function(x) range(pretty(x)))
useOuterStrips(xyplot(
resp ~ time | subject * var,
data = temp,
as.table = TRUE,
scales = list(
alternating = 1, tck = c(1, 0),
y = list(relation = 'free', rot = 0, limits
= rep(ylimits, each =
3))
)
))
This is a matrix of variables on subjects, where it makes sense to have
panel-specific y-axes because of the differing variable ranges. In fact, it
makes sense to have row-specific y-axes, because of the similarity of
intra-variable inter-subject response. However the graphic as presented
gives per-panel y-axis annotation - I'd like to drop the 2nd and 3rd columns
of y-axes labels, so the panels become a contiguous array.
Is this possible?
Thanks,
Jim Price.
Cardiome Pharma. Corp.
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Re: [R] Speed up sum of outer products?
Hey thanks alot guys !!! That really speeds things up !!! I didn't know %*%
and crossprod, could operate on matrices. I think you've saved me hours in
calculation time. Thanks again.
system.time({C=matrix(0,50,50);for(i in 1:n)C = C + (X[i,] %o% X[i,])})
user system elapsed
0.450.000.90
system.time({C1 = t(X) %*% X})
user system elapsed
0.020.000.05
system.time({C2 = crossprod(X)})
user system elapsed
0.020.000.02
--
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Re: [R] Finding pairs with least magnitude difference from mean
No, that's not what I meant, but maybe I didn't understand the question. What I suggested would involve sorting y, not x: sort the *distances*. If you want to minimize the sd of a subset of numbers, you sort the numbers and find a subset that is clumped together. If the numbers are a function of pairs, you compute the function for all pairs of numbers, and find a subset that's clumped together. Anyway, it's an idea, not a theorem, so proof is left as an exercise for the esteemed reader. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Hans W Borchers Sent: Monday, February 28, 2011 2:17 PM To: r-h...@stat.math.ethz.ch Subject: Re: [R] Finding pairs with least magnitude difference from mean rex.dwyer at syngenta.com writes: James, It seems the 2*mean(x) term is irrelevant if you are seeking to minimize sd. Then you want to sort the distances from smallest to largest. Then it seems clear that your five values will be adjacent in the list, since if you have a set of five adjacent values, exchanging any of them for one further away in the list will increase the sd. The only problem I see with this is that you can't use a number more than once. In any case, you need to compute the best five pairs beginning at position i in the sorted list, for 1=i=choose(n,2), then take the max over all i. There no R in my answer such as you'd notice, but I hope it helps just the same. Rex You probably mean something like the following: x - rnorm(10) y - outer(x, x, +) - (2 * mean(x)) o - order(x) sd(c(y[o[1],o[10]], y[o[2],o[9]], y[o[3],o[8]], y[o[4],o[7]], y[o[5],o[6]])) This seems reasonable, though you would have to supply a more stringent argument. I did two tests and it works alright. --Hans Werner __ 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. message may contain confidential information. If you are not the designated recipient, please notify the sender immediately, and delete the original and any copies. Any use of the message by you is prohibited. __ 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] bootstrap resampling - simplified
Hi:
On Tue, Mar 1, 2011 at 8:22 AM, Bodnar Laszlo EB_HU
laszlo.bod...@erstebank.hu wrote:
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on the
resampling part of it. I try to sum it up in a simplified way so that I
would not confuse anybody.
I have a small database consisting of 20 observations (basically numbers
from 1 to 20, I mean: 1, 2, 3, 4, 5, ... 18, 19, 20).
To check on the probability of this event happening, I ran the following:
bootmat - matrix(sample(1:20, 20, replace = TRUE), nrow = 1)
sum(apply(bootmat, 1, function(x) any(table(x) = 5)) )
[1] 492
It's about 0.05. A Q D 'solution' would be to oversample by at least 5%
(let's do 10% just to be on the safe side) and then pick out the first B of
these. In the above example, we could do 11000 samples instead, and pick out
the first 1 that meet the criterion:
bootmat - matrix(sample(1:20, 22, replace = TRUE), nrow = 11000)
badsamps - apply(bootmat, 1, function(x) any(tabulate(x) = 5))
bootfin - bootmat[-badsamps, ][1:1, ]
Time:
user system elapsed
0.280.000.28
(Note 1: Using table instead of tabulate took 4.22 seconds on my machine -
tabulate is much faster.)
(Note 2: In the call above, there were 539 bad samples, so the 5% ballpark
estimate seems plausible.)
This is a simple application of the accept-reject criterion. I don't know
how large 'many' is to you, but 10,000 seems to be a reasonable starting
point. I ran it again for 1,000,000 such samples, and the completion time
was
user system elapsed
36.740.31 37.15
so the processing time is of an order a bit larger than linear. If your
simulations are of this magnitude and are to be run repeatedly, you probably
need to write a function to improve the speed and to get rid of the waste
produced by a rejection sampling approach. If this is a one-off deal,
perhaps the above is sufficient.
HTH,
Dennis
I would like to resample this database many times for the bootstrap process
with the following conditions. Firstly, every resampled database should also
include 20 observations. Secondly, when selecting a number from the
above-mentioned 20 numbers, you can do this selection with replacement. The
difficult part comes now: one number can be selected only maximum 5 times.
In order to make this clear I show you a couple of examples. So the
resampled databases might be like the following ones:
(1st database) 1,2,1,2,1,2,1,2,1,2,3,3,3,3,3,4,4,4,4,4
4 different numbers are chosen (1, 2, 3, 4), each selected - for the
maximum possible - 5 times.
(2nd database) 1,8,8,6,8,8,8,2,3,4,5,6,6,6,6,7,19,1,1,1
Two numbers - 8 and 6 - selected 5 times (the maximum possible times),
number 1 selected 4 times, the others selected less than 4 times.
(3rd database) 1,1,2,2,3,3,4,4,9,9,9,10,10,13,10,9,3,9,2,1
Number 9 chosen for the maximum possible 5 times, number 10, 3, 2, 1 chosen
for 3 times, number 4 selected twice and number 13 selected only once.
...
Anybody knows how to implement my tricky condition into one of the R
functions - that one number can be selected only 5 times at most? Are 'boot'
and 'bootstrap' packages capable of managing this? I guess they are, I just
couldn't figure it out yet...
Thanks very much! Best regards,
Laszlo Bodnar
Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos és/vagy
jogilag, szakmailag vagy más módon védett információt tartalmazhat.
Amennyiben nem Ön a levél címzettje akkor a levél tartalmának közlése,
reprodukálása, másolása, vagy egyéb más úton történõ terjesztése,
felhasználása szigorúan tilos. Amennyiben tévedésbõl kapta meg ezt az
üzenetet kérjük azonnal értesítse az üzenet küldõjét. Az Erste Bank Hungary
Zrt. (EBH) nem vállal felelõsséget az információ teljes és pontos -
címzett(ek)hez történõ - eljuttatásáért, valamint semmilyen késésért,
kapcsolat megszakadásból eredõ hibáért, vagy az információ felhasználásából
vagy annak megbízhatatlanságából eredõ kárért.
Az üzenetek EBH-n kívüli küldõje vagy címzettje tudomásul veszi és
hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az EBH
folytonos munkamenetének biztosítása érdekében.
This e-mail and any attached files are confidential and/...{{dropped:19}}
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[R] inefficient ifelse() ?
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f() and g() for
all values first, and presumably then just picks left or right results
based on t%%2. uggh... wouldn't it make more sense to evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
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Re: [R] inefficient ifelse() ?
Try this:
mapply(function(x, f)f(x), split(t, t %% 2), list(g, f))
On Tue, Mar 1, 2011 at 4:19 PM, ivo welch ivo...@gmail.com wrote:
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f() and g() for
all values first, and presumably then just picks left or right results
based on t%%2. uggh... wouldn't it make more sense to evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
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--
Henrique Dallazuanna
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25° 25' 40 S 49° 16' 22 O
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[R] glht() used with coxph()
Hi, I am experimenting with using glht() from multcomp package together with coxph(), and glad to find that glht() can work on coph object, for example: (fit-coxph(Surv(stop, status0)~treatment,bladder1)) coxph(formula = Surv(stop, status 0) ~ treatment, data = bladder1) coef exp(coef) se(coef) zp treatmentpyridoxine -0.063 0.9390.161 -0.391 0.70 treatmentthiotepa -0.159 0.8530.168 -0.947 0.34 Likelihood ratio test=0.91 on 2 df, p=0.635 n= 294 glht(fit,linfct=mcp(treatment='Tukey')) General Linear Hypotheses Multiple Comparisons of Means: Tukey Contrasts Linear Hypotheses: Estimate pyridoxine - placebo == 0 -0.06303 thiotepa - placebo == 0-0.15885 thiotepa - pyridoxine == 0 -0.09582 However, once I added a strata term in the formula of coxph(), then glht() can't work anymore: (fit-coxph(Surv(stop, status0)~treatment+strata(enum),bladder1)) coxph(formula = Surv(stop, status 0) ~ treatment + strata(enum), data = bladder1) coef exp(coef) se(coef) zp treatmentpyridoxine 0.188 1.210.170 1.11 0.27 treatmentthiotepa -0.210 0.810.172 -1.22 0.22 Likelihood ratio test=4.39 on 2 df, p=0.111 n= 294 glht(fit,linfct=mcp(treatment='Tukey')) Error in glht.matrix(model = list(coefficients = c(0.187792527684977, : âncol(linfct)â is not equal to âlength(coef(model))â Can anyone suggest why strata would make coxph object ineligible for glht()? Or how to make it work? Thanks John [[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] expression help
Hello, I am trying to write math-type on a plot. Due to space limitations on the plot, I want 2 short expressions written on top of each other. It is certainly possible to write them in two separate calls, but that involves fine-tuning locations and a lot of trial and error (and I'm trying to write a general purpose function). Here's where I've gotten to: plot(0:1,0:1,xaxt=n) axis(side=1,at=.3,expression(paste(IFN-, gamma, \n, TNF-, alpha))) axis(side=1,at=.6,a label\n2nd line) What I am trying to do is illustrated by the the non-expression axis label (a label\n2nd line). The \n forces a new line when I'm not using expressions, but doesn't work for my real example. I have googled for general documentation on expressions, but the only thing I've been able to find is the help for plotmath(), so if someone can point me to more complete documentation that would also be helpful. thanks, Daryl SCHARP, FHCRC, UW Biostatistics __ 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] Export R dataframes to excel
I'm trying to do this in several ways but havent had any result. Im asked to install python, or perl etc. Can anybody suggest a direct, easy and understandable way? Every help would be appreciated. Thx. -- View this message in context: http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330399.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] odbcConnectExcel2007 creates corrupted files
I tried creating a .xlsx file using odbcConnectExcel2007 and adding a worksheet with sqlSave. This seems to work, I am even able to query the worksheet, but when I try opening the file in Excel I get the following message: Excel cannot open the file 'test.xlx' because the file format or file extension is not valid. Verify that the file has not been corrupted and that the file extension matches the format of the file. Is this a known issue? Or just user error? The RODBC manual seemed to indicate that sqlSave worked fine with Excel, however it did not mention Excel 2007. I am running Excel 2007 and R 2.12.1 on Windows XP. Below is my example code. $ library(RODBC) $ $ # This doesn't work $ # Connect to an previously non-existent Excel file $ out - odbcConnectExcel2007(test.xlsx, readOnly=FALSE) $ test - data.frame(x=1:10, y=rnorm(10)) $ sqlSave(out, test) $ sqlTables(out) TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS 1 C:\\Documents and Settings\\G69974\\My Documents\\test.xlsxNA test$ SYSTEM TABLENA 2 C:\\Documents and Settings\\G69974\\My Documents\\test.xlsx NA testTABLENA $ sqlFetch(out, test) x y 1 1 0.5832882 2 2 0.4387569 3 3 -0.6444048 4 4 -1.0013450 5 5 1.0324718 6 6 -0.7844128 7 7 -1.6789266 8 8 0.1402672 9 9 0.8650061 10 10 -0.0420201 $ close(out) $ # Opening test.xlsx now fails $ $ # This works $ out - odbcConnectExcel(test.xls, readOnly=FALSE) $ test - data.frame(x=1:10, y=rnorm(10)) $ sqlSave(out, test) $ sqlTables(out) TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS 1 C:\\Documents and Settings\\G69974\\My Documents\\testNA test$ SYSTEM TABLENA 2 C:\\Documents and Settings\\G69974\\My Documents\\testNA testTABLENA $ sqlFetch(out, test) x y 1 1 0.5955787 2 2 1.0517528 3 3 0.3884892 4 4 -2.1408813 5 5 -0.7081686 6 6 0.1511828 7 7 2.0560555 8 8 -0.5801912 9 9 -0.6988058 10 10 -0.1237739 $ close(out) $ # Opening test.xls now works __ 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] Export R dataframes to excel
write.table() using the sep=, and file extension as .csv works great to pull directly into excel. ?write.table Without more detail as to the problem, it is difficult to give a more specific answer. Adrian -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of maxsilva Sent: Tuesday, March 01, 2011 2:11 PM To: r-help@r-project.org Subject: [R] Export R dataframes to excel I'm trying to do this in several ways but havent had any result. Im asked to install python, or perl etc. Can anybody suggest a direct, easy and understandable way? Every help would be appreciated. Thx. -- View this message in context: http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330399.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-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] Data type problem when extract data from SQLite to R by using RSQLite
On Tue, Mar 1, 2011 at 10:06 AM, chen jia chen_1...@fisher.osu.edu wrote: Hi Seth, Thanks so much for identifying the problem and explaining everything. I think the first solution that you suggest--make sure the schema has well defined types--would work the best for me. But, I have one question about how to implement it, which is more about sqlite itself. First, I found out that the columns that don't have the expected data types in the table annual_data3 are created by aggregate functions in a separate table. These columns are later combined with other columns that do. I read the link that you provide, http://www.sqlite.org/datatype3.html. One paragraph says When grouping values with the GROUP BY clause values with different storage classes are considered distinct, except for INTEGER and REAL values which are considered equal if they are numerically equal. No affinities are applied to any values as the result of a GROUP by clause. If I understand it correctly, the columns created by aggregate functions with a GROUP by clause do not have any expected data types. My solution is to use CREATE TABLE clause to declare the expected datatype and then insert the values of columns created by the aggregate functions with the GROUP by clause. However, this solution requires a CREATE TABLE cause every time the aggregate function and the GROUP by clause is used. My question is: Is this the best way to make sure that the columns as a result of a GROUP by clause have the expected data types? Thanks. That might be a good question to post to the SQLite user's list :-) I don't have an answer off the top of my head. My reading of the SQLite docs would lead me to expect that a GROUP BY clause would not change/remove type if the column being grouped contains all the same declared type affinity. + seth Best, Jia On Tue, Mar 1, 2011 at 1:16 AM, Seth Falcon s...@userprimary.net wrote: Hi Jia, On Mon, Feb 28, 2011 at 6:57 PM, chen jia chen_1...@fisher.osu.edu wrote: The .schema of table annual_data3 is sqlite .schema annual_data3 CREATE TABLE annual_data3( PERMNO INT, DATE INT, CUSIP TEXT, EXCHCD INT, SICCD INT, SHROUT INT, PRC REAL, RET REAL, ... pret_var, pRET_sd, nmret, pya_var, [snip] Is there a reason that you've told SQLite the expected data type for only some of the columns? Interestingly, I find that the problem I reported does not for columns labeled real in the schema info. For example, the type of column RET never changes no matter what the first observation is. Yes, that is expected and I think it is the solution to your problem: setup your schema so that all columns have a declared type. For some details on SQLite's type system see http://www.sqlite.org/datatype3.html. RSQLite currently maps NA values to NULL in the database. Pulling data out of a SELECT query, RSQLite uses the sqlite3_column_type SQLite API to determine the data type and map it to an R type. If NULL is encountered, then the schema is inspected using sqlite3_column_decltype to attempt to obtain a type. If that fails, the data is mapped to a character vector at the R level. The type selection is done once after the first row has been fetched. To work around this you can: - make sure your schema has well defined types (which will help SQLite perform its operations); - check whether the returned column has the expected type and convert if needed at the R level. - remove NA/NULL values from the db or decide on a different way of encoding them (e.g you might be able to use -1 in the db in some situation to indicate missing). Your R code would then need to map these to proper NA. Hope that helps. + seth -- Seth Falcon | @sfalcon | http://userprimary.net/ -- 700 Fisher Hall 2100 Neil Ave. Columbus, Ohio 43210 http://www.fisher.osu.edu/~chen_1002/ -- Seth Falcon | @sfalcon | http://userprimary.net/ __ 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] Export R dataframes to excel
You can copy it with the following function and then paste into Excel...
copy = function (df, buffer.kb=256) {
write.table(df, file=paste(clipboard-,buffer.kb,sep=),
sep=\t, na='', quote=FALSE, row.names=FALSE)
}
From: maxsilva mmsil...@uc.cl
To:r-help@r-project.org
Date: 2/Mar/2011 8:50a
Subject: [R] Export R dataframes to excel
I'm trying to do this in several ways but havent had any result. Im asked to
install python, or perl etc. Can anybody suggest a direct, easy and
understandable way? Every help would be appreciated.
Thx.
--
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Re: [R] Export R dataframes to excel
maxsilva wrote: Thx, but im looking for a more direct solution... my problem is very simple, I have a dataframe and I want to create a standard excel spreadsheet. My dataframe could be something like this More or less the same question was answered several hours ago. See http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windowss=excel as Gabor Grothendieck suggested. /Berend -- View this message in context: http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330518.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] Export R dataframes to excel
Or ?write.csv which excel will import On 1-Mar-11, at 12:17 PM, Berend Hasselman wrote: maxsilva wrote: Thx, but im looking for a more direct solution... my problem is very simple, I have a dataframe and I want to create a standard excel spreadsheet. My dataframe could be something like this More or less the same question was answered several hours ago. See http://rwiki.sciviews.org/doku.php?id=tips:data- io:ms_windowss=excel as Gabor Grothendieck suggested. /Berend -- View this message in context: http://r.789695.n4.nabble.com/Export- R-dataframes-to-excel-tp3330399p3330518.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. Why does the universe go to all the bother of existing? -- Stephen Hawking #define QUESTION ((bb) || !(bb)) -- William Shakespeare Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@uw.edu donaldmcken...@fs.fed.us __ 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] Lattice: useOuterStrips and axes
On 2011-03-01 10:29, Jim Price wrote:
Consider the following:
library(lattice)
library(latticeExtra)
temp- expand.grid(
subject = factor(paste('Subject', 1:3)),
var = factor(paste('Variable', 1:3)),
time = 1:10
)
temp$resp- rnorm(nrow(temp), 10 * as.numeric(temp$var), 1)
ylimits- by(temp$resp, temp$var, function(x) range(pretty(x)))
useOuterStrips(xyplot(
resp ~ time | subject * var,
data = temp,
as.table = TRUE,
scales = list(
alternating = 1, tck = c(1, 0),
y = list(relation = 'free', rot = 0, limits
= rep(ylimits, each =
3))
)
))
This is a matrix of variables on subjects, where it makes sense to have
panel-specific y-axes because of the differing variable ranges. In fact, it
makes sense to have row-specific y-axes, because of the similarity of
intra-variable inter-subject response. However the graphic as presented
gives per-panel y-axis annotation - I'd like to drop the 2nd and 3rd columns
of y-axes labels, so the panels become a contiguous array.
Is this possible?
Looks like you want the combineLimits() function in latticeExtra.
Specifically,
p - [your code]
combineLimits(p, margin.y = 1, extend = FALSE)
Peter Ehlers
Thanks,
Jim Price.
Cardiome Pharma. Corp.
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Re: [R] inefficient ifelse() ?
thanks, Henrique. did you mean
as.vector(t(mapply(function(x, f)f(x), split(t, ((t %% 2)==0)),
list(f, g ?
otherwise, you get a matrix.
its a good solution, but unfortunately I don't think this can be used
to redefine ifelse(cond,ift,iff) in a way that is transparent. the
ift and iff functions will always be evaluated before the function
call happens, even with lazy evaluation. :-(
I still think that it makes sense to have a smarter vectorized %if% in
a vectorized language like R. just my 5 cents.
/iaw
Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)
On Tue, Mar 1, 2011 at 2:33 PM, Henrique Dallazuanna www...@gmail.com wrote:
Try this:
mapply(function(x, f)f(x), split(t, t %% 2), list(g, f))
On Tue, Mar 1, 2011 at 4:19 PM, ivo welch ivo...@gmail.com wrote:
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f() and g() for
all values first, and presumably then just picks left or right results
based on t%%2. uggh... wouldn't it make more sense to evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
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25° 25' 40 S 49° 16' 22 O
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Re: [R] Speed up sum of outer products?
...and this is where we cue the informative article on least squares
calculations in R by Doug Bates:
http://cran.r-project.org/doc/Rnews/Rnews_2004-1.pdf
HTH,
Dennis
On Tue, Mar 1, 2011 at 10:52 AM, AjayT ajaytal...@googlemail.com wrote:
Hey thanks alot guys !!! That really speeds things up !!! I didn't know %*%
and crossprod, could operate on matrices. I think you've saved me hours in
calculation time. Thanks again.
system.time({C=matrix(0,50,50);for(i in 1:n)C = C + (X[i,] %o% X[i,])})
user system elapsed
0.450.000.90
system.time({C1 = t(X) %*% X})
user system elapsed
0.020.000.05
system.time({C2 = crossprod(X)})
user system elapsed
0.020.000.02
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Re: [R] expression help
Hi: 1. expression() in plotmath ignores control characters such as \n. 2. The workaround, discussed a couple of times on this list (hence in the archives), is to use the atop function, so try something like plot(0:1,0:1,xaxt=n) axis(side=1,at=.3,expression(paste(IFN-, gamma, \n, TNF-, alpha))) axis(side=1,at=.6,expression(atop(a label, 2nd line))) It appears you'll have to make some vertical adjustments, but that's the basic workaround. HTH, Dennis On Tue, Mar 1, 2011 at 11:32 AM, Daryl Morris dar...@uw.edu wrote: Hello, I am trying to write math-type on a plot. Due to space limitations on the plot, I want 2 short expressions written on top of each other. It is certainly possible to write them in two separate calls, but that involves fine-tuning locations and a lot of trial and error (and I'm trying to write a general purpose function). Here's where I've gotten to: plot(0:1,0:1,xaxt=n) axis(side=1,at=.3,expression(paste(IFN-, gamma, \n, TNF-, alpha))) axis(side=1,at=.6,a label\n2nd line) What I am trying to do is illustrated by the the non-expression axis label (a label\n2nd line). The \n forces a new line when I'm not using expressions, but doesn't work for my real example. I have googled for general documentation on expressions, but the only thing I've been able to find is the help for plotmath(), so if someone can point me to more complete documentation that would also be helpful. thanks, Daryl SCHARP, FHCRC, UW Biostatistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regression with many independent variables
You can use ^2 to get all 2 way interactions and ^3 to get all 3 way interactions, e.g.: lm(Sepal.Width ~ (. - Sepal.Length)^2, data=iris) The lm.fit function is what actually does the fitting, so you could go directly there, but then you lose the benefits of using . and ^. The Matrix package has ways of dealing with sparse matricies, but I don't know if that would help here or not. You could also just create x'x and x'y matricies directly since the variables are 0/1 then use solve. A lot depends on what you are doing and what questions you are trying to answer. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: Matthew Douglas [mailto:matt.dougla...@gmail.com] Sent: Tuesday, March 01, 2011 1:09 PM To: Greg Snow Cc: r-help@r-project.org Subject: Re: [R] Regression with many independent variables Hi Greg, Thanks for the help, it works perfectly. To answer your question, there are 339 independent variables but only 10 will be used at one time . So at any given line of the data set there will be 10 non zero entries for the independent variables and the rest will be zeros. One more question: 1. I still want to find a way to look at the interactions of the independent variables. the regression would look like this: y = b12*X1X2 + b23*X2X3 +...+ bk-1k*Xk-1Xk so I think the regression in R would look like this: lm(MARGIN, P235:P236+P236:P237+,weights = Poss, data = adj0708), my problem is that since I have technically 339 independent variables, when I do this regression I would have 339 Choose 2 = approx 57000 independent variables (a vast majority will be 0s though) so I dont want to have to write all of these out. Is there a way to do this quickly in R? Also just a curious question that I cant seem to find to online: is there a more efficient model other than lm() that is better for very sparse data sets like mine? Thanks, Matt On Mon, Feb 28, 2011 at 4:30 PM, Greg Snow greg.s...@imail.org wrote: Don't put the name of the dataset in the formula, use the data argument to lm to provide that. A single period (.) on the right hand side of the formula will represent all the columns in the data set that are not on the left hand side (you can then use - to remove any other columns that you don't want included on the RHS). For example: lm(Sepal.Width ~ . - Sepal.Length, data=iris) Call: lm(formula = Sepal.Width ~ . - Sepal.Length, data = iris) Coefficients: (Intercept) Petal.Length Petal.Width Speciesversicolor 3.0485 0.1547 0.6234 - 1.7641 Speciesvirginica -2.1964 But, are you sure that a regression model with 339 predictors will be meaningful? -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Matthew Douglas Sent: Monday, February 28, 2011 1:32 PM To: r-help@r-project.org Subject: [R] Regression with many independent variables Hi, I am trying use lm() on some data, the code works fine but I would like to use a more efficient way to do this. The data looks like this (the data is very sparse with a few 1s, -1s and the rest 0s): head(adj0708) MARGIN Poss P235 P247 P703 P218 P430 P489 P83 P307 P337 1 64.28571 29 0 0 0 0 0 0 0 0 0 0 0 0 0 2 -100.0 6 0 0 0 0 0 0 0 1 0 0 0 0 0 3 100.0 4 0 0 0 0 0 0 0 1 0 0 0 0 0 4 -33.3 7 0 0 0 0 0 0 0 0 0 0 0 0 0 5 200.0 2 0 0 0 0 0 0 0 0 0 0 -1 0 0 6 -83.3 12 0 -1 0 0 0 0 0 0 0 0 0 0 0 adj0708 is actually a 35657x341 data set. Each column after Poss is an independent variable, the dependent variable is MARGIN and it is weighted by Poss The regression is below: fit.adj0708 - lm( adj0708$MARGIN~adj0708$P235 + adj0708$P247 + adj0708$P703 + adj0708$P430 + adj0708$P489 + adj0708$P218 + adj0708$P605 + adj0708$P337 + + adj0708$P510,weights=adj0708$Poss) I have two questions: 1. Is there a way to to condense how I write the independent variables in the lm(), instead of having such a long line of code (I have 339 independent variables to be exact)? 2. I would like to pair the data to look a regression of the interactions between two independent variables. I think it would look something like this fit.adj0708 - lm( adj0708$MARGIN~adj0708$P235:adj0708$P247 + adj0708$P703:adj0708$P430 + adj0708$P489:adj0708$P218 + adj0708$P605:adj0708$P337 +
Re: [R] Export R dataframes to excel
This appeared today on the r-bloggers site and might be useful for you.
http://www.r-bloggers.com/release-of-xlconnect-0-1-3/
cheers
i
--- On Tue, 1/3/11, Steve Taylor steve.tay...@aut.ac.nz wrote:
From: Steve Taylor steve.tay...@aut.ac.nz
Subject: Re: [R] Export R dataframes to excel
To: r-help@r-project.org, maxsilva mmsil...@uc.cl
Date: Tuesday, 1 March, 2011, 20:15
You can copy it with the following
function and then paste into Excel...
copy = function (df, buffer.kb=256) {
write.table(df,
file=paste(clipboard-,buffer.kb,sep=),
sep=\t, na='', quote=FALSE,
row.names=FALSE)
}
From: maxsilva mmsil...@uc.cl
To:r-help@r-project.org
Date: 2/Mar/2011 8:50a
Subject: [R] Export R dataframes to excel
I'm trying to do this in several ways but havent had any
result. Im asked to
install python, or perl etc. Can anybody suggest a
direct, easy and
understandable way? Every help would be appreciated.
Thx.
--
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http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330399.html
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mailing list
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)-project.org/posting-guide.html
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reproducible code.
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[R] error in saved .csv
Help me please! I would like to be saved a data table: write.csv(random.t1, place, dec=,, append = T, quote = FALSE, sep = , qmethod = double, eol = \n, row.names=F) It's OK! But the rows of file 1,1,21042,-4084.87179487179,2457.66483516483,-582.275562799881 2,2,23846,-6383.86480186479,-3409.98451548449,-3569.72145340269 and no 1 21042 - 4084.87179487179 2457.66483516483 Not proportional... What's the problem??? 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] inefficient ifelse() ?
Hi:
As far as I can see, the problem has to do with the way you wrote f and g:
f(2)
f for 2
[1] 4
g(5)
g for 5
[1] 15
You output an unsaved character string with a numeric result, but the
character string is not part of the return object since it is neither saved
as a name or an attribute. If you save the output to an object, it returns
the numeric outcome:
s - f(2)
s
[1] 4
names(s)
NULL
attributes(s)
NULL
However,
f - function(x) 2^x
g - function(x) 3^x
ifelse(t %% 2 == 0, f(t), g(t))
[1] 3.00e+00 4.00e+00 2.70e+01 1.60e+01 2.43e+02
[6] 6.40e+01 2.187000e+03 2.56e+02 1.968300e+04 1.024000e+03
[11] 1.771470e+05 4.096000e+03 1.594323e+06 1.638400e+04 1.434891e+07
[16] 6.553600e+04 1.291402e+08 2.621440e+05 1.162261e+09 1.048576e+06
[21] 1.046035e+10 4.194304e+06 9.414318e+10 1.677722e+07 8.472886e+11
[26] 6.710886e+07 7.625597e+12 2.684355e+08 6.863038e+13 1.073742e+09
is the same as
ifelse(t %% 2 == 0, 2^t, 3^t)
[1] 3.00e+00 4.00e+00 2.70e+01 1.60e+01 2.43e+02
[6] 6.40e+01 2.187000e+03 2.56e+02 1.968300e+04 1.024000e+03
[11] 1.771470e+05 4.096000e+03 1.594323e+06 1.638400e+04 1.434891e+07
[16] 6.553600e+04 1.291402e+08 2.621440e+05 1.162261e+09 1.048576e+06
[21] 1.046035e+10 4.194304e+06 9.414318e+10 1.677722e+07 8.472886e+11
[26] 6.710886e+07 7.625597e+12 2.684355e+08 6.863038e+13 1.073742e+09
which is the vector result you apparently wanted. If you also want to output
the character string with the value, you could either return a list or
rewrite the print method of ifelse() to accommodate it.
HTH,
Dennis
On Tue, Mar 1, 2011 at 12:36 PM, ivo welch ivo.we...@gmail.com wrote:
thanks, Henrique. did you mean
as.vector(t(mapply(function(x, f)f(x), split(t, ((t %% 2)==0)),
list(f, g ?
otherwise, you get a matrix.
its a good solution, but unfortunately I don't think this can be used
to redefine ifelse(cond,ift,iff) in a way that is transparent. the
ift and iff functions will always be evaluated before the function
call happens, even with lazy evaluation. :-(
I still think that it makes sense to have a smarter vectorized %if% in
a vectorized language like R. just my 5 cents.
/iaw
Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)
On Tue, Mar 1, 2011 at 2:33 PM, Henrique Dallazuanna www...@gmail.com
wrote:
Try this:
mapply(function(x, f)f(x), split(t, t %% 2), list(g, f))
On Tue, Mar 1, 2011 at 4:19 PM, ivo welch ivo...@gmail.com wrote:
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f() and g() for
all values first, and presumably then just picks left or right results
based on t%%2. uggh... wouldn't it make more sense to evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
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[R] [R-pkgs] Major update to rms package
A new version of rms is now available on CRAN for Linux and Windows (Mac will probably be available very soon). Largest changes include latex methods for validate.* and adding the capability to force a subset of variables to be included in all backwards stepdown models (single model or validation by resampling). Recent updates: * In survplot.rms, fixed bug (curves were undefined if conf='bands' and labelc was FALSE) * In survfit.cph, fixed bug by which n wasn't always defined * In cph, put survival::: on exact fit call * Quit ignoring zlim argument in bplot; added xlabrot argument * Added caption argument for latex.anova.rms * Changed predab to not print summaries of variables selected if bw=TRUE * Changed predab to pass force argument to fastbw * fastbw: implemented force argument * Added force argument to validate.lrm, validate.bj, calibrate.default, calibrate.cph, calibrate.psm, validate.bj, validate.cph, validate.ols * print.validate: added B argument to limit how many resamples are printed summarizing variables selected if BW=TRUE * print.calibrate, print.calibrate.default: added B argument * Added latex method for results produced by validate functions * Fixed survest.cph to convert summary.survfit std.err to log S(t) scale * Fixed val.surv by pulling surv object from survest result * Clarified in predict.lrm help file that doesn't always use the first intercept * lrm.fit, lrm: linear predictor stored in fit object now uses first intercept and not middle one (NOT DOWNWARD COMPATIBLE but makes predict work when using stored linear.predictors) * Fixed argument consistency with validate methods More information is at http://biostat.mc.vanderbilt.edu/Rrms -- Frank E Harrell Jr Professor and Chairman School of Medicine Department of Biostatistics Vanderbilt University ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Export R dataframes to excel
Thx, but im looking for a more direct solution... my problem is very simple, I have a dataframe and I want to create a standard excel spreadsheet. My dataframe could be something like this id sex weight 1 M 5'8 2 F6'2 3 F5'5 4 M 5'7 5 F6'3 -- View this message in context: http://r.789695.n4.nabble.com/Export-R-dataframes-to-excel-tp3330399p3330491.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] Regression with many independent variables
Hi Greg, Thanks for the help, it works perfectly. To answer your question, there are 339 independent variables but only 10 will be used at one time . So at any given line of the data set there will be 10 non zero entries for the independent variables and the rest will be zeros. One more question: 1. I still want to find a way to look at the interactions of the independent variables. the regression would look like this: y = b12*X1X2 + b23*X2X3 +...+ bk-1k*Xk-1Xk so I think the regression in R would look like this: lm(MARGIN, P235:P236+P236:P237+,weights = Poss, data = adj0708), my problem is that since I have technically 339 independent variables, when I do this regression I would have 339 Choose 2 = approx 57000 independent variables (a vast majority will be 0s though) so I dont want to have to write all of these out. Is there a way to do this quickly in R? Also just a curious question that I cant seem to find to online: is there a more efficient model other than lm() that is better for very sparse data sets like mine? Thanks, Matt On Mon, Feb 28, 2011 at 4:30 PM, Greg Snow greg.s...@imail.org wrote: Don't put the name of the dataset in the formula, use the data argument to lm to provide that. A single period (.) on the right hand side of the formula will represent all the columns in the data set that are not on the left hand side (you can then use - to remove any other columns that you don't want included on the RHS). For example: lm(Sepal.Width ~ . - Sepal.Length, data=iris) Call: lm(formula = Sepal.Width ~ . - Sepal.Length, data = iris) Coefficients: (Intercept) Petal.Length Petal.Width Speciesversicolor 3.0485 0.1547 0.6234 -1.7641 Speciesvirginica -2.1964 But, are you sure that a regression model with 339 predictors will be meaningful? -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Matthew Douglas Sent: Monday, February 28, 2011 1:32 PM To: r-help@r-project.org Subject: [R] Regression with many independent variables Hi, I am trying use lm() on some data, the code works fine but I would like to use a more efficient way to do this. The data looks like this (the data is very sparse with a few 1s, -1s and the rest 0s): head(adj0708) MARGIN Poss P235 P247 P703 P218 P430 P489 P83 P307 P337 1 64.28571 29 0 0 0 0 0 0 0 0 0 0 0 0 0 2 -100.0 6 0 0 0 0 0 0 0 1 0 0 0 0 0 3 100.0 4 0 0 0 0 0 0 0 1 0 0 0 0 0 4 -33.3 7 0 0 0 0 0 0 0 0 0 0 0 0 0 5 200.0 2 0 0 0 0 0 0 0 0 0 0 -1 0 0 6 -83.3 12 0 -1 0 0 0 0 0 0 0 0 0 0 0 adj0708 is actually a 35657x341 data set. Each column after Poss is an independent variable, the dependent variable is MARGIN and it is weighted by Poss The regression is below: fit.adj0708 - lm( adj0708$MARGIN~adj0708$P235 + adj0708$P247 + adj0708$P703 + adj0708$P430 + adj0708$P489 + adj0708$P218 + adj0708$P605 + adj0708$P337 + + adj0708$P510,weights=adj0708$Poss) I have two questions: 1. Is there a way to to condense how I write the independent variables in the lm(), instead of having such a long line of code (I have 339 independent variables to be exact)? 2. I would like to pair the data to look a regression of the interactions between two independent variables. I think it would look something like this fit.adj0708 - lm( adj0708$MARGIN~adj0708$P235:adj0708$P247 + adj0708$P703:adj0708$P430 + adj0708$P489:adj0708$P218 + adj0708$P605:adj0708$P337 + ,weights=adj0708$Poss) but there will be 339 Choose 2 combinations, so a lot of independent variables! Is there a more efficient way of writing this code. Is there a way I can do this? Thanks, Matt __ 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] Logistic Stepwise Criterion
Dear R-help members, I'd like to run a binomial logistic stepwise regression with ten explanatory variables and as many interaction terms as R can handle. I'll come up with the right R command sooner or later, but my real question is whether and how the criterion for the evaluation of the different models can be set to be the probability of the residual deviance in the Chi-Square distribution (which would be more informative of overall model fit than AIC). Thanks in advance for all your help. Kind regards, Mano Gabor TOTH MA Political Science Central European University [[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] Pairwise T-Tests and Dunnett's Test (possibly using multcomp)
Hello Everyone,
I've been learning to use R in my spare time over the past several months. I've
read about 7-8 books on the subject. Lately I've been testing what I've learned
by trying to replicate the analyses from some of my SAS books. This helps me
make sure I know how to use R properly and also helps me to understand how the
two programs are similar and different.
Below is an attempt to replicate some SAS analyses. This was a success in that
I was able to reproduce the results using R. I have a feeling though that it
may not represent an ideal approach. So I was hoping a few people might be
willing to look at what I've done and to suggest improvements or alternatives.
One thing I'm struggling with is setting the reference level. I inserted a
comment about this in the R code.
I've also pasted in the original SAS code in case some people are also SAS
users and might find this helpful or interesting.
Thanks,
Paul
R Code:
##
Chapter 6: One-Way ANOVA
##
Example 6.1: One-Way ANOVA using HAM-A Scores in GAD
connection - textConnection(
101 LO 21 104 LO 18
106 LO 19 110 LO .
112 LO 28 116 LO 22
120 LO 30 121 LO 27
124 LO 28 125 LO 19
130 LO 23 136 LO 22
137 LO 20 141 LO 19
143 LO 26 148 LO 35
152 LO . 103 HI 16
105 HI 21 109 HI 31
111 HI 25 113 HI 23
119 HI 25 123 HI 18
127 HI 20 128 HI 18
131 HI 16 135 HI 24
138 HI 22 140 HI 21
142 HI 16 146 HI 33
150 HI 21 151 HI 17
102 PB 22 107 PB 26
108 PB 29 114 PB 19
115 PB . 117 PB 33
118 PB 37 122 PB 25
126 PB 28 129 PB 26
132 PB . 133 PB 31
134 PB 27 139 PB 30
144 PB 25 145 PB 22
147 PB 36 149 PB 32
)
gad - data.frame(scan(connection,
list(patno=0,dosegrp=,hama=0),na.strings=.))
gad
Summary Statistics
summaryfn - function(x) data.frame(
mean=mean(x,na.rm=T),std.dev=sd(x,na.rm=T),n=sum(!is.na(x)))
by(gad$hama,gad$dosegrp,summaryfn)
ANOVA
model1 - aov(hama~dosegrp,data=gad)
summary.aov(model1)
Multiple Comparisons
library(multcomp)
Pairwise T-Tests
cht - glht(model1, linfct = mcp(dosegrp = Tukey))
summary(cht, test = univariate())
Dunnett's Test
#Not sure how to set the reference group to placebo
#cht - glht(model1, linfct = mcp(dosegrp = Dunnett))
#summary(cht, test = univariate())
model2 - aov(hama ~ dosegrp - 1, data = gad)
K - rbind(c(1, 0, -1), c(0, 1, -1))
rownames(K) - c(HI vs PL, LO vs PL)
colnames(K) - names(coef(model2))
summary(glht(model2, linfct = K))
Treatment vs. Placebo
#SAS produces an F-test but this is equivalent
K - rbind(c(0.5, 0.5, -1))
rownames(K) - c(TR vs PL)
colnames(K) - names(coef(model2))
summary(glht(model2, linfct = K))
SAS code:
*---*;
** Chapter 6: One-Way ANOVA;
/* Example 6.1: HAM-A Scores in GAD */
DATA GAD;
INPUT PATNO DOSEGRP $ HAMA @@;
DATALINES;
101 LO 21 104 LO 18
106 LO 19 110 LO .
112 LO 28 116 LO 22
120 LO 30 121 LO 27
124 LO 28 125 LO 19
130 LO 23 136 LO 22
137 LO 20 141 LO 19
143 LO 26 148 LO 35
152 LO . 103 HI 16
105 HI 21 109 HI 31
111 HI 25 113 HI 23
119 HI 25 123 HI 18
127 HI 20 128 HI 18
131 HI 16 135 HI 24
138 HI 22 140 HI 21
142 HI 16 146 HI 33
150 HI 21 151 HI 17
102 PB 22 107 PB 26
108 PB 29 114 PB 19
115 PB . 117 PB 33
118 PB 37 122 PB 25
126 PB 28 129 PB 26
132 PB . 133 PB 31
134 PB 27 139 PB 30
144 PB 25 145 PB 22
147 PB 36 149 PB 32
;
PROC SORT DATA = GAD; BY DOSEGRP;
PROC MEANS MEAN STD N DATA = GAD;
BY DOSEGRP;
VAR HAMA;
TITLE1 'One-Way ANOVA';
TITLE2 'EXAMPLE 6.1: HAM-A Scores in GAD';
RUN;
PROC GLM DATA = GAD;
CLASS DOSEGRP;
MODEL HAMA = DOSEGRP;
MEANS DOSEGRP/T DUNNETT('PB');
CONTRAST 'ACTIVE vs. PLACEBO' DOSEGRP 0.5 0.5 -1;
RUN;
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Re: [R] Lattice: useOuterStrips and axes
Thank you, that's exactly what I needed. -- View this message in context: http://r.789695.n4.nabble.com/Lattice-useOuterStrips-and-axes-tp3330338p3330613.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 understand output from R's polr function (ordered logistic regression)?
I am new to R, ordered logistic regression, and polr. The Examples section at the bottom of the help page for polrhttp://stat.ethz.ch/R-manual/R-patched/library/MASS/html/polr.html(that fits a logistic or probit regression model to an ordered factor response) shows options(contrasts = c(contr.treatment, contr.poly)) house.plr - polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing) pr - profile(house.plr) plot(pr) pairs(pr) - What information does pr contain? The help page on profilehttp://stat.ethz.ch/R-manual/R-patched/library/stats/html/profile.htmlis generic, and gives no guidance for polr. - What is plot(pr) showing? I see six graphs. Each has an X axis that is numeric, although the label is an indicator variable (looks like an input variable that is an indicator for an ordinal value). Then the Y axis is tau which is completely unexplained. - What is pairs(pr) showing? It looks like a plot for each pair of input variables, but again I see no explanation of the X or Y axes. - How can one understand if the model gave a good fit? summary(house.plr)shows Residual Deviance 3479.149 and AIC (Akaike Information Criterion?) of 3495.149. Is that good? In the case those are only useful as relative measures (i.e. to compare to another model fit), what is a good absolute measure? Is the residual deviance approximately chi-squared distributed? Can one use % correctly predicted on the original data or some cross-validation? What is the easiest way to do that? - How does one apply and interpret anova on this model? The docs say There are methods for the standard model-fitting functions, including predict, summary, vcov, anova. However, running anova(house.plr) results in anova is not implemented for a single polr object - How does one interpret the t values for each coefficient? Unlike some model fits, there are no P values here. I realize this is a lot of questions, but it makes sense to me to ask as one bundle (how do I use this thing?) rather than 7 different questions. Any information appreciated. [[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 understand output from R's polr function (ordered logistic regression)?
Also posted as http://stats.stackexchange.com/questions/7720/how-to-understand-output-from-rs-polr-function-ordered-logistic-regression . Also, I read section 7.3 of Modern Applied Statistics with S by Venables and Ripley (who wrote polr?), and I can still not answer many of these questions. On Tue, Mar 1, 2011 at 3:25 PM, Dan Frankowski dfran...@gmail.com wrote: I am new to R, ordered logistic regression, and polr. The Examples section at the bottom of the help page for polrhttp://stat.ethz.ch/R-manual/R-patched/library/MASS/html/polr.html(that fits a logistic or probit regression model to an ordered factor response) shows options(contrasts = c(contr.treatment, contr.poly)) house.plr - polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing) pr - profile(house.plr) plot(pr) pairs(pr) - What information does pr contain? The help page on profilehttp://stat.ethz.ch/R-manual/R-patched/library/stats/html/profile.htmlis generic, and gives no guidance for polr. - What is plot(pr) showing? I see six graphs. Each has an X axis that is numeric, although the label is an indicator variable (looks like an input variable that is an indicator for an ordinal value). Then the Y axis is tau which is completely unexplained. - What is pairs(pr) showing? It looks like a plot for each pair of input variables, but again I see no explanation of the X or Y axes. - How can one understand if the model gave a good fit? summary(house.plr)shows Residual Deviance 3479.149 and AIC (Akaike Information Criterion?) of 3495.149. Is that good? In the case those are only useful as relative measures (i.e. to compare to another model fit), what is a good absolute measure? Is the residual deviance approximately chi-squared distributed? Can one use % correctly predicted on the original data or some cross-validation? What is the easiest way to do that? - How does one apply and interpret anova on this model? The docs say There are methods for the standard model-fitting functions, including predict, summary, vcov, anova. However, running anova(house.plr)results in anova is not implemented for a single polr object - How does one interpret the t values for each coefficient? Unlike some model fits, there are no P values here. I realize this is a lot of questions, but it makes sense to me to ask as one bundle (how do I use this thing?) rather than 7 different questions. Any information appreciated. [[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] splitting and stacking matrices
Dear R users, I am having some difficulty arranging some matrices and wondered if anyone could help out. As an example, consider the following matrix: a - matrix(1:32, nrow = 4, ncol = 8) a [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]159 13 17 21 25 29 [2,]26 10 14 18 22 26 30 [3,]37 11 15 19 23 27 31 [4,]48 12 16 20 24 28 32 I would like it to look like the following matrix: [,1] [,2] [,3] [,4] [1,]159 13 [2,]26 10 14 [3,]37 11 15 [4,]48 12 16 [5,] 17 21 25 29 [6,] 18 22 26 30 [7,] 19 23 27 31 [8,] 20 24 28 32 I can achieve this using the following: a1 - a[, 1:4] a2 - a[, 5:8] b - rbind(a1, a2) However, my initial matrix often has a varibale number of columns (in multiples of 4, and I still want to split the columns into blocks of 4 and stack these). I have considered working out how many blocks the matrix must be split into using: no.blocks - ncol(a)/4. My problem is then implementing this information to actually split the matrix up and then stack it. Any guidance on this would be much appreciated. Regards Darcy Webber __ 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] inefficient ifelse() ?
An ifelse-like function that only evaluated
what was needed would be fine, but it would
have to be different from ifelse itself. The
trick is to come up with a good parameterization.
E.g., how would it deal with things like
ifelse(is.na(x), mean(x, na.rm=TRUE), x)
or
ifelse(x1, log(x), runif(length(x),-1,0))
or
ifelse(x1, log(x), -seq_along(x))
Would it reject such things? Deciding that the
x in mean(x,na.rm=TRUE) should be replaced by
x[is.na(x)] would be wrong. Deciding that
runif(length(x)) should be replaced by runif(sum(x1))
seems a bit much to expect. Replacing seq_along(x) with
seq_len(sum(x1)) is wrong. It would be better to
parameterize the new function so it wouldn't have to
think about those cases.
Would you want it to depend only on a logical
vector or perhaps also on a factor (a vectorized
switch/case function)?
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of ivo welch
Sent: Tuesday, March 01, 2011 12:36 PM
To: Henrique Dallazuanna
Cc: r-help
Subject: Re: [R] inefficient ifelse() ?
thanks, Henrique. did you mean
as.vector(t(mapply(function(x, f)f(x), split(t, ((t %% 2)==0)),
list(f, g ?
otherwise, you get a matrix.
its a good solution, but unfortunately I don't think this can be used
to redefine ifelse(cond,ift,iff) in a way that is transparent. the
ift and iff functions will always be evaluated before the function
call happens, even with lazy evaluation. :-(
I still think that it makes sense to have a smarter vectorized %if% in
a vectorized language like R. just my 5 cents.
/iaw
Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)
On Tue, Mar 1, 2011 at 2:33 PM, Henrique Dallazuanna
www...@gmail.com wrote:
Try this:
mapply(function(x, f)f(x), split(t, t %% 2), list(g, f))
On Tue, Mar 1, 2011 at 4:19 PM, ivo welch ivo...@gmail.com wrote:
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f()
and g() for
all values first, and presumably then just picks left or
right results
based on t%%2. uggh... wouldn't it make more sense to
evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
__
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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
__
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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] inefficient ifelse() ?
On Wed, Mar 2, 2011 at 9:36 AM, ivo welch ivo.we...@gmail.com wrote: thanks, Henrique. did you mean as.vector(t(mapply(function(x, f)f(x), split(t, ((t %% 2)==0)), list(f, g ? otherwise, you get a matrix. its a good solution, but unfortunately I don't think this can be used to redefine ifelse(cond,ift,iff) in a way that is transparent. the ift and iff functions will always be evaluated before the function call happens, even with lazy evaluation. :-( I still think that it makes sense to have a smarter vectorized %if% in a vectorized language like R. just my 5 cents. Ivo, There is no guarantee in general that f(x[3,5,7]) is the same as f(x)[3,5,7] -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland __ 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] inefficient ifelse() ?
yikes. you are asking me too much.
thanks everybody for the information. I learned something new.
my suggestion would be for the much smarter language designers (than
I) to offer us more or less blissfully ignorant users another
vector-related construct in R. It could perhaps be named %if% %else%,
analogous to if else (with naming inspired by %in%, and with
evaluation only of relevant parts [just as if else for scalars]), with
different outcomes in some cases, but with the advantage of typically
evaluating only half as many conditions as the ifelse() vector
construct. %if% %else% may work only in a subset of cases, but when
it does work, it would be nice to have. it would probably be my first
goto function, with ifelse() use only as a fallback.
of course, I now know how to fix my specific issue. I was just
surprised that my first choice, ifelse(), was not as optimized as I
had thought.
best,
/iaw
On Tue, Mar 1, 2011 at 5:13 PM, William Dunlap wdun...@tibco.com wrote:
An ifelse-like function that only evaluated
what was needed would be fine, but it would
have to be different from ifelse itself. The
trick is to come up with a good parameterization.
E.g., how would it deal with things like
ifelse(is.na(x), mean(x, na.rm=TRUE), x)
or
ifelse(x1, log(x), runif(length(x),-1,0))
or
ifelse(x1, log(x), -seq_along(x))
Would it reject such things? Deciding that the
x in mean(x,na.rm=TRUE) should be replaced by
x[is.na(x)] would be wrong. Deciding that
runif(length(x)) should be replaced by runif(sum(x1))
seems a bit much to expect. Replacing seq_along(x) with
seq_len(sum(x1)) is wrong. It would be better to
parameterize the new function so it wouldn't have to
think about those cases.
Would you want it to depend only on a logical
vector or perhaps also on a factor (a vectorized
switch/case function)?
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of ivo welch
Sent: Tuesday, March 01, 2011 12:36 PM
To: Henrique Dallazuanna
Cc: r-help
Subject: Re: [R] inefficient ifelse() ?
thanks, Henrique. did you mean
as.vector(t(mapply(function(x, f)f(x), split(t, ((t %% 2)==0)),
list(f, g ?
otherwise, you get a matrix.
its a good solution, but unfortunately I don't think this can be used
to redefine ifelse(cond,ift,iff) in a way that is transparent. the
ift and iff functions will always be evaluated before the function
call happens, even with lazy evaluation. :-(
I still think that it makes sense to have a smarter vectorized %if% in
a vectorized language like R. just my 5 cents.
/iaw
Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com)
On Tue, Mar 1, 2011 at 2:33 PM, Henrique Dallazuanna
www...@gmail.com wrote:
Try this:
mapply(function(x, f)f(x), split(t, t %% 2), list(g, f))
On Tue, Mar 1, 2011 at 4:19 PM, ivo welch ivo...@gmail.com wrote:
dear R experts---
t - 1:30
f - function(t) { cat(f for, t, \n); return(2*t) }
g - function(t) { cat(g for, t, \n); return(3*t) }
s - ifelse( t%%2==0, g(t), f(t))
shows that the ifelse function actually evaluates both f()
and g() for
all values first, and presumably then just picks left or
right results
based on t%%2. uggh... wouldn't it make more sense to
evaluate only
the relevant parts of each vector and then reassemble them?
/iaw
Ivo Welch
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Re: [R] error in saved .csv
I am not sure what you are saying your problem is? Is the format incorrect? BTW, notice that write.csv does not have a 'sep' parameter. Maybe you should be using write.table. On Tue, Mar 1, 2011 at 4:36 PM, Tamas Barjak tamas.barja...@gmail.com wrote: Help me please! I would like to be saved a data table: write.csv(random.t1, place, dec=,, append = T, quote = FALSE, sep = , qmethod = double, eol = \n, row.names=F) It's OK! But the rows of file 1,1,21042,-4084.87179487179,2457.66483516483,-582.275562799881 2,2,23846,-6383.86480186479,-3409.98451548449,-3569.72145340269 and no 1 21042 - 4084.87179487179 2457.66483516483 Not proportional... What's the problem??? 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 Data Munger Guru 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] error in saved .csv
On Tue, Mar 1, 2011 at 1:36 PM, Tamas Barjak tamas.barja...@gmail.com wrote: Help me please! I would like to be saved a data table: write.csv(random.t1, place, dec=,, append = T, quote = FALSE, sep = , qmethod = double, eol = \n, row.names=F) It's OK! But the rows of file 1,1,21042,-4084.87179487179,2457.66483516483,-582.275562799881 2,2,23846,-6383.86480186479,-3409.98451548449,-3569.72145340269 and no 1 21042 - 4084.87179487179 2457.66483516483 Not proportional... What's the problem??? If I understand you correctly, you want a text file where the separator is a white space. You cannot get that with write.csv - you have to use write.table(). The function write.csv does not allow you to change the sep argument. Here's what the help file says: ‘write.csv’ and ‘write.csv2’ provide convenience wrappers for writing CSV files. They set ‘sep’, ‘dec’ and ‘qmethod’, and ‘col.names’ to ‘NA’ if ‘row.names = TRUE’ and ‘TRUE’ otherwise. ‘write.csv’ uses ‘.’ for the decimal point and a comma for the separator. ‘write.csv2’ uses a comma for the decimal point and a semicolon for the separator, the Excel convention for CSV files in some Western European locales. These wrappers are deliberately inflexible: they are designed to ensure that the correct conventions are used to write a valid file. Attempts to change ‘append’, ‘col.names’, ‘sep’, ‘dec’ or ‘qmethod’ are ignored, with a warning. __ 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] Logistic Stepwise Criterion
The probability OF the residual deviance is zero. The significance level for the residual deviance according to its asymptotic Chi-squared distribution is a possible criterion, but a silly one. If you want to minimise that, just fit no variables at all. That's the best you can do. If you want to maximise it, just minimise the deviance itself, which means include all possible variables in the regression, together with as many interactions as you can as well. (Incidently R doesn't have restrictions on how many interaction terms it can handle, those are imposed my your computer.) I suggest you think again about what criterion you really want to use. Somehow you need to balance fit in the training sample against some complexity measure. AIC and BIC are commonly used criteria, but not the only ones. I suggest you start with these and see if either does the kind of job you want. Stepwise regression with interaction terms can be a bit tricky if you want to impose the marginality constraints, but that is a bigger issue. Bill Venables. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Mano Gabor Toth Sent: Wednesday, 2 March 2011 6:44 AM To: r-help@r-project.org Subject: [R] Logistic Stepwise Criterion Dear R-help members, I'd like to run a binomial logistic stepwise regression with ten explanatory variables and as many interaction terms as R can handle. I'll come up with the right R command sooner or later, but my real question is whether and how the criterion for the evaluation of the different models can be set to be the probability of the residual deviance in the Chi-Square distribution (which would be more informative of overall model fit than AIC). Thanks in advance for all your help. Kind regards, Mano Gabor TOTH MA Political Science Central European University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to prove the MLE estimators are normal distributed?
Dear List, I'm now working on MLE and OSL estimators.I just noticed that the textbook argues they are joint normal distributed.But how to prove the conclusion? Thanks for your time in advance! Best, Ning __ 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] error in saved .csv
Yes, the format is incorrect. I have already tried the write.table, but it didn't work. 2011/3/1 jim holtman jholt...@gmail.com I am not sure what you are saying your problem is? Is the format incorrect? BTW, notice that write.csv does not have a 'sep' parameter. Maybe you should be using write.table. On Tue, Mar 1, 2011 at 4:36 PM, Tamas Barjak tamas.barja...@gmail.com wrote: Help me please! I would like to be saved a data table: write.csv(random.t1, place, dec=,, append = T, quote = FALSE, sep = , qmethod = double, eol = \n, row.names=F) It's OK! But the rows of file 1,1,21042,-4084.87179487179,2457.66483516483,-582.275562799881 2,2,23846,-6383.86480186479,-3409.98451548449,-3569.72145340269 and no 1 21042 - 4084.87179487179 2457.66483516483 Not proportional... What's the problem??? 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 Data Munger Guru 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.
Re: [R] splitting and stacking matrices
On Tue, Mar 1, 2011 at 4:55 PM, Darcy Webber darcy.web...@gmail.com wrote: Dear R users, I am having some difficulty arranging some matrices and wondered if anyone could help out. As an example, consider the following matrix: a - matrix(1:32, nrow = 4, ncol = 8) a [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1 5 9 13 17 21 25 29 [2,] 2 6 10 14 18 22 26 30 [3,] 3 7 11 15 19 23 27 31 [4,] 4 8 12 16 20 24 28 32 I would like it to look like the following matrix: [,1] [,2] [,3] [,4] [1,] 1 5 9 13 [2,] 2 6 10 14 [3,] 3 7 11 15 [4,] 4 8 12 16 [5,] 17 21 25 29 [6,] 18 22 26 30 [7,] 19 23 27 31 [8,] 20 24 28 32 I can achieve this using the following: a1 - a[, 1:4] a2 - a[, 5:8] b - rbind(a1, a2) However, my initial matrix often has a varibale number of columns (in multiples of 4, and I still want to split the columns into blocks of 4 and stack these). I have considered working out how many blocks the matrix must be split into using: no.blocks - ncol(a)/4. My problem is then implementing this information to actually split the matrix up and then stack it. Any guidance on this would be much appreciated. Regards Darcy Webber Try converting to a 3d array, swapping the last two dimensions and reconstituting it as a matrix: matrix(aperm(array(a, c(4, 4, 2)), c(1, 3, 2)), nc = 4) [,1] [,2] [,3] [,4] [1,]159 13 [2,]26 10 14 [3,]37 11 15 [4,]48 12 16 [5,] 17 21 25 29 [6,] 18 22 26 30 [7,] 19 23 27 31 [8,] 20 24 28 32 -- 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.
