[R] Legend and Main Title positioning
Hi folks, can anybody give me a hint how to solve the following problem? I have several plots in one window like this: layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE)) plot(rnorm(100)) plot(rnorm(200)) plot(rnorm(300)) plot(rnorm(400)) Now, I'd like to create a legend below each plot and generate a common title. How can I do that? Antje __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] complex(?) reshaping question
Hello, I have a problem with data reshaping. Here's my data DF - structure(list(idvar1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(patient1, patient2 ), class = factor), idvar2 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(dob, hog), class = factor), eka1 = structure(c(NA, NA, NA, NA, NA, NA, NA, NA, 2L, 1L, NA, NA, NA, NA), .Label = c(as, df), class = factor), eka2 = structure(c(NA, NA, NA, NA, NA, NA, NA, NA, 2L, 1L, NA, NA, NA, NA), .Label = c(as, df), class = factor), eka3 = structure(c(NA, NA, NA, NA, NA, NA, NA, NA, 2L, 1L, NA, NA, NA, NA), .Label = c(as, df), class = factor), toka1 = structure(c(NA, NA, NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, NA, 2L), .Label = c(as, fg), class = factor), toka2 = structure(c(NA, NA, NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, NA, 2L), .Label = c(as, fg), class = factor), toka3 = structure(c(NA, NA, NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, NA, 2L), .Label = c(as, fg), class = factor), kol1 = structure(c(NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, 1L, 2L, NA, NA), .Label = c(hj, ht), class = factor), kol2 = structure(c(NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, 1L, 2L, NA, NA), .Label = c(hj, ht), class = factor), kol3 = structure(c(NA, NA, NA, NA, 1L, 2L, NA, NA, NA, NA, 1L, 2L, NA, NA), .Label = c(hj, th), class = factor)), .Names = c(idvar1, idvar2, eka1, eka2, eka3, toka1, toka2, toka3, kol1, kol2, kol3), class = data.frame, row.names = c(NA, -14L)) I'm trying to omit NA:s in this DF and produce a reduced DF. The problem is that I cannot completely omit NA rows. I tried library(reshape) g - melt(DF, id=c(idvar1, idvar2)) g - na.omit(g) reshape(g, idvar=c(idvar1, idvar2), timevar=variable, direction=wide) But this drops the second row. This is the resulting DF I should be getting res - structure(list(idvar1 = structure(c(1L, 1L, 2L, 2L), .Label = c(patient1, patient2), class = factor), idvar2 = structure(c(1L, 1L, 2L, 2L), .Label = c(dob, hog), class = factor), eka1 = structure(c(NA, NA, 2L, 1L), .Label = c(as, df), class = factor), eka2 = structure(c(NA, NA, 2L, 1L), .Label = c(as, df), class = factor), eka3 = structure(c(NA, NA, 2L, 1L), .Label = c(as, df), class = factor), toka1 = structure(c(1L, 2L, NA, 2L), .Label = c(as, fg), class = factor), toka2 = structure(c(1L, 2L, NA, 2L), .Label = c(as, fg), class = factor), toka3 = structure(c(1L, 2L, NA, 2L), .Label = c(as, fg), class = factor), kol1 = structure(c(1L, 2L, 1L, 2L), .Label = c(hj, ht), class = factor), kol2 = structure(c(1L, 2L, 1L, 2L), .Label = c(hj, ht), class = factor), kol3 = structure(c(1L, 2L, 1L, 2L), .Label = c(hj, th), class = factor)), .Names = c(idvar1, idvar2, eka1, eka2, eka3, toka1, toka2, toka3, kol1, kol2, kol3), class = data.frame, row.names = c(NA, -4L)) Any ideas how to proceed? Many thanks Lauri __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] setwd and list.files / linux
Hey Marc, sorry I forgot to mention it. I am the owner of the path, which is in my home directory. I can change the read/write/execute privileges. All users have - at least - read privileges for all files and the directory: hpc36 40: ls -la *.tped -rw-r--r-- 1 cremer C020 205487231 Dec 3 17:35 chr10.tped -rw-r--r-- 1 cremer C020 192067641 Dec 3 17:37 chr11.tped -rw-r--r-- 1 cremer C020 192243483 Dec 3 17:38 chr12.tped etc A workaround I found now is to start R in the respective directory from the command line, which will set working directory to current directory. list.files will work then too: list.file(.,pattern=\\.tped) The problem now is to start a script via R CMD BATCH from a different directory, because setwd() will not work: datapath-'/nfs/home/m/cremer/gaw/data/traws_chrs/' getwd() [1] /nfs/home/m/cremer/gaw/code/plink setwd(datapath) Error in setwd(datapath) : cannot change working directory Execution halted Lars Original-Nachricht Datum: Thu, 04 Dec 2008 10:58:16 -0600 Von: Marc Schwartz [EMAIL PROTECTED] An: Lars Beckmann [EMAIL PROTECTED] CC: r-help@r-project.org Betreff: Re: [R] setwd and list.files / linux on 12/04/2008 10:38 AM Lars Beckmann wrote: Hey, two questions regarding setwd and list.files: I try to use the function list.files: datapath-'/nfs/home/m/cremer/gaw/data/traws_chrs' fl-list.files(datapath, pattern=\\.tped) but I get the following error message: Warning message: In list.files(datapath, pattern = \\.tped) : list.files: '/nfs/home/m/cremer/gaw/data/traws_chrs/' is not a readable directory fl character(0) The use of setwd also does not work: setwd(datapath) Error in setwd(datapath) : cannot change working directory datapath exists and it doesn't matter whether I use single quotes ' or double quotes Any help? Lars You do not have read privileges to the path in question. Either you or your SysAdmin will have to grant the appropriate privileges so that you can read the files in the path. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sensationsangebot verlängert: GMX FreeDSL - Telefonanschluss + DSL für nur 16,37 Euro/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K1308T4569a __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 needed
I need some help in comparing t values avalaible from two data frames. I have two data frames with each containing a column for t valuse For example one data frame looks like as follows IDlogFCt P.Value adj.P.Val B 39 a39 -1.737037118 -2.861784118 0.004212637 0.3120295 -2.504407 1 a1 -1.66000 -2.734864780 0.006240591 0.3120295 -2.729203 53 a53 1.443725119 2.378549988 0.017380880 0.4517324 -3.305806 52 a52 -1.323602459 -2.180646836 0.029209548 0.4517324 -3.591351 14 a14 -1.288769209 -2.123258747 0.033732176 0.4517324 -3.669516 58 a58 1.286212422 2.119046418 0.034086544 0.4517324 -3.675172 75 a75 -1.268523811 -2.089904273 0.036626400 0.4517324 -3.713989 I have another one like it but the order of the ID column is different as shown below IDlogFCt P.Value adj.P.Val B 76 a1 0.157096693 -0.258818201 0.795775514 0.9146845 -5.075035 63 a53 -0.146205781 -0.240875326 0.809651746 0.9200588 -5.077872 88 a58 0.108707988 0.179097379 0.857861240 0.9638890 -5.086083 23 a52 -0.086871304 -0.143121247 0.886194425 0.9757266 -5.089752 98 a53 0.085552023 0.140947720 0.887911243 0.9757266 -5.089947 90 a14 0.068456451 0.112782613 0.910202903 0.9868965 -5.092209 91 a75 -0.052087970 -0.085815394 0.931613169 0.9868965 -5.093904 I have to plot the t values obtained for each id against each other. Since the order of the IDs in the two data frames is different , I am finding it difficult to plot the t values. Can anyone help in arranging the t values in the increasing order of the ID values like 1 a1 .. 2 a2 .. 2 a3 .. Thank you -- View this message in context: http://www.nabble.com/help-needed-tp20849087p20849087.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] help needed
gauravbhatti wrote: ... I have to plot the t values obtained for each id against each other. Since the order of the IDs in the two data frames is different , I am finding it difficult to plot the t values. Can anyone help in arranging the t values in the increasing order of the ID values like 1 a1 .. 2 a2 .. 2 a3 .. Hi gauravbhatti, Try this (assuming your data frame is named gvb): gvb[order(as.character(gvb[,1])),] Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Seasonality in time series
Hi, I?m looking for a package which includes a test for seasonality in time series. Any help and input would be greatly appreciated. Thanks, Matthias __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Changing 'record' option in open graphics device
On Thu, 4 Dec 2008, Andrew Hooker wrote: Hi, I am wondering if there is a way to change the value of the record option in a graphics device that is already open (and accepts this option). I don't want to open a new device with, for example dev.new(record=T), but just want to change the settings of the current device. This can be done by pointing and clicking on the history tab of a graphics device in windows, but I would like to do this in a script. 1) AFAIK only one device, windows(), accepts this. 2) That device has no way to do this from R code. Best regards, Andy Andrew Hooker, Ph.D. Associate Professor of Pharmacometrics Div. of Pharmacokinetics and Drug Therapy Dept. of Pharmaceutical Biosciences Uppsala University Box 591, 751 24, Uppsala, Sweden Phone: +46 18 471 4355 Mobile: +46 701 679 048 http://www.farmbio.uu.se/research.php?avd=5 www.farmbio.uu.se/research.php?avd=5 [[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. PLEASE do: double-spaced text is unfriendly and is undoubtedly caused by sending HTML when asked not to. -- Brian D. Ripley, [EMAIL PROTECTED] 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] Lexical Permutation Algorithm in R
Hi all
Here is a rather naive implementation of the SEPA algorithm for generating
lexical permutations:
lexperm3 - function(x, n=length(x)) {
perms - list()
k - 1
perms[[k]] - x
k - k + 1
for (y in 1:(factorial(n)-1)) {
i - n-1
while (x[i] x[i+1] i 0) {
i - i - 1
}
# i is largest index st x[i] x[i+1]
j - n
# find min{ x[j], st n=j=i+1 and x[j]x[i] }
while (x[j] = x[i] j i) {
j - j - 1
}
# swap x[i] and x[j]
tmp - x[i];x[i] - x[j]; x[j] - tmp
# now sort everything from x[i+1]..x[n]
# by reversing the elements within
p - i + 1
q - n
while (p q) {
tmp - x[p]; x[p] - x[q]; x[q] - tmp
p - p + 1
q - q - 1
}
perms[[k]] - x
k - k + 1
}
perms
}
This, as you can imagine, is severely slow. I would like to speed up this
function if possible, which I guess would involve vectorizing the inner
loop..does anyone have any ideas about how to improve this code's runtime?
One small potential optimization I tried was to shorten the sort by reverse
ordering near the end of the inner loop : I tried replacing it with rev() and
sort(decreasing=TRUE) over a partial subset of the x vector: however rev() was
much slower, and I dont think sort() supports lexicographic ordering, so that
didnt work.
Thanks
rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
***
The Royal Bank of Scotland plc. Registered in Scotland No 90312. Registered
Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:25}}
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Re: [R] Seasonality in time series
Here's something that may help you get started: library(ts) ?decompose ?stl Thanks, -Girish On Dec 5, 1:55 pm, Matthias [EMAIL PROTECTED] wrote: Hi, I?m looking for a package which includes a test for seasonality in time series. Any help and input would be greatly appreciated. Thanks, Matthias __ [EMAIL PROTECTED] mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://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] Strplit code
Dear Wacek,
I've thought a bit more about this problem, and recall that I originally
wrote Strsplit() [and replacements for sub() and gsub(), which were not then
in S-PLUS] for the version of the car package that I released for S-PLUS,
because other functions in the package used these. The strings involved were
small, so performance issues weren't that important, although of course it's
better to have a more efficient solution.
Although I no longer have an installed copy of S-PLUS to confirm this, I
believe that gregexepr() is still not present in S-PLUS (though I think that
strsplit() is in the latest version). If that's the case, then your function
wouldn't work at all in the context of the original posting, which asked for
a solution in S-PLUS. You could make your code work in S-PLUS, and probably
still have it more efficient than mine, by writing a replacement for
gregexpr().
-Original Message-
From: Wacek Kusnierczyk [mailto:[EMAIL PROTECTED]
Sent: December-04-08 7:29 AM
To: John Fox
Cc: R help
Subject: Re: [R] Strplit code
John Fox wrote:
Dear Wacek,
Wrong is a bit strong, I think -- limited to single-pattern characters
is
more accurate.
nothing is ever wrong if seen from an appropriate perspective. for
example, there is nothing wrong in that many core functions in r deparse
some, but not all, of the argument expressions, without any obvious
pattern -- when you get used to it and learn each single case by heart,
it's perfectly correct.
Moreover, it isn't hard to make the function work with
multiple-character matches as well:
which you probably should have done before posting the flawed version.
Indeed. Had I anticipated the possibility of multiple-character splits I
would have done so.
John
Strsplit - function(x, split){
if (length(x) 1) {
return(lapply(x, Strsplit, split)) # vectorization
}
result - character(0)
if (nchar(x) == 0) return(result)
posn - regexpr(split, x)
if (posn = 0) return(x)
c(result, substring(x, 1, posn - 1),
Recall(substring(x, posn + attr(posn, match.length),
nchar(x)), split)) # recursion
}
On the other hand, your function is much more efficient.
just one order of magnitude in my tests. might not be completely fool
proof, though.
vQ
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Re: [R] Lexical Permutation Algorithm in R
Rory
there are several packages that perform this.
I would use permn() of the combinat library, then, if
lexicographical order is important, sort it explicitly.
HTH
rksh
[EMAIL PROTECTED] wrote:
Hi all
Here is a rather naive implementation of the SEPA algorithm for generating
lexical permutations:
lexperm3 - function(x, n=length(x)) {
perms - list()
k - 1
perms[[k]] - x
k - k + 1
for (y in 1:(factorial(n)-1)) {
i - n-1
while (x[i] x[i+1] i 0) {
i - i - 1
}
# i is largest index st x[i] x[i+1]
j - n
# find min{ x[j], st n=j=i+1 and x[j]x[i] }
while (x[j] = x[i] j i) {
j - j - 1
}
# swap x[i] and x[j]
tmp - x[i];x[i] - x[j]; x[j] - tmp
# now sort everything from x[i+1]..x[n]
# by reversing the elements within
p - i + 1
q - n
while (p q) {
tmp - x[p]; x[p] - x[q]; x[q] - tmp
p - p + 1
q - q - 1
}
perms[[k]] - x
k - k + 1
}
perms
}
This, as you can imagine, is severely slow. I would like to speed up this
function if possible, which I guess would involve vectorizing the inner
loop..does anyone have any ideas about how to improve this code's runtime?
One small potential optimization I tried was to shorten the sort by reverse
ordering near the end of the inner loop : I tried replacing it with rev() and
sort(decreasing=TRUE) over a partial subset of the x vector: however rev() was much
slower, and I dont think sort() supports lexicographic ordering, so that didnt work.
Thanks
rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
***
The Royal Bank of Scotland plc. Registered in Scotland No 90312. Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:25}}
__
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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.
--
Robin K. S. Hankin
Uncertainty Analyst
University of Cambridge
19 Silver Street
Cambridge CB3 9EP
01223-764877
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Re: [R] Strplit code
John Fox wrote: Dear Wacek, I've thought a bit more about this problem, and recall that I originally wrote Strsplit() [and replacements for sub() and gsub(), which were not then in S-PLUS] for the version of the car package that I released for S-PLUS, because other functions in the package used these. The strings involved were small, so performance issues weren't that important, although of course it's better to have a more efficient solution. right. the speedup is not due to any substantial algorithmic difference, but rather in that your code is r code, while mine uses gregexpr, which is i assume is precompiled from c code or the like. about 'wrong' and 'flawed', again: what i meant is that as a suggestion for how strsplit, in general, could be written, it doesn't meet the challenge, and should at least issue a warning if the split pattern specifies non-1-length splits. otherwise, it could be perfectly fit for a student's exercise. Although I no longer have an installed copy of S-PLUS to confirm this, I believe that gregexepr() is still not present in S-PLUS (though I think that strsplit() is in the latest version). If that's the case, then your function wouldn't work at all in the context of the original posting, which asked for a solution in S-PLUS. You could make your code work in S-PLUS, and probably still have it more efficient than mine, by writing a replacement for gregexpr(). possibly, and then it's my code that is wrong and flawed ;) i haven't used s-plus for ages, jumping to the global frame instead of lexical scoping scared me away. vQ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] constructing a relative path
Dear R-users, I want to construct a relative path in R, but I am not able to do it or to find a function that does it. My solution is somewhat awkward as it uses string manipulation. The problem: I have the following folder structure. project_folder | |- data |- scripts |- output My working directory is the scripts folder. Now I want to construct a relative path to the data folder. Is there an easy way or a function? TIA Mark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] constructing a relative path
On Fri, 5 Dec 2008, Mark Heckmann wrote: Dear R-users, I want to construct a relative path in R, but I am not able to do it or to find a function that does it. My solution is somewhat awkward as it uses string manipulation. The problem: I have the following folder structure. project_folder | |- data |- scripts |- output My working directory is the scripts folder. Now I want to construct a relative path to the data folder. Is there an easy way or a function? file.path(.., data) should work on any R platform (actually ../data will on any current R platform). -- Brian D. Ripley, [EMAIL PROTECTED] 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] Trouble with gridBase and inset plots
Dear All,
I ma having a trouble in generating a figure containing 3 insets with
the gridBase package.
I always get an error message of the kind:
Error in gridPLT() : Figure region too small and/or viewport too large
No matter which parameters I choose. The plots works nicely with two
insets only, but when I try adding the third one, my troubles begin.
I am probably doing something wrong in the generation of the 3rd inset
and I paste below everything I do in this (a bit complicated) figure.
Any suggestion is welcome.
Cheers
Lorenzo
pdf(./post-processing-plots/exploratory_research_figure_2.pdf)
par( mar = c(4.5,5, 2, 1) + 0.1)
plot(time[1:time_end],tot_num_150[1:time_end]/1e6,type=b,lwd=2,col=blue,lty=2,
xlab=expression(paste(tau,[s])),
ylab=expression(paste(N[,
cm^{-3},])),cex.lab=1.6,ylim=range(c(7.4e7,1.43e8)),yaxt=n,cex.axis=1.4)
#lines(time[1],ini_pop/1e6, p,col=red,lwd=2,lty=1,pch=5 )
lines(time[time_end],8.25e7, p,col=red,lwd=2,lty=1,pch=5)
axis(side=2, at=c( 7.4e7, 9.6e7, 1.18e8, 1.4e8),
labels=expression(7.4%*%10^7, 9.6%*%10^7,
1.18%*%10^8,1.4%*%10^8),cex.lab=1.6,cex.axis=1.4)
#axis(side=1,cex.axis=1.4)
## lines(time[1:time_end],
N_approx[1:time_end],col=red,type=b,lwd=2,lty=1,pch=4)
## lines(time[1:time_end],
N_approx2[1:time_end],col=black,type=b,lwd=2,lty=1,pch=2)
## lines(time[1:time_end],
N_approx_beta1[1:time_end,2],col=brown,type=b,lwd=2,lty=1,pch=5)
legend(topright,cex=1.2, c(expression(Simulation),
expression(Outlet measurement)),
lwd=c(2,2),lty=c(2,0),pch = c(1,5),col=c(blue, red),box.lwd=0,box.lty=0,
,xjust = 1, yjust = 1)
# abline(v=time[12],lwd=2,pch=2,lty=2)
lines(c(time[14],time[14]), c(0,1.2e8),lwd=2,lty=2,pch=2)
# legend(-0.2,9.2e7,cex=1.2,c(expression(numerical result for a
5m-long pipe)),bty=n)
arrows(0.8, 9e7, time[14], tot_num_150[14]/1e6, length = 0.15,lwd=2)
text(0.8,8.8e7,cex=1.2,Final concentration for a)
text(0.8,8.5e7,cex=1.2,6.5m long transfer tube (LAT))
text(0.8,8e7,cex=1.2,Final concentration for a)
text(0.68,7.7e7,cex=1.2,9m long transfer tube)
text(0.8,7.4e7,cex=1.2,(VELA))
text(1.3,1.24e8,cex=1.2,Residence time for)
text(1.3,1.21e8,cex=1.2,a 6.5m long transfer tube)
arrows(1.1, 7.7e7, time[21], 7.7e7, length = 0.,lwd=2)
arrows( time[21], 7.7e7, time[21], tot_num_150[21]/1e6, length = 0.15,lwd=2)
par( mar = c(0.,0., 0., 0.) )
#1st inset
vp - baseViewports()
pushViewport(vp$inner,vp$figure,vp$plot)
pushViewport(viewport(x=-0.0,y=1.04,width=.4,height=.4,just=c(0,1)))
par(fig=gridPLT(),new=F)
#grid.rect(gp=gpar(lwd=0,col=red))
plot(D_mean,data_150[1, ]/log_factor*log(10)/1e6,l,
pch=1,col=black, lwd=2,xlab=,ylab=
,cex.axis=1.,cex.lab=1.,log=x,xaxt=n,yaxt=n,
ylim=range(c(0, 2.4e8)))
## axis(side=2, at=c( 0, 0.6e8, 1.2e8, 1.8e8, 2.4e8),
## labels=expression(0, 6%*%10^7, 1.2%*%10^8,
1.8%*%10^8,2.4%*%10^8),cex.lab=1.4,cex.axis=1.2)
#2nd inset
#vp - baseViewports()
pushViewport(vp$inner,vp$figure,vp$plot)
pushViewport(viewport(x=0.5,y=0.65,width=.4,height=.4,just=c(0,1)))
par(fig=gridPLT(),new=F)
#grid.rect(gp=gpar(lwd=0,col=red))
plot(D_mean,data_150[21, ]/log_factor*log(10)/1e6,l,lwd=2,
pch=1,col=black,xlab=,ylab=, log=x
,cex.axis=1.4,cex.lab=1.6,xaxt=n,yaxt=n,ylim=range(c(0, 2.4e8)))
#3rd inset
#vp - baseViewports()
pushViewport(vp$inner,vp$figure,vp$plot)
pushViewport(viewport(x=0.25,y=0.7,width=.4,height=.4,just=c(0,1)))
par(fig=gridPLT(),new=T)
#grid.rect(gp=gpar(lwd=0,col=red))
plot(D_mean,data_150[14, ]/log_factor*log(10)/1e6,l,lwd=2,
pch=1,col=black,xlab=,ylab=, log=x
,cex.axis=1.4,cex.lab=1.6,xaxt=n,yaxt=n,ylim=range(c(0, 2.4e8)))
popViewport(3)
dev.off()
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[R] Logical inconsistency
Dear colleagues Please could someone kindly explain the following inconsistencies I've discovered when performing logical calculations in R: 8.8 - 7.8 1 TRUE 8.3 - 7.3 1 TRUE Thank you, Emma Jane [[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] Logical inconsistency
On 12/5/2008 7:23 AM, emma jane wrote: Dear colleagues Please could someone kindly explain the following inconsistencies I've discovered when performing logical calculations in R: 8.8 - 7.8 1 TRUE 8.3 - 7.3 1 TRUE See R FAQ 7.31 (http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f). Thank you, Emma Jane [[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. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Logical inconsistency
Dear Emma, On Fri, 5 Dec 2008 04:23:53 -0800 (PST) emma jane [EMAIL PROTECTED] wrote: Please could someone kindly explain the following inconsistencies I've discovered__when performing logical calculations in R: 8.8 - 7.8 1 TRUE 8.3 - 7.3 1 TRUE Gladly: FAQ 7.31 http://cran.at.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f HTH. Cheers, Berwin === Full address = Berwin A TurlachTel.: +65 6516 4416 (secr) Dept of Statistics and Applied Probability+65 6516 6650 (self) Faculty of Science FAX : +65 6872 3919 National University of Singapore 6 Science Drive 2, Blk S16, Level 7 e-mail: [EMAIL PROTECTED] Singapore 117546http://www.stat.nus.edu.sg/~statba __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 retrieve a method
Hi there, I am interested in the inner workings of wilcox.test: wilcox.test function (x, ...) UseMethod(wilcox.test) environment: namespace:stats how can I get at the code, if it is R-code? For Methods one should be able to learn what extension to use, but here default or such do not help. Is there a wayplot.default to learn which different versions of plot there are except trying plot.lm, ... best -c -- Dr. Christian W. Hoffmann, Swiss Federal Research Institute WSL Zuercherstrasse 111, CH-8903 Birmensdorf, Switzerland Tel +41-44-7392-277 (office), -111(exchange), -215 (fax) [EMAIL PROTECTED], www.wsl.ch/personal_homepages/hoffmann/ Jene, die grundlegende Freiheit aufgeben wuerden, um eine geringe voruebergehende Sicherheit zu erwerben, verdienen weder Freiheit noch Sicherheit. - Benjamin Franklin -- Christian W. Hoffmann, Rigiblickstrasse 15 b, CH-8915Hausen am Albis, Switzerland, Tel +41-44-7640853, [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] levels update
Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) # still gives me a,b,c, though the subset does not contain entries with c anymore I guess, the solution is rather simple, but I cannot find it. Antje __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Quantile Regression for longitudinal data
Hi all, does anybody know about R implementations for quantile regression for longitudinal data? I am just aware of a very basic version of R. Koenker's approach using fixed effects. Thanks in advance Armin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Comparing survival curves with survdiff strata help
ExpeRts,
I'm trying to compare three survival curves using the function survdiff in
the survival package. Following is my code and corresponding error message.
survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata)
Error in survdiff(Surv(st_months, status) ~ strata(BOR), data = mydata) :
No groups to test
When I check the strata of the variable. I get . . .
strata(BOR)
Levels: BOR=PD BOR=PR BOR=SD
So I know there are levels associated with the BOR variable, but I can't seem
to get the strata statement to work within survdiff. I checked ?strata and
(unless I am not understanding) the help, it seems to be working as I get the
following output
levels(strata(BOR))
[1] BOR=PD BOR=PR BOR=SD
I can get the survival tables and KM plot with no problem, but I can't seem to
be able to compare the survival curves. I can get an overall comparison if I
remove the strata statement
survdiff(Surv(st_months, status) ~ BOR, data=mydata)
Call:
survdiff(formula = Surv(st_months, status) ~ BOR, data = mydata)
n=35, 5 observations deleted due to missingness.
N Observed Expected (O-E)^2/E (O-E)^2/V
BOR=PD 13 13 6.51 6.464 8.87
BOR=PR 10 1018.48 3.888 10.59
BOR=SD 12 1210.01 0.395 0.61
Chisq= 13.5 on 2 degrees of freedom, p= 0.00117
But when I use survdiff(Surv(st_months, status) ~ strata(BOR), data=mydata) it
doesn't want to work. Any suggestions?
Many thanks for your assistance!
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute - Webb Lab
333 Bostwick Avenue NE
Grand Rapids, MI 49503
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Re: [R] levels update
try this: df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) [1] a b c my.sub$X1 - factor(my.sub$X1) levels(my.sub$X1) [1] a b On Fri, Dec 5, 2008 at 7:50 AM, Antje [EMAIL PROTECTED] wrote: Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) # still gives me a,b,c, though the subset does not contain entries with c anymore I guess, the solution is rather simple, but I cannot find it. Antje __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] complex(?) reshaping question
I'm trying to omit NA:s in this DF and produce a reduced DF. The problem is that I cannot completely omit NA rows. I tried library(reshape) g - melt(DF, id=c(idvar1, idvar2)) g - na.omit(g) You're missing an id variable: DF$idvar3 - 1:2 g - melt(DF, id=c(idvar1, idvar2, idvar3), na.rm = T) cast(g, idvar1 + idvar2 + idvar3 ~ variable) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] levels update
I do the following for a subsetted dataframe:
cleanfactors - function(mydf){
outdf-mydf
for (i in 1:dim(mydf)[2]){
if (is.factor(mydf[,i]))
outdf[,i]-factor(mydf[,i])
}
outdf
}
Antje wrote:
Hello,
I hope this question is not too stupid. I would like to know how to
update levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
# still gives me a,b,c, though the subset does not contain entries
with c anymore
I guess, the solution is rather simple, but I cannot find it.
Antje
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Re: [R] How to retrieve a method
On Fri, 5 Dec 2008, Christian Hoffmann wrote: Hi there, I am interested in the inner workings of wilcox.test: wilcox.test function (x, ...) UseMethod(wilcox.test) environment: namespace:stats how can I get at the code, if it is R-code? For Methods one should be able to learn what extension to use, but here default or such do not help. Is there a wayplot.default to learn which different versions of plot there are except trying plot.lm, ... methods(wilcox.test) getS3method(wilcox.test, default) Uwe Ligges wrote an article in R-News about this in issue 6(4), although this particular usage (the most general for S3 methods) seems not to be there. [Note for the cognescenti: the S3 method for function foo() of class bar need not be foo.bar() if no such function is visible and registration was used. I'm not aware of any public package that does that though.] best -c -- Dr. Christian W. Hoffmann, Swiss Federal Research Institute WSL Zuercherstrasse 111, CH-8903 Birmensdorf, Switzerland Tel +41-44-7392-277 (office), -111(exchange), -215 (fax) [EMAIL PROTECTED], www.wsl.ch/personal_homepages/hoffmann/ Jene, die grundlegende Freiheit aufgeben wuerden, um eine geringe voruebergehende Sicherheit zu erwerben, verdienen weder Freiheit noch Sicherheit. - Benjamin Franklin -- Christian W. Hoffmann, Rigiblickstrasse 15 b, CH-8915Hausen am Albis, Switzerland, Tel +41-44-7640853, [EMAIL PROTECTED] -- Brian D. Ripley, [EMAIL PROTECTED] 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.
Re: [R] levels update
On Fri, Dec 5, 2008 at 6:50 AM, Antje [EMAIL PROTECTED] wrote: Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) # still gives me a,b,c, though the subset does not contain entries with c anymore I guess, the solution is rather simple, but I cannot find it. You might find it easier just to work with character vectors: options(stringsAsFactors = FALSE) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] levels update
On Fri, 5 Dec 2008, jim holtman wrote: try this: df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) [1] a b c my.sub$X1 - factor(my.sub$X1) I find my.sub$X1 - my.sub$X1[drop=TRUE] a lot more self-explanatory. See ?[.factor. However, if you find yourself wanting to do this, ask why you have a factor (rather than a character vector) in the first place. levels(my.sub$X1) [1] a b On Fri, Dec 5, 2008 at 7:50 AM, Antje [EMAIL PROTECTED] wrote: Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) # still gives me a,b,c, though the subset does not contain entries with c anymore I guess, the solution is rather simple, but I cannot find it. Antje __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] 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.
Re: [R] Logical inconsistency
Berwin A Turlach wrote: Dear Emma, On Fri, 5 Dec 2008 04:23:53 -0800 (PST) emma jane [EMAIL PROTECTED] wrote: Please could someone kindly explain the following inconsistencies I've discovered__when performing logical calculations in R: 8.8 - 7.8 1 TRUE 8.3 - 7.3 1 TRUE Gladly: FAQ 7.31 http://cran.at.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f well, this answer the question only partially. this explains why a system with finite precision arithmetic, such as r, will fail to be logically correct in certain cases. it does not explain why r, a language said to isolate a user from the underlying implementational choices, would have to fail this way. there is, in principle, no problem in having a high-level language perform the computation in a logically consistent way. for example, bc is an arbitrary precision calculator language, and has no problem with examples as the above: bc 8.8 - 7.8 1 # 0, meaning 'no' bc 8.3 - 7.3 1 # 0, meaning 'no' bc 8.8 - 7.8 == 1 # 1, meaning 'yes' the fact that r (and many others, including matlab and sage, perhaps not mathematica) does not perform logically here is a consequence of its implementation of floating point arithmetic. the faq you were pointed to, and its referring to the goldberg's article, show that r does not successfully isolate a user from details of the lower-level implementation. vQ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Cartesian Product Of Character Vectors
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to get
*apply to paste the column values together.
Thanks
Rory
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[R] Loop swith String replacement
Hi All,
I'm trying to split my dataset, into multiple datasets that i'll analyse
later, i wanted to do this dynamically as i might need to rerun the code
later.
I was looking at doing this via a loop, (Are other methods more appropriate?
Would a function be better?)
However i'm not sure in R how do do string replacement within the loop in
order to create unique dataset names based on the number of 'Groups' i have
Many thanks
Mike
my code is as follows:
no.groups -names(table(Conv$Group))
for (i in length(Conv$no.groups))
{
groupi - subset(Conv, Conv$Group == i)
}
Metal Secs cost Income stable Group
1 Chrome 6014 3.3458 1 2
2 Chrome 5110 1.8561 0 1
3 Chrome 2412 0.6304 0 1
4 Chrome 38 8 3.4183 1 2
5 Chrome 2512 2.7852 1 3
6 Chrome 6712 2.3866 1 1
7 Chrome 4012 4.2857 0 1
8 Chrome 5610 9.3205 1 1
9 Chrome 3212 3.8797 1 3
10 Chrome 7516 2.7031 1 3
11 Chrome 4615 11.2307 1 2
12 Chrome 5212 8.6696 1 2
13 Chrome 2212 1.7443 0 2
14 Chrome 6012 0.2253 0 2
15 Chrome 2414 4.3348 1 3
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[R] Changing order of x factor in interaction plot
Hi all, I would like to ask one simple question. I wanna know how you can ask R to change the order of x factor when you draw interaction plot. I find that R puts factors in an alphabetic order (e.g., dog, lamb, and monkey etc). But I simply want to control the order. How can I do this? Aloha, Kota __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Changing order of x factor in interaction plot
Check out the 'levels' option on 'factor' On Fri, Dec 5, 2008 at 8:32 AM, Kota Hattori [EMAIL PROTECTED] wrote: Hi all, I would like to ask one simple question. I wanna know how you can ask R to change the order of x factor when you draw interaction plot. I find that R puts factors in an alphabetic order (e.g., dog, lamb, and monkey etc). But I simply want to control the order. How can I do this? Aloha, Kota __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] legend at fixed distance form the bottom
Hi the list I would like to add a legend under a graph but at a fixed distance from the graphe. Is it possible ? More precisely, here is my code : --- 8 symboles - c(3,4,5,6) dn - rbind(matrix(rnorm(20),,5),matrix(rnorm(20,2),,5)) listSymboles - rep(symboles,each=2) matplot(t(dn),pch=listSymboles,type=b) legend(bottom, pch = unique(listSymboles), legend = c(ane, cheval, poney, mule), inset = c(0,-0.175), horiz = TRUE, xpd = NA) --- 8 But when I change the size of the graph, the legend is misplaced. Instead, I try to put some text in xlab, but I do not know how to get the +, x , V and other symbol. Does anyone got a solution ? Thanks a lot. Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get Greenhouse-Geisser epsilons from anova?
Dear Mr. Daalgard. thank you very much for your reply, it helped me to progress a bit. The following works fine: dd - expand.grid(C = 1:7, B= c(r, l), A= c(c, f)) myma - as.matrix(myma) #myma is a 12 by 28 list mlmfit - lm(myma~1) mlmfit0 - update(mlmfit, ~0) anova(mlmfit, mlmfit0, X= ~C+B, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of A. anova(mlmfit, mlmfit0, X= ~A+C, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of B. However, I can not figure out how this works for the other effects. If I try: anova(mlmfit, mlmfit0, X= ~A+B, M = ~A+C+B, idata = dd, test=Spherical) I get: Fehler in function (object, ..., test = c(Pillai, Wilks, Hotelling-Lawley, : residuals have rank 1 4 I also don't know how I can calculate the various interactions.. My read is I should change the second argument mlmfit0, too, but I can't figure out how... Do you know what to do? Thank you very much! Peter Dalgaard schrieb: Skotara wrote: Dear all, I apologize for my basic question. I try to calculate an anova for repeated measurements with 3 factors (A,B,C) having 2, 2, and 7 levels. or with an additional fourth between subjects factor D. Everything works fine using aov(val ~ A*B*C + Error(subject/ (A*B*C) ) ) or aov(val ~ (D*A*B*C) + Error(subject/(A*B*C)) + D ) val, A, B, C, D and subject are columns in a data.frame. How can I get the estimated Greenhouse-Geisser and Huynh-Feldt epsilons? I know Peter Dalgaard described it in R-News Vol. 7/2, October 2007. However, unfortunately I am not able to apply that using my data... Why? It is supposed to work. You just need to work out the X and M specification for the relevant error strata and set test=Spherical for anova.mlm, or work out the T contrast matrix explicitly if that suits your temper better. Furthermore, I am still confused of how SPSS calculates the epsilons since it is mentioned that perhaps there are any errors in SPSS?? I would be glad if anyone could help me! I am looking forward to hearing from you! Thank you! Nils __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] applying a function several times
I am sorry. I am not sure if the mail a send before to this list was rejected
because of header (subject).
I've changed it. The first maybe was not appropriate.
I did a function (sec_conop) whose arguments are syndic,
well and wellconop.
sec_conop(syndic='01syndic.txt',well='well-1.csv',wellconop='well-1.dat');closeAllConnections()
This function takes well and syndic, matching between
them and then it does some transformations. The result is exported to
wellconop.
I will apply this function to one hundred different wells.
Therefore, for each well I use, the wellconop argument will change too.
For intance if well is now well-2.csv, the function will be
sec_conop(syndic='01syndic.txt',well='well-2.csv',wellconop='well-2.dat');closeAllConnections()
I am trying to apply this function automatically to all
well I have, but I do not find the way.
The last I tried, for three different wells, was :
wells-data.frame(funct=rep('sec_conop(',3),syndic=c('01syndic.txt','01syndic.txt','01syndic.txt'),well=c('well-1.csv','well-2.csv','well-3-1.csv'),wellconop=c('well-1.dat','well-2.dat','well-3.dat'))
funct_3wells-paste(wells$funct,',wells$syndic,', ,
,', wells$well,',
, , ' ,wells$wellconop,',),;,closeAllConnections(),sep='')
lapply(funct_3wells,as.formula)
This way works partially because the results in wellconop
are truncated.
Has anyone any suggestion?
Thanks in advance
Carlos
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Re: [R] Legend and Main Title positioning
layout(matrix(c(1,2,3,4), nrow = 2, byrow = TRUE))
plot(rnorm(100))
plot(rnorm(200))
plot(rnorm(300))
plot(rnorm(400))
Now, I'd like to create a legend below each plot and generate a common
title.
How can I do that?
If you are laying plots out in grids like this then lattice graphics are
generally the way to go, but here's a solution based upon base graphics.
The trick is to include extra potting space in your layout for the
legends. The code is messy, since it requires you to manually specify
which cell of the layout to plot into, but I'm sure guven some thought you
can automate this.
#4 space for plots, 4 for legends
layout(matrix(1:8, nrow = 4, byrow = TRUE), heights=rep(c(3,1),4))
#Check the layout looks suitable
layout.show(8)
#Avoid clipping problems, and create space for your title
par(xpd=TRUE, oma=c(0,0,2,0))
#First plot
plot(rnorm(100))
#Move down and plot the first legend
par(mfg=c(2,1))
legend(0,0, legend=foo, pch=1)
#Repeat for the other plots and legends
par(mfg=c(1,2))
plot(rnorm(200))
par(mfg=c(2,2))
legend(0,0, legend=bar, pch=1)
par(mfg=c(3,1))
plot(rnorm(300))
par(mfg=c(4,1))
legend(0,0, legend=baz, pch=1)
par(mfg=c(3,2))
plot(rnorm(400))
par(mfg=c(4,2))
legend(0,0, legend=quux, pch=1)
#Title for all the plots
title(main=4 plots, outer=TRUE)
Regards,
Richie.
Mathematical Sciences Unit
HSL
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Re: [R] comparing SAS and R survival analysis with time-dependent covariates
This query of why do SAS and S give different answers for Cox models comes
up every so often. The two most common reasons are that
a. they are using different options for the ties
b. the SAS and S data sets are slightly different.
You have both errors.
First, make sure I have the same data set by reading a common file, and then
compare the results.
tmt54% more sdata.txt
1 0.0 0.5 0 0
1 0.5 3.0 1 1
2 0.0 1.0 0 0
2 1.0 1.5 1 1
3 0.0 6.0 0 0
4 0.0 8.0 0 1
5 0.0 1.0 0 0
5 1.0 8.0 1 0
6 0.0 21.0 0 1
7 0.0 3.0 0 0
7 3.0 11.0 1 1
tmt55% more test.sas
options linesize=80;
data trythis;
infile 'sdata.txt';
input id start end delir outcome;
proc phreg data=trythis;
model (start, end)*outcome(0)=delir/ ties=discrete;
proc phreg data=trythis;
model (start, end)*outcome(0)=delir/ ties=efron;
tmt56% more test.r
trythis - read.table('sdata.txt',
col.names=c(id, start, end, delir, outcome))
coxph(Surv(start, end, outcome) ~ delir, data=trythis, ties='exact')
coxph(Surv(start, end, outcome) ~ delir, data=trythis, ties='efron')
-
I now get comparable answers. Note that Cox's exact partial likelihood is
the correct form to use for discrete time data. I labeled this as the 'exact'
method and SAS as the 'discrete' method. The exact marginal likelihood of
Prentice et al, which SAS calls the 'exact' method is not implemented in S.
As to which package is more reliable, I can only point to a set of formal
test
cases that are found in Appendix E of the book by Therneau and Grambsch. These
are small data sets where the coefficients, log-likelihood, residuals, etc have
all been worked out exactly in closed form. R gets all of these test cases
right, SAS gets almost all.
Terry Therneau
-
Svetlan Eden wrote
Dear R-help,
I was comparing SAS (I do not know what version it is) and R (version
2.6.0 (2007-10-03) on Linux) survival analyses with time-dependent
covariates. The results differed significantly so I tried to understand
on a short example where I went wrong. The following example shows that
even when argument 'method' in R function coxph and argument 'ties' in
SAS procedure phreg are the same, the results of Cox regr. are
different. This seems to happen when there are ties in the
events/covariates times.
My question is what software, R or SAS, is more reliable for the
survival analysis with time-dependent covariates or if you could point
out a problem in the following example.
...
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Re: [R] Plotting a kriging on a map
Rodrigo, This is an old and quite basic Krig, my data was continuous measurements in lat, long so binned first. library(geoR) counts-bins2d(long, lat,bin=c(0.1,0.1),plot=FALSE, nlevels=15, color.palette=heat.colors, xaxs='i', yaxs='i', las=1, main='') countsgeo-as.geodata(counts) xrange-seq(range(countsgeo$coords[,1])[1],range(countsgeo$coords[,1]) [2], by=bin[1]) yrange-seq(range(countsgeo$coords[,2])[1],range(countsgeo$coords[,2]) [2], by=bin[2]) loci-expand.grid(x=range(counts,y=yrange) #krig grid kc - krige.conv(countsgeo, loc=loci, krige=krige.control(cov.pars=c(1, .25))) #can fine tune color.palette-rainbow #use heat etc. contour(kc,asp=1, col=col- c(#F,color.palette(length(levels)-1)) , add=TRUE) #makes 0 counts white You could also use image or other 3d graphics. Hope that Helps, Jon Loehrke Graduate Research Assistant Department of Fisheries Oceanography School for Marine Science and Technology University of Massachusetts 200 Mill Road, Suite 325 Fairhaven, MA 02719 [EMAIL PROTECTED] [[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] determine assortative mixing in networks (igraph-library)
Dear all, I'm looking for a method to quantify assortative mixing in undirected graphs, preferably something that is compatible with the igraph-library for network analysis. Is anybody aware of an already existing function/library for this or an easy way to implement this functionality in R? Many thanks, Rainer WikipediaWictionaryChambers (UK)Google imagesGoogle defineThe Free DictionaryJoin exampleWordNetGoogleUrban DictionaryAnswers.comrhymezone.comMerriam-Webster0 wvcidfjoguarm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Cartesian Product Of Character Vectors
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem
to get *apply to paste the column values together.
Thanks
Rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
***
The Royal Bank of Scotland plc. Registered in Scotland No 90312.
Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:
25}}
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Re: [R] Cartesian Product Of Character Vectors
Hi David
Perfect - if I specify sep=, it gives me exactly what I need.
Cheers
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: 05 December 2008 14:16
To: WINSTON, Rory, GBM
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to
get *apply to paste the column values together.
Thanks
Rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
**
* The Royal Bank of Scotland plc. Registered in Scotland
No 90312.
Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:
25}}
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cartesian Product Of Character Vectors
Sorry, I spoke too soon...
interaction() only works for sequences of equal length. Anyone know a method
that works with unequal-length vectors?
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of WINSTON, Rory,
GBM
Sent: 05 December 2008 14:20
To: [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Hi David
Perfect - if I specify sep=, it gives me exactly what I need.
Cheers
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: 05 December 2008 14:16
To: WINSTON, Rory, GBM
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to
get *apply to paste the column values together.
Thanks
Rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
**
* The Royal Bank of Scotland plc. Registered in Scotland
No 90312.
Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:
25}}
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[R] read.gal3.R
Hello everybody, I'm trying to perform a ESDA on some data about per capita GDP among European regions and I'm trying to learn how to make R and GeoDa interact. I'm reading that in order to import a GAL file from GeoDa to R it's best to source into R the read.gal3.R file. But where can I get it from? Thank you, Annachiara Saguatti [[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] levels update
Thanks a lot!!! the drop thing was exactly what I was looking for (I already used it some time ago but forgot about it). Thanks to everybody else too. Antje Prof Brian Ripley schrieb: On Fri, 5 Dec 2008, jim holtman wrote: try this: df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) [1] a b c my.sub$X1 - factor(my.sub$X1) I find my.sub$X1 - my.sub$X1[drop=TRUE] a lot more self-explanatory. See ?[.factor. However, if you find yourself wanting to do this, ask why you have a factor (rather than a character vector) in the first place. levels(my.sub$X1) [1] a b On Fri, Dec 5, 2008 at 7:50 AM, Antje [EMAIL PROTECTED] wrote: Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) - c(X1,X2,X3) my.sub - subset(df, X1 == a | X1 == b) levels(my.sub$X1) # still gives me a,b,c, though the subset does not contain entries with c anymore I guess, the solution is rather simple, but I cannot find it. Antje __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] levels update
I hope this question is not too stupid. I would like to know how to
update
levels after subsetting data from a data.frame.
df - data.frame(factor(c(a,a,c,b,b)), c(4,5,6,7,8),
c(9,1,2,3,4))
names(df) - c(X1,X2,X3)
my.sub - subset(df, X1 == a | X1 == b)
levels(my.sub$X1)
# still gives me a,b,c, though the subset does not contain entries
with
c anymore
Two questions in one afternon; aren't I good to you!
levels(my.sub$X1[,drop=TRUE])
[1] a b
levels(factor(my.sub$X1))
[1] a b
Regards,
Richie.
Mathematical Sciences Unit
HSL
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] setwd and list.files / linux
Hi Lars, I am having trouble replicating this problem, using the default file/folder permissions on my Fedora 10 system. When creating a new folder, I get: mkdir Test drwxrwxr-x 2 marcs marcs 4096 2008-12-05 08:17 Test When creating a new file, I get: touch Test.txt -rw-rw-r-- 1 marcs marcs 0 2008-12-05 08:17 Test.txt Even when modifying the file permissions to be similar to yours: chmod 644 Test.txt -rw-r--r-- 1 marcs marcs 0 2008-12-05 08:17 Test.txt I can still read the file from a different folder. It is likely that your 'umask', which defines the default file/folder permissions on your system is different than mine. For me, as a regular user, it is 002. This value is usually defined in /etc/bashrc: $ grep umask /etc/bashrc umask 002 umask 022 The first value is for a regular user, the second for 'root'. It is likely that your umask is 022 based upon the file permissions that you list, as that would result in a default permission of 644. This might mean that the default umask on RHEL is different than on Fedora, or perhaps your default umask has been changed in ~/.bashrc. I suspect that you might have a folder permission issue, rather than a file permission issue and I think that would be consistent with the behavior that you are observing. I would check the permissions on each folder in the path that you are trying to access to see if something is amiss there. Perhaps try using setwd() on each folder in the datapath tree, starting at 'the top', adding a new lower level folder each time, until it fails. That might provide some guidance. HTH, Marc on 12/05/2008 02:46 AM Lars Beckmann wrote: Hey Marc, sorry I forgot to mention it. I am the owner of the path, which is in my home directory. I can change the read/write/execute privileges. All users have - at least - read privileges for all files and the directory: hpc36 40: ls -la *.tped -rw-r--r-- 1 cremer C020 205487231 Dec 3 17:35 chr10.tped -rw-r--r-- 1 cremer C020 192067641 Dec 3 17:37 chr11.tped -rw-r--r-- 1 cremer C020 192243483 Dec 3 17:38 chr12.tped etc A workaround I found now is to start R in the respective directory from the command line, which will set working directory to current directory. list.files will work then too: list.file(.,pattern=\\.tped) The problem now is to start a script via R CMD BATCH from a different directory, because setwd() will not work: datapath-'/nfs/home/m/cremer/gaw/data/traws_chrs/' getwd() [1] /nfs/home/m/cremer/gaw/code/plink setwd(datapath) Error in setwd(datapath) : cannot change working directory Execution halted Lars Original-Nachricht Datum: Thu, 04 Dec 2008 10:58:16 -0600 Von: Marc Schwartz [EMAIL PROTECTED] An: Lars Beckmann [EMAIL PROTECTED] CC: r-help@r-project.org Betreff: Re: [R] setwd and list.files / linux on 12/04/2008 10:38 AM Lars Beckmann wrote: Hey, two questions regarding setwd and list.files: I try to use the function list.files: datapath-'/nfs/home/m/cremer/gaw/data/traws_chrs' fl-list.files(datapath, pattern=\\.tped) but I get the following error message: Warning message: In list.files(datapath, pattern = \\.tped) : list.files: '/nfs/home/m/cremer/gaw/data/traws_chrs/' is not a readable directory fl character(0) The use of setwd also does not work: setwd(datapath) Error in setwd(datapath) : cannot change working directory datapath exists and it doesn't matter whether I use single quotes ' or double quotes Any help? Lars You do not have read privileges to the path in question. Either you or your SysAdmin will have to grant the appropriate privileges so that you can read the files in the path. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cartesian Product Of Character Vectors
Got it...I completely overlooked the collapse argument to paste():
apply(expand.grid(x,y),1,function(x) paste(x,collapse=))
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: WINSTON, Rory, GBM
Sent: 05 December 2008 14:30
To: WINSTON, Rory, GBM; [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: RE: [R] Cartesian Product Of Character Vectors
Sorry, I spoke too soon...
interaction() only works for sequences of equal length. Anyone know a method
that works with unequal-length vectors?
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of WINSTON, Rory,
GBM
Sent: 05 December 2008 14:20
To: [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Hi David
Perfect - if I specify sep=, it gives me exactly what I need.
Cheers
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: 05 December 2008 14:16
To: WINSTON, Rory, GBM
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to
get *apply to paste the column values together.
Thanks
Rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
**
* The Royal Bank of Scotland plc. Registered in Scotland
No 90312.
Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
[R] Sink does not send graphs to sink file
Wiindows XP
R 2.7
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there anyway that I
can have both the text and graphic output of my analyses appear in a file?
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)
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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] How to get Greenhouse-Geisser epsilons from anova?
Dear Nils, You might also take a look at the Anova() function in the car package, which though less flexible than anova() should get you the tests that you want more simply. ?Anova has an example of a repeated-measures ANOVA with two within-subject and two between-subject factors. I hope that this helps, John -- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Skotara Sent: December-05-08 8:40 AM To: Peter Dalgaard; r-help@r-project.org Subject: Re: [R] How to get Greenhouse-Geisser epsilons from anova? Dear Mr. Daalgard. thank you very much for your reply, it helped me to progress a bit. The following works fine: dd - expand.grid(C = 1:7, B= c(r, l), A= c(c, f)) myma - as.matrix(myma) #myma is a 12 by 28 list mlmfit - lm(myma~1) mlmfit0 - update(mlmfit, ~0) anova(mlmfit, mlmfit0, X= ~C+B, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of A. anova(mlmfit, mlmfit0, X= ~A+C, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of B. However, I can not figure out how this works for the other effects. If I try: anova(mlmfit, mlmfit0, X= ~A+B, M = ~A+C+B, idata = dd, test=Spherical) I get: Fehler in function (object, ..., test = c(Pillai, Wilks, Hotelling-Lawley, : residuals have rank 1 4 I also don't know how I can calculate the various interactions.. My read is I should change the second argument mlmfit0, too, but I can't figure out how... Do you know what to do? Thank you very much! Peter Dalgaard schrieb: Skotara wrote: Dear all, I apologize for my basic question. I try to calculate an anova for repeated measurements with 3 factors (A,B,C) having 2, 2, and 7 levels. or with an additional fourth between subjects factor D. Everything works fine using aov(val ~ A*B*C + Error(subject/ (A*B*C) ) ) or aov(val ~ (D*A*B*C) + Error(subject/(A*B*C)) + D ) val, A, B, C, D and subject are columns in a data.frame. How can I get the estimated Greenhouse-Geisser and Huynh-Feldt epsilons? I know Peter Dalgaard described it in R-News Vol. 7/2, October 2007. However, unfortunately I am not able to apply that using my data... Why? It is supposed to work. You just need to work out the X and M specification for the relevant error strata and set test=Spherical for anova.mlm, or work out the T contrast matrix explicitly if that suits your temper better. Furthermore, I am still confused of how SPSS calculates the epsilons since it is mentioned that perhaps there are any errors in SPSS?? I would be glad if anyone could help me! I am looking forward to hearing from you! Thank you! Nils __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cartesian Product Of Character Vectors
Try this also:
with(expand.grid(x,y), paste(Var1, Var2, sep = ))
On Fri, Dec 5, 2008 at 1:08 PM, [EMAIL PROTECTED] wrote:
Got it...I completely overlooked the collapse argument to paste():
apply(expand.grid(x,y),1,function(x) paste(x,collapse=))
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: WINSTON, Rory, GBM
Sent: 05 December 2008 14:30
To: WINSTON, Rory, GBM; [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: RE: [R] Cartesian Product Of Character Vectors
Sorry, I spoke too soon...
interaction() only works for sequences of equal length. Anyone know a
method that works with unequal-length vectors?
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of WINSTON, Rory, GBM
Sent: 05 December 2008 14:20
To: [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Hi David
Perfect - if I specify sep=, it gives me exactly what I need.
Cheers
Rory Winston
RBS Global Banking Markets
Office: +44 20 7085 4476
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: 05 December 2008 14:16
To: WINSTON, Rory, GBM
Cc: r-help@r-project.org
Subject: Re: [R] Cartesian Product Of Character Vectors
Does this satisfy?
levels(interaction(x,y))
[1] aaa.1 bbb.1 ccc.1 aaa.2 bbb.2 ccc.2 aaa.3 bbb.3
ccc.3
--
David Winsemius
On Dec 5, 2008, at 8:12 AM, [EMAIL PROTECTED] [EMAIL PROTECTED]
wrote:
Hi all
(I'm sure this question has been asked before, but I cant find it).
If I have two character vectors:
x - c(aaa,bbb,ccc)
y - c(1,2,3)
How can I get the cartesian product of the string values?
expand.grid(x,y)
Gives me a data frame with separate columns...however, I cant seem to
get *apply to paste the column values together.
Thanks
Rory
Rory Winston
RBS Global Banking Markets
280 Bishopsgate, London, EC2M 4RB
Office: +44 20 7085 4476
**
* The Royal Bank of Scotland plc. Registered in Scotland
No 90312.
Registered Office: 36 St Andrew Square, Edinburgh EH2 2YB.
Authorised and regulated by the Financial Services Authority
This e-mail message is confidential and for use by the=2...{{dropped:
25}}
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Sink does not send graphs to sink file
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there
anyway that I can have both the text and graphic output of my
analyses appear in a file?
You can create a latex document with text, graphs and R-code using
?Sweave. If you prefer to write to Open document format , there is an
Odfweave package.
The other alternative is simply writing your graphs to files (or one file
with all the graphs), e.g.
pdf(test.pdf)
plot(1:10)
hist(rnorm(100))
dev.off()
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
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[R] systematically matching the numbers in two sequences
I am having trouble writing a code for matching two pairs of sequences with differing lengths: for example sequence1= 1,2,3,4,5,6,7 sequence2=1,2,3,4,5,6,7,8,9,10 I want to create several new pairs of sequences in several dataframes such that: 1st dataframe is 1,10 (start of sequence1, end of sequence 2) 2nd dataframe is 1, 9 2, 10 3rd dataframe is 1, 8 2, 9 3, 10 etc etc. I realise this may involve a complicated loop and am really struggling to make a start on it. -- View this message in context: http://www.nabble.com/systematically-matching-the-numbers-in-two-sequences-tp20856673p20856673.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] Sink does not send graphs to sink file
I am using sink() to send the results of my analyses to a text file. Unfortunately my graphs do not become part of the file. Is there anyway that I can have both the text and graphic output of my analyses appear in a file? Well, how would you expect the graphs to be represented in a text file? So the short answer is no you can't do that. The longer and probably more helpful answer is: Have a look at Sweave(). Sweave is a system which allows you to integrate R-code with LaTeX formatting -- including the possibility to insert graphs. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 2D density tophat
In case anyone other than me was interested, a pretty efficient
circular tophat can be made using the fields function fields.rdist.near:
CircHat=function (x, y, h=1, gridres = c((max(x)-min(x))/25,(max(y)-
min(y))/25), lims = c(range(x), range(y)),density=FALSE)
{
require(fields)
nx - length(x)
ny - length(y)
n=c(1+(lims[2]-lims[1])/gridres[1],1+(lims[4]-lims[3])/gridres[2])
if (length(y) != nx)
stop(data vectors must be the same length)
if (any(!is.finite(x)) || any(!is.finite(y)))
stop(missing or infinite values in the data are not allowed)
if (any(!is.finite(lims)))
stop(only finite values are allowed in 'lims')
gx - seq(lims[1], lims[2], by = gridres[1])
gy - seq(lims[3], lims[4], by = gridres[2])
fullgrid=expand.grid(gx,gy)
if (missing(h))
h - c(bandwidth.nrd(x), bandwidth.nrd(y))
temp
=
table
(fields
.rdist
.near
(as
.matrix
(fullgrid
),as.matrix(cbind(x,y)),mean.neighbor=ceiling(length(x)*pi*h^2/
((lims[2]-lims[1])*(lims[4]-lims[3]))),delta=h)$ind[,1])
pad=rep(0,length(gx)*length(gy))
pad[as.numeric(names(temp))]=as.numeric(temp)
z - matrix(pad, length(gx), length(gy))
if(density){z=z/(nx*pi*h^2)}
return=list(x = gx, y = gy, z = z)
}
It works in more or less the same way as kde2d, but by default it
returns counts not densities
Aaron
On 1 Dec 2008, at 11:46, Aaron Robotham wrote:
Hello R users,
I have successfully created a square (or more generally, rectangular)
tophat smoothing routine based on altering the already available
KDE2D. I would be keen to implement a circular tophat routine also,
however this appears to be much more difficult to write efficiently (I
have a routine, but it's very slow). I tried to create one based on
using crossdist (in spatstat) to create a distance matrix between my
data and the sampling grid, but it doesn't take a particularly large
amount of data (or hi-res grid) for memory to be a big problem. The 2D
density routines I have been able to find either don't support a
simple tophat, or don't use the absolute distances between the
sampling grid and the data. Should anyone know of more general 2D
density routines that might support circular tophats, or know of a
simple and efficient method for creating them, I would be very
grateful.
Thanks for your time,
Aaron
PS: I tried sending this on Friday originally, but as far as I know
that didn't work, so should another post appear from me asking the
same thing I apologise in advance.
[[alternative HTML version deleted]]
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Re: [R] systematically matching the numbers in two sequences
Try this: s1 - 1:7 s2 - 1:10 lapply(seq(length(s1)), function(i)cbind(head(s1, i), tail(s2, i))) On Fri, Dec 5, 2008 at 1:44 PM, emj83 [EMAIL PROTECTED] wrote: I am having trouble writing a code for matching two pairs of sequences with differing lengths: for example sequence1= 1,2,3,4,5,6,7 sequence2=1,2,3,4,5,6,7,8,9,10 I want to create several new pairs of sequences in several dataframes such that: 1st dataframe is 1,10 (start of sequence1, end of sequence 2) 2nd dataframe is 1, 9 2, 10 3rd dataframe is 1, 8 2, 9 3, 10 etc etc. I realise this may involve a complicated loop and am really struggling to make a start on it. -- View this message in context: http://www.nabble.com/systematically-matching-the-numbers-in-two-sequences-tp20856673p20856673.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to calculate the distance between two density functions
Dear all, I wrote the following code to calculate the density functions for two data sets, respectively. den_str -density(str_data$Similarity); den_non_str -density(nonstr_data$Similarity); However, I would like to knowing the difference between den_str and den_non_str, that is, the difference between the region under the curve of the den_str and the region under the curve of the den_non_str. How to do? Thank you for help. Jia-Ming [[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] systematically matching the numbers in two sequences
I am having trouble writing a code for matching two pairs of sequences with differing lengths: for example sequence1= 1,2,3,4,5,6,7 sequence2=1,2,3,4,5,6,7,8,9,10 I want to create several new pairs of sequences in several dataframes such that: 1st dataframe is 1,10 (start of sequence1, end of sequence 2) 2nd dataframe is 1, 9 2, 10 3rd dataframe is 1, 8 2, 9 3, 10 etc etc. I realise this may involve a complicated loop and am really struggling to make a start on it. -- View this message in context: http://www.nabble.com/systematically-matching-the-numbers-in-two-sequences-tp20855728p20855728.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] Problems with lattice-histograms or png within loops
Dear R-users,
I have a question concerning the use of lattice plots within for-loops.
I want to create a png file containing a lattice histogram which works out
fine (part 1).
When I loop the whole code, the graphic file does not contain anything (part
2).
I can fix it by wrapping the histogram function into a print command (part
3).
Why is that so? Why can I not loop it directly?
TIA,
Mark
attach(iris)
### part 1
png(filename = graphic_1.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
### part 2
for (i in c(1))
{
png(filename = graphic_2.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
}
### part 3
for (i in c(1))
{
png(filename = graphic_3.png)
print(histogram( ~ Sepal.Length | Species, data = iris))
dev.off()
}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Strplit code
[R] Strplit code pomchip at free.fr pomchip at free.fr Wed Dec 3 20:52:21 CET 2008 Dear R-users, The strsplit function does not exist in S-plus and I would like to use it. How could I reproduce the function in Splus or access to its source code? Thank you in advance, Sebastien strsplit() was added to S+ 8.0 (May 2007). At the same time we changed the default regular expression style from 'basic' (a.k.a. 'obsolete') to 'extended' and we added the string functions sub(), gsub(), and sprintf(). S+ 8.1 is now available (as of November 2008). Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.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] Problems with lattice-histograms or png within loops
I have a question concerning the use of lattice plots within for-loops.
I want to create a png file containing a lattice histogram which works
out
fine (part 1).
When I loop the whole code, the graphic file does not contain anything
(part
2).
I can fix it by wrapping the histogram function into a print command
(part
3).
Why is that so? Why can I not loop it directly?
TIA,
Mark
attach(iris)
### part 1
png(filename = graphic_1.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
### part 2
for (i in c(1))
{
png(filename = graphic_2.png)
histogram( ~ Sepal.Length | Species, data = iris)
dev.off()
}
### part 3
for (i in c(1))
{
png(filename = graphic_3.png)
print(histogram( ~ Sepal.Length | Species, data = iris))
dev.off()
}
This is FAQ 7.22:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f
The help page on ?histogram also mentions that the function returns an
object of class 'trellis' and the 'print' method (usually called by
default) will plot it on an appropriate plotting device
It's a good idea to get into the habit of calling print any time you
create a lattice graph. If you reorder your code like
h1 - histogram( ~ Sepal.Length | Species, data = iris)
png(filename = graphic_h1.png)
print(h1)
dev.off()
then you won't end up with an open device if drawing the plot fails.
Regards,
Richie.
Mathematical Sciences Unit
HSL
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Re: [R] comparing SAS and R survival analysis with time-dependent covariates
Thank you so much, this was very helpful.
Svetlana
Terry Therneau wrote:
This query of why do SAS and S give different answers for Cox models comes
up every so often. The two most common reasons are that
a. they are using different options for the ties
b. the SAS and S data sets are slightly different.
You have both errors.
First, make sure I have the same data set by reading a common file, and then
compare the results.
tmt54% more sdata.txt
1 0.0 0.5 0 0
1 0.5 3.0 1 1
2 0.0 1.0 0 0
2 1.0 1.5 1 1
3 0.0 6.0 0 0
4 0.0 8.0 0 1
5 0.0 1.0 0 0
5 1.0 8.0 1 0
6 0.0 21.0 0 1
7 0.0 3.0 0 0
7 3.0 11.0 1 1
tmt55% more test.sas
options linesize=80;
data trythis;
infile 'sdata.txt';
input id start end delir outcome;
proc phreg data=trythis;
model (start, end)*outcome(0)=delir/ ties=discrete;
proc phreg data=trythis;
model (start, end)*outcome(0)=delir/ ties=efron;
tmt56% more test.r
trythis - read.table('sdata.txt',
col.names=c(id, start, end, delir, outcome))
coxph(Surv(start, end, outcome) ~ delir, data=trythis, ties='exact')
coxph(Surv(start, end, outcome) ~ delir, data=trythis, ties='efron')
-
I now get comparable answers. Note that Cox's exact partial likelihood is
the correct form to use for discrete time data. I labeled this as the 'exact'
method and SAS as the 'discrete' method. The exact marginal likelihood of
Prentice et al, which SAS calls the 'exact' method is not implemented in S.
As to which package is more reliable, I can only point to a set of formal test
cases that are found in Appendix E of the book by Therneau and Grambsch. These
are small data sets where the coefficients, log-likelihood, residuals, etc have
all been worked out exactly in closed form. R gets all of these test cases
right, SAS gets almost all.
Terry Therneau
-
Svetlan Eden wrote
Dear R-help,
I was comparing SAS (I do not know what version it is) and R (version
2.6.0 (2007-10-03) on Linux) survival analyses with time-dependent
covariates. The results differed significantly so I tried to understand
on a short example where I went wrong. The following example shows that
even when argument 'method' in R function coxph and argument 'ties' in
SAS procedure phreg are the same, the results of Cox regr. are
different. This seems to happen when there are ties in the
events/covariates times.
My question is what software, R or SAS, is more reliable for the
survival analysis with time-dependent covariates or if you could point
out a problem in the following example.
...
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Re: [R] legend at fixed distance form the bottom
It is best to create the graphics device at the final size desired, then do the plotting and add the legend. For getting a fixed distance, look at the function grconvertY for one possibility. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Christophe Genolini Sent: Friday, December 05, 2008 6:40 AM To: r-help@r-project.org Subject: [R] legend at fixed distance form the bottom Hi the list I would like to add a legend under a graph but at a fixed distance from the graphe. Is it possible ? More precisely, here is my code : --- 8 symboles - c(3,4,5,6) dn - rbind(matrix(rnorm(20),,5),matrix(rnorm(20,2),,5)) listSymboles - rep(symboles,each=2) matplot(t(dn),pch=listSymboles,type=b) legend(bottom, pch = unique(listSymboles), legend = c(ane, cheval, poney, mule), inset = c(0,-0.175), horiz = TRUE, xpd = NA) --- 8 But when I change the size of the graph, the legend is misplaced. Instead, I try to put some text in xlab, but I do not know how to get the +, x , V and other symbol. Does anyone got a solution ? Thanks a lot. Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to calculate the distance between two density functions
A similar question was posed and answered: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/119793.html Two aspects needed to be addressed ... specifying the same domain, and getting the x-values to line up prior to the subtraction (or whatever function is desired). What are you going to do when the two functions cross? d1 - dnorm(seq(-2,2,by=.1)) d2 - dnorm(seq(-2,2,by=.1), mean=2) plot(seq(-2,2,by=.1),d1) lines(seq(-2,2,by=.1),d2) or d4 - dnorm(seq(-4,4,by=.1)) d5 - dnorm(seq(-4,4,by=.1), sd=5) plot(seq(-4,4,by=.1),d4) lines(seq(-4,4,by=.1),d5) -- David Winsemius On Dec 5, 2008, at 8:59 AM, Chang Jia-Ming wrote: Dear all, I wrote the following code to calculate the density functions for two data sets, respectively. den_str -density(str_data$Similarity); den_non_str -density(nonstr_data$Similarity); However, I would like to knowing the difference between den_str and den_non_str, that is, the difference between the region under the curve of the den_str and the region under the curve of the den_non_str. How to do? Thank you for help. Jia-Ming [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with wavCWTPeaks
I cannot understand the following error printed out when I try to get the extrema of my time series. I would appreciate some suggestion as I really cannot interpret the error. I might not be using a proper set of parameters in calling such functions. I am learning by doing ... aa.peak - wavCWTPeaks (aa.tree) Error in `row.names-.data.frame`(`*tmp*`, value = c(1, 0)) : invalid 'row.names' length My data vector is as follows: aa [1] -0.852625404 -0.886941142 -0.920699269 -0.953858240 -0.986377090 [6] -1.018215725 -1.049335086 -1.079697274 -1.109265628 -1.138004782 [11] -1.165880733 -1.192860894 -1.218914146 -1.244010899 -1.268123116 [16] -1.291224432 -1.313290096 -1.334297111 -1.354224193 -1.373051833 [21] -1.390762317 -1.407339716 -1.422769965 -1.437040775 -1.450141701 [26] -1.462064157 -1.472801336 -1.482348272 -1.490701780 -1.497860461 [31] -1.503824640 -1.508596384 -1.512179425 -1.514579128 -1.515802478 [36] -1.515858035 -1.514755845 -1.512507449 -1.509125815 -1.504625250 [41] -1.499021421 -1.492331231 -1.484572809 -1.475765432 -1.465929450 [46] -1.455086269 -1.443258266 -1.430468724 -1.416741774 -1.402102340 [51] -1.386576077 -1.370189287 -1.352968891 -1.334942359 -1.316137624 [56] -1.296583049 -1.276307350 -1.255339556 -1.233708931 -1.211444912 [61] -1.188577082 -1.165135088 -1.141148591 -1.116647219 -1.091660532 [66] -1.066217945 -1.040348675 -1.014081694 -0.987445689 -0.960469010 [71] -0.933179631 -0.905605109 -0.82541 -0.849708278 -0.821438088 [76] -0.792987122 -0.764379898 -0.735640266 -0.706791393 -0.677855738 [81] -0.648855033 -0.619810274 -0.590741693 -0.561668758 -0.532610156 [86] -0.503583784 -0.474606745 -0.445695338 -0.416865062 -0.388130609 [91] -0.359505865 -0.331003918 -0.302637056 -0.274416782 -0.246353810 [96] -0.218458088 -0.190738802 -0.163204394 -0.135862573 -0.108720337 [101] -0.081783987 -0.055059152 -0.028550808 -0.002263304 0.023799613 [106] 0.049634774 0.075239551 0.100611836 0.125750005 0.150652886 [111] 0.175319722 0.199750141 0.223944116 0.247901927 0.271624125 [116] 0.295111492 0.318365000 0.341385774 0.364175049 0.386734130 [121] 0.409064353 0.431167045 0.453043480 0.474694850 0.496122212 [126] 0.517326465 0.538308303 0.559068185 0.579606300 0.599922532 [131] 0.620016432 0.639887183 0.659533578 0.678953988 0.698146340 [136] 0.716367156 0.734459833 0.752428741 0.770277372 0.788008310 [141] 0.805623188 0.823122667 0.840506396 0.857773019 0.874920094 [146] 0.891944162 0.908840641 0.925603903 0.942227205 0.958702728 [151] 0.975021541 0.991173645 1.007147946 1.022932278 1.038513449 [156] 1.053877190 1.069008243 1.083890313 1.098506181 1.112837647 [161] 1.126865608 1.140570088 1.153930270 1.166924526 1.179530477 [166] 1.191725035 1.203484451 1.214784382 1.225599912 1.235905642 [171] 1.245675755 1.254884072 1.263504095 1.271509107 1.278872248 [176] 1.285566555 1.291565077 1.296840902 1.301367293 1.305117743 [181] 1.308066011 1.310186290 1.311453212 1.311841963 1.311328350 [186] 1.309888930 1.307500999 1.304142758 1.299793348 1.294432925 [191] 1.288042752 1.280605267 1.272104143 1.262524364 1.251852272 [196] 1.240075658 1.227183764 1.213167403 1.198018950 1.181732402 [201] 1.164303440 1.145729407 1.126009401 1.105144212 1.083136426 [206] 1.059990384 1.035712188 1.010309729 0.983792617 0.956172242 [211] 0.927461713 0.897675875 0.866831225 0.834945955 0.802039881 [216] 0.768134393 0.733252445 0.697418506 0.660658502 0.622999781 [221] 0.584471057 0.545102363 0.504924994 0.463971453 0.422275398 [226] 0.379871580 0.336795792 0.293084805 0.248776314 0.203908880 [231] 0.158521870 0.112655403 0.066350293 0.019647991 -0.027409468 [236] -0.074779520 -0.122419124 -0.170284817 -0.218332766 -0.266518818 [241] -0.314798550 I convert it into a time series and then I get the CWT coefficients. Then I build the tree that exhibits only 3 branches (see attached plot) aats - ts (aa, deltat =1/30, start = 0.0) aa.cwt - wavCWT(aats) aa.tree - wavCWTTree (aa.cwt) I can get the data for each of the 3 branches: aa.tree[[1]] $itime [1] 135 135 134 133 132 130 128 126 123 122 122 122 122 123 126 $iscale [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 $time [1] 4.47 4.47 4.43 4.40 4.37 4.30 4.23 4.17 [9] 4.07 4.03 4.03 4.03 4.03 4.07 4.17 $scale [1] 0.0333 0.0667 0.1000 0.1333 0.1667 0.2000 [7] 0.2333 0.2667 0.3000 0. 0.3667 0.4000 [13] 0.4333 0.4667 0.5000 $extrema [1] 0.001143350 0.004153357 0.009778222 0.018315375 0.030294895 0.046225772 [7] 0.066645451 0.092064228 0.122788334 0.158757578 0.199797960 0.245925327 [13] 0.297235272 0.354056031 0.417519487 aa.tree[[2]] $itime [1] 202 202 202 202 202 202 202 202 202 202 205 $iscale [1] 1 2 3 4 5 6 7 8 9 10 11
Re: [R] Sink does not send graphs to sink file
2008/12/5 Philipp Pagel [EMAIL PROTECTED]: I am using sink() to send the results of my analyses to a text file. Unfortunately my graphs do not become part of the file. Is there anyway that I can have both the text and graphic output of my analyses appear in a file? Well, how would you expect the graphs to be represented in a text file? So the short answer is no you can't do that. Splus used to have (does it still?) an ascii-text based graphics device. Fine for simple things, I'm not sure how well fancy lattice graphics would work on it. If someone has a few hours to waste, they might be able to build something similar for R using aalib: http://aa-project.sourceforge.net/aalib/ Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sink does not send graphs to sink file
I think Sweave and/or odfWeave are the real answer, though. Obviously, a bigger and more elaborate kettle of fish. -- Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Barry Rowlingson Sent: Friday, December 05, 2008 9:21 AM To: Philipp Pagel Cc: r-help@r-project.org Subject: Re: [R] Sink does not send graphs to sink file 2008/12/5 Philipp Pagel [EMAIL PROTECTED]: I am using sink() to send the results of my analyses to a text file. Unfortunately my graphs do not become part of the file. Is there anyway that I can have both the text and graphic output of my analyses appear in a file? Well, how would you expect the graphs to be represented in a text file? So the short answer is no you can't do that. Splus used to have (does it still?) an ascii-text based graphics device. Fine for simple things, I'm not sure how well fancy lattice graphics would work on it. If someone has a few hours to waste, they might be able to build something similar for R using aalib: http://aa-project.sourceforge.net/aalib/ Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sink does not send graphs to sink file
Others mentioned Sweave, which is the best way to go if you have a planned set
of routines to do. If you are more just playing and want a transcript of what
you are doing (unpreplanned) that includes graphics in the final version, then
look at etxtStart from the TeachingDemos package. You will need to
postproccess the file using enscript (and optionally ps2pdf), but then you will
end up with a file including both the text and the graphics.
Note: using the current version (new one due soon) with R2.8 generates some
warnings that can be ignored.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of John Sorkin
Sent: Friday, December 05, 2008 8:15 AM
To: r-help@r-project.org
Subject: [R] Sink does not send graphs to sink file
Wiindows XP
R 2.7
I am using sink() to send the results of my analyses to a text file.
Unfortunately my graphs do not become part of the file. Is there anyway
that I can have both the text and graphic output of my analyses appear
in a file?
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
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[R] Running R Script on a Sequence of Files
Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] [R-pkgs] RMySQL 0.7-2 now available on CRAN
Dear R users, RMySQL 0.7-2 is now available on CRAN: http://cran.r-project.org/web/packages/RMySQL/index.html From the NEWS file: * New maintainer is Jeffrey Horner [EMAIL PROTECTED]. * We no longer distribute libmysql.dll. This library is now found either by reading the MYSQL_HOME environment variable or by reading the windows registry entries. * Removed dependence on MySQL C function load_defaults() as it was meant for command-line tools, not for (re)connecting to a database. * Fixed \r issue with dbWriteTable(). * Tests have been added and depend on proper values set in the environment variables MYSQL_DATABASE, MYSQL_USER, and MYSQL_PASSWD. Best, Jeff ___ R-packages mailing list [EMAIL PROTECTED] 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] Running R Script on a Sequence of Files
Have you thought about using one of the Python/R interface modules? http://www.omegahat.org/RSPython/ http://rpy.sourceforge.net/ Admittedly, I have not had much success in getting these to work on my machine, but I know others who have. Kyle H. Ambert Graduate Student Department of Medical Informatics Clinical Epidemiology Oregon Health Science University [EMAIL PROTECTED] On Fri, Dec 5, 2008 at 10:01 AM, Chris Poliquin [EMAIL PROTECTED]wrote: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
This is almost a macro problem. It could be done in SAS language using the WPS product (660 USD) I think. It is a familiar problem and I would be quite interested in the result. Is there any concept of Macros in R or a package to do the same. Regards, Ajay On Fri, Dec 5, 2008 at 11:31 PM, Chris Poliquin [EMAIL PROTECTED]wrote: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
2008/12/5 Chris Poliquin [EMAIL PROTECTED]: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? I can't believe the two 'solutions' already posted. It's easy: ?list.files Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
R has quite a few functions to get and manipulate filenames to facilitate exactly what you want to do. See ?files and especially the links at the end to the file name manipulation functions. e.g. dir(pathname) lists all file names in the directory pathname. ?list.files gives details. -- Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Chris Poliquin Sent: Friday, December 05, 2008 10:02 AM To: r-help@r-project.org Subject: [R] Running R Script on a Sequence of Files Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Help with wavCWTPeaks
#my suggestion would be to use the morlet wavelet as opposed to the mexican hat wavelet (default). aa - (structure(list(X.0.85 = c(-1.02, -1.17, -1.29, -1.39, -1.46, -1.5, -1.52, -1.5, -1.46, -1.39, -1.3, -1.19, -1.07, -0.93, -0.79, -0.65, -0.5, -0.36, -0.22, -0.08, 0.05, 0.18, 0.3, 0.41, 0.52, 0.62, 0.72, 0.81, 0.89, 0.98, 1.05, 1.13, 1.19, 1.25, 1.29, 1.31, 1.31, 1.29, 1.24, 1.16, 1.06, 0.93, 0.77, 0.58, 0.38, 0.16, -0.07, -0.31, -0.89, -1.05, -1.19, -1.31, -1.41, -1.47, -1.51, -1.51, -1.49, -1.44, -1.37, -1.28, -1.17, -1.04, -0.91, -0.76, -0.62, -0.47, -0.33, -0.19, -0.06, 0.08, 0.2, 0.32, 0.43, 0.54, 0.64, 0.73, 0.82, 0.91, 0.99, 1.07, 1.14, 1.2, 1.25, 1.29, 1.31, 1.31, 1.28, 1.23, 1.15, 1.04, 0.9, 0.73, 0.55, 0.34, 0.11, -0.12, -0.92, -1.08, -1.22, -1.33, -1.42, -1.48, -1.51, -1.51, -1.48, -1.43, -1.35, -1.26, -1.14, -1.01, -0.88, -0.74, -0.59, -0.45, -0.3, -0.16, -0.03, 0.1, 0.22, 0.34, 0.45, 0.56, 0.66, 0.75, 0.84, 0.93, 1.01, 1.08, 1.15, 1.21, 1.26, 1.3, 1.31, 1.3, 1.27, 1.21, 1.13, 1.01, 0.87, 0.7, 0.5, 0.29, 0.07, -0.17, -0.95, -1.11, -1.24, -1.35, -1.44, -1.49, -1.51, -1.51, -1.48, -1.42, -1.33, -1.23, -1.12, -0.99, -0.85, -0.71, -0.56, -0.42, -0.27, -0.14, 0, 0.13, 0.25, 0.36, 0.47, 0.58, 0.68, 0.77, 0.86, 0.94, 1.02, 1.1, 1.17, 1.23, 1.27, 1.3, 1.31, 1.3, 1.26, 1.2, 1.11, 0.98, 0.83, 0.66, 0.46, 0.25, 0.02, -0.22, -0.99, -1.14, -1.27, -1.37, -1.45, -1.5, -1.52, -1.5, -1.47, -1.4, -1.32, -1.21, -1.09, -0.96, -0.82, -0.68, -0.53, -0.39, -0.25, -0.11, 0.02, 0.15, 0.27, 0.39, 0.5, 0.6, 0.7, 0.79, 0.87, 0.96, 1.04, 1.11, 1.18, 1.24, 1.28, 1.31, 1.31, 1.29, 1.25, 1.18, 1.08, 0.96, 0.8, 0.62, 0.42, 0.2, -0.03, -0.27)), .Names = X.0.85, class = data.frame, row.names = c(NA, -240L))) library(wmtsa) library(fields) aats - ts (aa, deltat =1/30, start = 0.0) aa.cwt - wavCWT(aats, wavelet=morlet, col=tim.colors(100)) aa.tree - wavCWTTree (aa.cwt) On Fri, Dec 5, 2008 at 12:15 PM, [EMAIL PROTECTED] wrote: I cannot understand the following error printed out when I try to get the extrema of my time series. I would appreciate some suggestion as I really cannot interpret the error. I might not be using a proper set of parameters in calling such functions. I am learning by doing ... aa.peak - wavCWTPeaks (aa.tree) Error in `row.names-.data.frame`(`*tmp*`, value = c(1, 0)) : invalid 'row.names' length My data vector is as follows: aa [1] -0.852625404 -0.886941142 -0.920699269 -0.953858240 -0.986377090 [6] -1.018215725 -1.049335086 -1.079697274 -1.109265628 -1.138004782 [11] -1.165880733 -1.192860894 -1.218914146 -1.244010899 -1.268123116 [16] -1.291224432 -1.313290096 -1.334297111 -1.354224193 -1.373051833 [21] -1.390762317 -1.407339716 -1.422769965 -1.437040775 -1.450141701 [26] -1.462064157 -1.472801336 -1.482348272 -1.490701780 -1.497860461 [31] -1.503824640 -1.508596384 -1.512179425 -1.514579128 -1.515802478 [36] -1.515858035 -1.514755845 -1.512507449 -1.509125815 -1.504625250 [41] -1.499021421 -1.492331231 -1.484572809 -1.475765432 -1.465929450 [46] -1.455086269 -1.443258266 -1.430468724 -1.416741774 -1.402102340 [51] -1.386576077 -1.370189287 -1.352968891 -1.334942359 -1.316137624 [56] -1.296583049 -1.276307350 -1.255339556 -1.233708931 -1.211444912 [61] -1.188577082 -1.165135088 -1.141148591 -1.116647219 -1.091660532 [66] -1.066217945 -1.040348675 -1.014081694 -0.987445689 -0.960469010 [71] -0.933179631 -0.905605109 -0.82541 -0.849708278 -0.821438088 [76] -0.792987122 -0.764379898 -0.735640266 -0.706791393 -0.677855738 [81] -0.648855033 -0.619810274 -0.590741693 -0.561668758 -0.532610156 [86] -0.503583784 -0.474606745 -0.445695338 -0.416865062 -0.388130609 [91] -0.359505865 -0.331003918 -0.302637056 -0.274416782 -0.246353810 [96] -0.218458088 -0.190738802 -0.163204394 -0.135862573 -0.108720337 [101] -0.081783987 -0.055059152 -0.028550808 -0.002263304 0.023799613 [106] 0.049634774 0.075239551 0.100611836 0.125750005 0.150652886 [111] 0.175319722 0.199750141 0.223944116 0.247901927 0.271624125 [116] 0.295111492 0.318365000 0.341385774 0.364175049 0.386734130 [121] 0.409064353 0.431167045 0.453043480 0.474694850 0.496122212 [126] 0.517326465 0.538308303 0.559068185 0.579606300 0.599922532 [131] 0.620016432 0.639887183 0.659533578 0.678953988 0.698146340 [136] 0.716367156 0.734459833 0.752428741 0.770277372 0.788008310 [141] 0.805623188 0.823122667 0.840506396 0.857773019 0.874920094 [146] 0.891944162 0.908840641 0.925603903 0.942227205 0.958702728 [151] 0.975021541 0.991173645 1.007147946 1.022932278 1.038513449 [156] 1.053877190 1.069008243 1.083890313 1.098506181 1.112837647 [161] 1.126865608 1.140570088 1.153930270 1.166924526 1.179530477 [166] 1.191725035 1.203484451 1.214784382 1.225599912 1.235905642 [171] 1.245675755 1.254884072 1.263504095 1.271509107 1.278872248 [176] 1.285566555 1.291565077 1.296840902 1.301367293 1.305117743 [181] 1.308066011
Re: [R] Running R Script on a Sequence of Files
Thanks, Barry. I'll use that in the future. ---Kyle. On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson [EMAIL PROTECTED] wrote: 2008/12/5 Chris Poliquin [EMAIL PROTECTED]: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? I can't believe the two 'solutions' already posted. It's easy: ?list.files Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
Use dir to get the names and then lapply over them with a
custom anonymous function where L is a list of the returned
values:
# assumes file names are those in
# current directory that end in .dat
filenames - dir(pattern = \\.dat$)
L - lapply(filenames, function(x) {
DF - read.table(x, ...whatever...)
somefunction(DF)
})
Now L is a list of the returned 900 values. Alternately you could use
a loop.
On Fri, Dec 5, 2008 at 1:01 PM, Chris Poliquin [EMAIL PROTECTED] wrote:
Hi,
I have about 900 files that I need to run the same R script on. I looked
over the R Data Import/Export Manual and couldn't come up with a way to
read in a sequence of files.
The files all have unique names and are in the same directory. What I want
to do is:
1) Create a list of the file names in the directory (this is really what I
need help with)
2) For each item in the list...
a) open the file with read.table
b) perform some analysis
c) append some results to an array or save them to another file
3) Next File
My initial instinct is to use Python to rename all the files with numbers
1:900 and then read them all, but the file names contain some information
that I would like to keep intact and having to keep a separate database of
original names and numbers seems inefficient. Is there a way to have R read
all the files in a directory one at a time?
- Chris
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
I can't believe the two 'solutions' already posted. It's easy: Me neither. ?list.files That's what I would use, too. If the OP is on a UNIX platform, run the R-script in a loop in the shell is an alternative. Something like this (bourne shell syntax): for datafile in *.csv ; do Rscript analyze.R $datafile done The R script (analyze.R) can use commandArgs() to read the filename argument. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan 85350 Freising, Germany http://mips.gsf.de/staff/pagel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
Is there a way to list only the files in a given directory without passing pattern=... to list.files()? On Fri, Dec 5, 2008 at 5:10 PM, Kyle. [EMAIL PROTECTED] wrote: Thanks, Barry. I'll use that in the future. ---Kyle. On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson [EMAIL PROTECTED] wrote: 2008/12/5 Chris Poliquin [EMAIL PROTECTED]: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? I can't believe the two 'solutions' already posted. It's easy: ?list.files Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R Script on a Sequence of Files
Try this: dir()[!file.info(dir())$isdir] On Fri, Dec 5, 2008 at 2:30 PM, Gustavo Carvalho [EMAIL PROTECTED] wrote: Is there a way to list only the files in a given directory without passing pattern=... to list.files()? On Fri, Dec 5, 2008 at 5:10 PM, Kyle. [EMAIL PROTECTED] wrote: Thanks, Barry. I'll use that in the future. ---Kyle. On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson [EMAIL PROTECTED] wrote: 2008/12/5 Chris Poliquin [EMAIL PROTECTED]: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? I can't believe the two 'solutions' already posted. It's easy: ?list.files Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Running R Script on a Sequence of Files
This is almost a macro problem. It could be done in SAS language using the WPS product (660 USD) I think. ... OUCH! Why do it the complicated way??? Check out ?dir, ?list.files, and then ?lapply for a simple start. Don't give up so soon! When it comes to R there is no need to punt - you can always keep possession of the ball ... :-) Cheers, Jagat -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ajay ohri Sent: Friday, December 05, 2008 12:59 PM To: Chris Poliquin Cc: r-help@r-project.org Subject: Re: [R] Running R Script on a Sequence of Files This is almost a macro problem. It could be done in SAS language using the WPS product (660 USD) I think. It is a familiar problem and I would be quite interested in the result. Is there any concept of Macros in R or a package to do the same. Regards, Ajay On Fri, Dec 5, 2008 at 11:31 PM, Chris Poliquin [EMAIL PROTECTED]wrote: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running R Script on a Sequence of Files
Thanks for the solution . I especially liked the analogy along with the code of course. Regards, Ajay www.decisionstats.com On Sat, Dec 6, 2008 at 1:23 AM, [EMAIL PROTECTED] wrote: This is almost a macro problem. It could be done in SAS language using the WPS product (660 USD) I think. ... OUCH! Why do it the complicated way??? Check out ?dir, ?list.files, and then ?lapply for a simple start. Don't give up so soon! When it comes to R there is no need to punt - you can always keep possession of the ball ... :-) Cheers, Jagat -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ajay ohri Sent: Friday, December 05, 2008 12:59 PM To: Chris Poliquin Cc: r-help@r-project.org Subject: Re: [R] Running R Script on a Sequence of Files This is almost a macro problem. It could be done in SAS language using the WPS product (660 USD) I think. It is a familiar problem and I would be quite interested in the result. Is there any concept of Macros in R or a package to do the same. Regards, Ajay On Fri, Dec 5, 2008 at 11:31 PM, Chris Poliquin [EMAIL PROTECTED]wrote: Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. [[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] Running R Script on a Sequence of Files
It seems that you have 900 files with the same parameters in each file (I might be reading more between the lines here than you inferred). However if this is the case, why not import each of the files into a common database and then link the database using ODBC connectivity options. If that is practical, you could then code a series of subsetting options to select the data you need for specific analysis, write reports, and then iteratively select the next set of records. I may be suggesting a very simple solution, so forgive me if this trivializes your problem too greatly. Steve Friedman Ph. D. Spatial Statistical Analyst Everglades and Dry Tortugas National Park 950 N Krome Ave (3rd Floor) Homestead, Florida 33034 [EMAIL PROTECTED] Office (305) 224 - 4282 Fax (305) 224 - 4147 Chris Poliquin [EMAIL PROTECTED] nn.eduTo Sent by: r-help@r-project.org [EMAIL PROTECTED] cc project.org Subject [R] Running R Script on a Sequence 12/05/2008 01:01 of Files PM EST Hi, I have about 900 files that I need to run the same R script on. I looked over the R Data Import/Export Manual and couldn't come up with a way to read in a sequence of files. The files all have unique names and are in the same directory. What I want to do is: 1) Create a list of the file names in the directory (this is really what I need help with) 2) For each item in the list... a) open the file with read.table b) perform some analysis c) append some results to an array or save them to another file 3) Next File My initial instinct is to use Python to rename all the files with numbers 1:900 and then read them all, but the file names contain some information that I would like to keep intact and having to keep a separate database of original names and numbers seems inefficient. Is there a way to have R read all the files in a directory one at a time? - Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding rows as arithmatic calculation on original rows
Dear R users, Suppose I have the following data.frame: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 c Double 22 33 44 d Double4 6 8 and I want to have new records: myID myType myNum1 myNum2 myNum3 e Single 12.5 18 23.5 f Double 13 19.5 28 where record e got its myNum1-3 as the average from record a and b, and record f got its myNum1-3 as the average from record c and d. and the final data.frame should be like the following: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 e Single 12.5 18 43.5 c Double 22 33 44 d Double4 6 8 fDouble 13 19.5 28 Any idea is appreciated. Thanks beforehand. Ferry [[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 calculate the distance between two density functions
In general, comparing two continuous densities is difficult because they can differ on a set of measure 0 (i.e., at a single point) and yet have the same distribution function. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Chang Jia-Ming Sent: Friday, December 05, 2008 8:00 AM To: r-help@r-project.org Subject: [R] How to calculate the distance between two density functions Dear all, I wrote the following code to calculate the density functions for two data sets, respectively. den_str -density(str_data$Similarity); den_non_str -density(nonstr_data$Similarity); However, I would like to knowing the difference between den_str and den_non_str, that is, the difference between the region under the curve of the den_str and the region under the curve of the den_non_str. How to do? Thank you for help. Jia-Ming [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random Forest weighting
Andy,
Thanks for your email.
I understand that by default, the sampsize variable will use the behavior
variable that we are classifying as the strata variable.
Then, I could set sampsize=c(no=89, yes=11). I implemented that but I got
99% classification error rate on the yes value. When I oversample on the yes
values by taking sampsize=c(50,50) I get more or less equal classification
error rate of 50% for both the yes and no values.
Is there some science to what extent we should oversample on the imbalanced
level?
Thanks for your reply and all your help.
Raghu
On 12/4/08, Liaw, Andy [EMAIL PROTECTED] wrote:
If I understand your situation correctly, you may be able to make use of
the strata and sampsize arguments in randomForest() to get bootstrap
samples that resemble the original data distribution. They allow you to
specify stratified samples using the strata variable.
Best,
Andy
From: Raghu Naik
Folks,
I have a query around weighting in Random Forest (RF). I know
that several
earlier emails in this group have raised this issue, but I
did not find an
answer to my query.
I am working on a dataset (dataset1) that consists of 4
million records that
can be reduced to a dataset (dataset2) of approximately 1500
unique records
with frequency counts that add up to the 4 million records
number as above.
Because of size issues, I cannot work with dataset1 in R and
therefore, I am
working with dataset2 .
Each record consists of whether or not a patient chose a
particular drug
based on 14 comorbidity (Yes / No) variables; I am using RF
to understand
the comorbidity drivers of drug adoption (yes/no) classification.
At full dataset level (dataset1), the drug adoption incidence
is ~11%. At
the reduced dataset dataset2 level, the drug adoption
incidence increases to
~38%.
My question is that, if am using the reduced dataset
(dataset2), how should
I inform RF that the adoption incidence at the full dataset
level was 11%.
Should that be used as a classwt prior with
classwt=c(Yes=.11, No=.89)? My
understanding is that RF does not allow case weighting.
Or can this be handled with the sampsize arguement through
oversampling?
What proportions should one use for this (e.g., sampsize=c(Yes=100,
No=100))?
I would appreciate any feedback or pointers to any earlier
thread that I may
have overlooked.
Regards,
Raghu
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Re: [R] alternative way to replicate()
Am 03.12.2008 um 09:06 schrieb Liviu Andronic: Dear all, I'm looking for an alternative way to replicate the 2, string for an x number of times, and end up with one string containing 2, x times. I can partly achieve this using replicate(). y - rep(2,, times=3) y JFTR: replicate() is a different function from rep(). See ?rep and ?replicate. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lme4, error in mer_finalize(ans)
Using lmer() on my data results in an error. The problem, I think, is my model specification. However, lm() works ok. I recreated this error with a more simple dataset. (See code below.) # word and letter recognition data # two within factors: # word length: 4, 5, 6 letters # letter position: 1-4 (in 4-letter words), 1-5 (in 5-letter words), 1-6 (in 6-letter words) # one dependent variable: # reaction time # make artificial data length - c(rep(4,4), rep(5,5), rep(6,6)) # independent variable word length length - factor(c(rep(length, 2))) pos - c(1:4, 1:5, 1:6) # independent variable letter position pos - factor(c(rep(pos, 2))) rt - c(rnorm(15, 200, sd=10), rnorm(15, 300, sd=15)) # dependent variable reaction time df - data.frame(subj=factor(c(rep(1:2, each=15))), length=length, pos=pos, rt=rt) # to use lmer from lme4 package library(lme4) # first fit a linear model with letter position nested in word length lm(rt ~ length + length:pos, data=df) # fit a mixed effects model, with subj (participant) as random effect lmer(rt ~ length + length:pos + (1 | subj), data=df) Using lmer() results in an error: Error in mer_finalize(ans) : Downdated X'X is not positive definite, 13. I don't experience any problems using lm(). Does anyone know where things go wrong? ~ Ben Meijering __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding rows as arithmatic calculation on original rows
This should get you close:
x - read.table(textConnection(myID myType myNum1 myNum2 myNum3
+ a Single 10 11 12
+ b Single 15 25 35
+ c Double 22 33 44
+ d Double4 6 8), header=TRUE)
closeAllConnections()
y - lapply(split(x, x$myType), function(.type){
+ .means - colMeans(.type[,3:5])
+ # create the new line for the data frame
+ .df - data.frame(myID='', myType=.type$myType[1], myNum1=.means[1],
+ myNum2=.means[2], myNum3=.means[3])
+ rbind(.type, .df) # append the line to the original dataframe
+ })
do.call(rbind, y) # you can add the names your want
myID myType myNum1 myNum2 myNum3
Double.3 c Double 22.0 33.0 44.0
Double.4 d Double4.06.08.0
Double.myNum1 Double 13.0 19.5 26.0
Single.1 a Single 10.0 11.0 12.0
Single.2 b Single 15.0 25.0 35.0
Single.myNum1 Single 12.5 18.0 23.5
On Fri, Dec 5, 2008 at 3:21 PM, Ferry [EMAIL PROTECTED] wrote:
Dear R users,
Suppose I have the following data.frame:
myID myType myNum1 myNum2 myNum3
a Single 10 11 12
b Single 15 25 35
c Double 22 33 44
d Double4 6 8
and I want to have new records:
myID myType myNum1 myNum2 myNum3
e Single 12.5 18 23.5
f Double 13 19.5 28
where record e got its myNum1-3 as the average from record a and b, and
record f got its myNum1-3 as the average from record c and d.
and the final data.frame should be like the following:
myID myType myNum1 myNum2 myNum3
a Single 10 11 12
b Single 15 25 35
e Single 12.5 18 43.5
c Double 22 33 44
d Double4 6 8
fDouble 13 19.5 28
Any idea is appreciated. Thanks beforehand.
Ferry
[[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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] lme4, error in mer_finalize(ans)
On Fri, Dec 5, 2008 at 3:44 PM, B. Meijering [EMAIL PROTECTED] wrote: Using lmer() on my data results in an error. The problem, I think, is my model specification. However, lm() works ok. I recreated this error with a more simple dataset. (See code below.) # word and letter recognition data # two within factors: # word length: 4, 5, 6 letters # letter position: 1-4 (in 4-letter words), 1-5 (in 5-letter words), 1-6 (in 6-letter words) # one dependent variable: # reaction time # first fit a linear model with letter position nested in word length lm(rt ~ length + length:pos, data=df) # fit a mixed effects model, with subj (participant) as random effect lmer(rt ~ length + length:pos + (1 | subj), data=df) Using lmer() results in an error: Error in mer_finalize(ans) : Downdated X'X is not positive definite, 13. I don't experience any problems using lm(). Does anyone know where things go wrong? That, admittedly obscure, error message relates to the fixed-effects specification rt ~ length + length:pos being rank deficient. If you look at the summary of the linear model fit you will see that there are 3 coefficients that are not determined because of singularities. The lm function detects the singularities and fits a lower-rank model. The lmer function is not as sophisticated. It just detects the singularities and quits. The length and the position are confounded. xtabs(~ len + pos, df) pos len 1 2 3 4 5 6 4 2 2 2 2 0 0 5 2 2 2 2 2 0 6 2 2 2 2 2 2 (By the way, I changed the name of the length variable to len as typing length makes me expect the function called length.) Even when you remove this confounding by creating the len:pos interaction separately as a factor, you will still get singularities because there is only one len:pos combination for len = 6. You will need to think of a way of parameterizing the fixed effects without the singularities. You can check for singularities in the summary of the lm fit. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding rows as arithmatic calculation on original rows
Thanks much Jim.
On Fri, Dec 5, 2008 at 2:05 PM, jim holtman [EMAIL PROTECTED] wrote:
This should get you close:
x - read.table(textConnection(myID myType myNum1 myNum2 myNum3
+ a Single 10 11 12
+ b Single 15 25 35
+ c Double 22 33 44
+ d Double4 6 8), header=TRUE)
closeAllConnections()
y - lapply(split(x, x$myType), function(.type){
+ .means - colMeans(.type[,3:5])
+ # create the new line for the data frame
+ .df - data.frame(myID='', myType=.type$myType[1], myNum1=.means[1],
+ myNum2=.means[2], myNum3=.means[3])
+ rbind(.type, .df) # append the line to the original dataframe
+ })
do.call(rbind, y) # you can add the names your want
myID myType myNum1 myNum2 myNum3
Double.3 c Double 22.0 33.0 44.0
Double.4 d Double4.06.08.0
Double.myNum1 Double 13.0 19.5 26.0
Single.1 a Single 10.0 11.0 12.0
Single.2 b Single 15.0 25.0 35.0
Single.myNum1 Single 12.5 18.0 23.5
On Fri, Dec 5, 2008 at 3:21 PM, Ferry [EMAIL PROTECTED] wrote:
Dear R users,
Suppose I have the following data.frame:
myID myType myNum1 myNum2 myNum3
a Single 10 11 12
b Single 15 25 35
c Double 22 33 44
d Double4 6 8
and I want to have new records:
myID myType myNum1 myNum2 myNum3
e Single 12.5 18 23.5
f Double 13 19.5 28
where record e got its myNum1-3 as the average from record a and b, and
record f got its myNum1-3 as the average from record c and d.
and the final data.frame should be like the following:
myID myType myNum1 myNum2 myNum3
a Single 10 11 12
b Single 15 25 35
e Single 12.5 18 43.5
c Double 22 33 44
d Double4 6 8
fDouble 13 19.5 28
Any idea is appreciated. Thanks beforehand.
Ferry
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Cartesian Product Of Character Vectors
Rory.WINSTON at rbs.com writes: Sorry, I spoke too soon... interaction() only works for sequences of equal length. Anyone know a method that works with unequal-length vectors? Something like c(outer(vec1,vec2,paste,sep=)) ? (totally untested) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Question about lrandom effects specification in lme4
On Thu, Dec 4, 2008 at 6:21 PM, Bert Gunter [EMAIL PROTECTED] wrote: Folks: Suppose I have 3 random effects, A,B, and C. Using the older lme() function (in nlme) it was possible (using the pdMat classes) to specify that they are uncorrelated with identical variances. Is it possible to do this with lmer? My understanding is that if I specify them as lmer( y ~ ... + (A|Grp) + (B|Grp) + (C|Grp)) then they are uncorrelated but have different variances. That's correct. At present there isn't an easy way to do what you want to do but I am working on modifications that would allow such changes. You would need to define a new reCovFac (random-effects covariance factor) class. In this case it would be straightforward because the covariance factor for the combined terms is a multiple of the identity. If you are feeling adventurous contact me off-list and I will describe how I think this could be done using the code in the allcoef branch of the SVN tree at R-forge. Installing that branch is not for the faint of heart. It is seriously broken right now but I'm confident I'll get it all back together in the not-too-distant future. Motivation: I'd like to use lmer instead of lme for fitting smoothing splines to longitudinal data. Cheers, Bert Gunter Genentech __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding rows as arithmatic calculation on original rows
Here is a solution using sqldf library(sqldf) DF2 - structure(list(myID = structure(1:4, .Label = c(a, b, c, d), class = factor), myType = structure(c(2L, 2L, 1L, 1L), .Label = c(Double, Single), class = factor), myNum1 = c(10, 15, 22, 4), myNum2 = c(11, 25, 33, 6), myNum3 = c(12, 35, 44, 8)), .Names = c(myID, myType, myNum1, myNum2, myNum3), row.names = c(NA, -4L), class = data.frame) sqldf(select 1 TotalLevel, '+' myID, myType, avg(myNum1) myNum1, avg(myNum2) myNum2, avg(myNum3) myNum3 from DF2 group by myType union select 0 TotalLevel, * from DF2 order by myType, TotalLevel, myID, method = raw)[-1] The output is (display in fixed font): myID myType myNum1 myNum2 myNum3 1c Double 22.0 33.0 44.0 2d Double4.06.08.0 3+ Double 13.0 19.5 26.0 4a Single 10.0 11.0 12.0 5b Single 15.0 25.0 35.0 6+ Single 12.5 18.0 23.5 On Fri, Dec 5, 2008 at 3:21 PM, Ferry [EMAIL PROTECTED] wrote: Dear R users, Suppose I have the following data.frame: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 c Double 22 33 44 d Double4 6 8 and I want to have new records: myID myType myNum1 myNum2 myNum3 e Single 12.5 18 23.5 f Double 13 19.5 28 where record e got its myNum1-3 as the average from record a and b, and record f got its myNum1-3 as the average from record c and d. and the final data.frame should be like the following: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 e Single 12.5 18 43.5 c Double 22 33 44 d Double4 6 8 fDouble 13 19.5 28 Any idea is appreciated. Thanks beforehand. Ferry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get Greenhouse-Geisser epsilons from anova?
Skotara wrote: Dear Mr. Daalgard. thank you very much for your reply, it helped me to progress a bit. The following works fine: dd - expand.grid(C = 1:7, B= c(r, l), A= c(c, f)) myma - as.matrix(myma) #myma is a 12 by 28 list mlmfit - lm(myma~1) mlmfit0 - update(mlmfit, ~0) anova(mlmfit, mlmfit0, X= ~C+B, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of A. anova(mlmfit, mlmfit0, X= ~A+C, M = ~A+C+B, idata = dd, test=Spherical), which tests the main effect of B. However, I can not figure out how this works for the other effects. If I try: anova(mlmfit, mlmfit0, X= ~A+B, M = ~A+C+B, idata = dd, test=Spherical) I get: Fehler in function (object, ..., test = c(Pillai, Wilks, Hotelling-Lawley, : residuals have rank 1 4 dd$C is not a factor with that construction. It works for me after dd$C - factor(dd$C) (The other message is nasty, though. It's slightly different in R-patched: anova(mlmfit, mlmfit0, X= ~A+B, M = ~A+C+B, idata = dd, test=Spherical) Error in solve.default(Psi, B) : system is computationally singular: reciprocal condition number = 2.17955e-34 but it shouldn't happen... Looks like it is a failure of the internal Thin.row function. Ick! ) I also don't know how I can calculate the various interactions.. My read is I should change the second argument mlmfit0, too, but I can't figure out how... The within interactions should be straightforward, e.g. M=~A*B*C X=~A*B*C-A:B:C etc. The within/between interactions are otained from the similar tests of the between factor(s) e.g. mlmfitD - lm(myma~D) and then anova(mlmfitD, mlmfit,) Do you know what to do? Thank you very much! Peter Dalgaard schrieb: Skotara wrote: Dear all, I apologize for my basic question. I try to calculate an anova for repeated measurements with 3 factors (A,B,C) having 2, 2, and 7 levels. or with an additional fourth between subjects factor D. Everything works fine using aov(val ~ A*B*C + Error(subject/ (A*B*C) ) ) or aov(val ~ (D*A*B*C) + Error(subject/(A*B*C)) + D ) val, A, B, C, D and subject are columns in a data.frame. How can I get the estimated Greenhouse-Geisser and Huynh-Feldt epsilons? I know Peter Dalgaard described it in R-News Vol. 7/2, October 2007. However, unfortunately I am not able to apply that using my data... Why? It is supposed to work. You just need to work out the X and M specification for the relevant error strata and set test=Spherical for anova.mlm, or work out the T contrast matrix explicitly if that suits your temper better. Furthermore, I am still confused of how SPSS calculates the epsilons since it is mentioned that perhaps there are any errors in SPSS?? I would be glad if anyone could help me! I am looking forward to hearing from you! Thank you! Nils __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] adding rows as arithmatic calculation on original rows
Here is a solution using doBy and Gabor's DF2 created below: library( doBy ) newrows - summaryBy ( myNum1 + myNum2 + myNum3 ~ myType , DF2, keep.names = TRUE ) newrows[,myID] - + rbind ( DF2, newrows) - Original message - From: Gabor Grothendieck [EMAIL PROTECTED] To: Ferry [EMAIL PROTECTED] Cc: r-help@r-project.org Date: Fri, 5 Dec 2008 17:50:42 -0500 Subject: Re: [R] adding rows as arithmatic calculation on original rows Here is a solution using sqldf library(sqldf) DF2 - structure(list(myID = structure(1:4, .Label = c(a, b, c, d), class = factor), myType = structure(c(2L, 2L, 1L, 1L), .Label = c(Double, Single), class = factor), myNum1 = c(10, 15, 22, 4), myNum2 = c(11, 25, 33, 6), myNum3 = c(12, 35, 44, 8)), .Names = c(myID, myType, myNum1, myNum2, myNum3), row.names = c(NA, -4L), class = data.frame) sqldf(select 1 TotalLevel, '+' myID, myType, avg(myNum1) myNum1, avg(myNum2) myNum2, avg(myNum3) myNum3 from DF2 group by myType union select 0 TotalLevel, * from DF2 order by myType, TotalLevel, myID, method = raw)[-1] The output is (display in fixed font): myID myType myNum1 myNum2 myNum3 1c Double 22.0 33.0 44.0 2d Double4.06.08.0 3+ Double 13.0 19.5 26.0 4a Single 10.0 11.0 12.0 5b Single 15.0 25.0 35.0 6+ Single 12.5 18.0 23.5 On Fri, Dec 5, 2008 at 3:21 PM, Ferry [EMAIL PROTECTED] wrote: Dear R users, Suppose I have the following data.frame: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 c Double 22 33 44 d Double4 6 8 and I want to have new records: myID myType myNum1 myNum2 myNum3 e Single 12.5 18 23.5 f Double 13 19.5 28 where record e got its myNum1-3 as the average from record a and b, and record f got its myNum1-3 as the average from record c and d. and the final data.frame should be like the following: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 e Single 12.5 18 43.5 c Double 22 33 44 d Double4 6 8 fDouble 13 19.5 28 Any idea is appreciated. Thanks beforehand. Ferry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] adding rows as arithmatic calculation on original rows
thanks Gabor, and rmailbox. On Fri, Dec 5, 2008 at 3:32 PM, [EMAIL PROTECTED] wrote: Here is a solution using doBy and Gabor's DF2 created below: library( doBy ) newrows - summaryBy ( myNum1 + myNum2 + myNum3 ~ myType , DF2, keep.names = TRUE ) newrows[,myID] - + rbind ( DF2, newrows) - Original message - From: Gabor Grothendieck [EMAIL PROTECTED] To: Ferry [EMAIL PROTECTED] Cc: r-help@r-project.org Date: Fri, 5 Dec 2008 17:50:42 -0500 Subject: Re: [R] adding rows as arithmatic calculation on original rows Here is a solution using sqldf library(sqldf) DF2 - structure(list(myID = structure(1:4, .Label = c(a, b, c, d), class = factor), myType = structure(c(2L, 2L, 1L, 1L), .Label = c(Double, Single), class = factor), myNum1 = c(10, 15, 22, 4), myNum2 = c(11, 25, 33, 6), myNum3 = c(12, 35, 44, 8)), .Names = c(myID, myType, myNum1, myNum2, myNum3), row.names = c(NA, -4L), class = data.frame) sqldf(select 1 TotalLevel, '+' myID, myType, avg(myNum1) myNum1, avg(myNum2) myNum2, avg(myNum3) myNum3 from DF2 group by myType union select 0 TotalLevel, * from DF2 order by myType, TotalLevel, myID, method = raw)[-1] The output is (display in fixed font): myID myType myNum1 myNum2 myNum3 1c Double 22.0 33.0 44.0 2d Double4.06.08.0 3+ Double 13.0 19.5 26.0 4a Single 10.0 11.0 12.0 5b Single 15.0 25.0 35.0 6+ Single 12.5 18.0 23.5 On Fri, Dec 5, 2008 at 3:21 PM, Ferry [EMAIL PROTECTED] wrote: Dear R users, Suppose I have the following data.frame: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 c Double 22 33 44 d Double4 6 8 and I want to have new records: myID myType myNum1 myNum2 myNum3 e Single 12.5 18 23.5 f Double 13 19.5 28 where record e got its myNum1-3 as the average from record a and b, and record f got its myNum1-3 as the average from record c and d. and the final data.frame should be like the following: myID myType myNum1 myNum2 myNum3 a Single 10 11 12 b Single 15 25 35 e Single 12.5 18 43.5 c Double 22 33 44 d Double4 6 8 fDouble 13 19.5 28 Any idea is appreciated. Thanks beforehand. Ferry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] Kaplan-Meier function from survfit
Hi All,
Please pardon me if I am missing something obvious here. How do I get
the Kaplan-Meier estimate function that is created by survfit and
plotted by the code.
fit - survfit(Surv(time, status) , data=aml)
plot(fit)
That is, I need a function that will give me the survival estimate at
a given time: \hat{S}(t).
Thanks in advance.
Ritwik Sinha
[EMAIL PROTECTED] | +12033042111 | http://ritwik.sinha.googlepages.com/
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Re: [R] Kaplan-Meier function from survfit
on 12/05/2008 09:10 PM Ritwik Sinha wrote:
Hi All,
Please pardon me if I am missing something obvious here. How do I get
the Kaplan-Meier estimate function that is created by survfit and
plotted by the code.
fit - survfit(Surv(time, status) , data=aml)
plot(fit)
That is, I need a function that will give me the survival estimate at
a given time: \hat{S}(t).
Thanks in advance.
Ritwik Sinha
See:
library(survival)
?summary.survfit
and note the 'times' argument.
fit - survfit(Surv(time, status) , data = aml)
summary(fit)
Call: survfit(formula = Surv(time, status), data = aml)
time n.risk n.event survival std.err lower 95% CI upper 95% CI
5 23 2 0.9130 0.0588 0.80491.000
8 21 2 0.8261 0.0790 0.68480.996
9 19 1 0.7826 0.0860 0.63100.971
12 18 1 0.7391 0.0916 0.57980.942
13 17 1 0.6957 0.0959 0.53090.912
18 14 1 0.6460 0.1011 0.47530.878
23 13 2 0.5466 0.1073 0.37210.803
27 11 1 0.4969 0.1084 0.32400.762
30 9 1 0.4417 0.1095 0.27170.718
31 8 1 0.3865 0.1089 0.22250.671
33 7 1 0.3313 0.1064 0.17650.622
34 6 1 0.2761 0.1020 0.13380.569
43 5 1 0.2208 0.0954 0.09470.515
45 4 1 0.1656 0.0860 0.05980.458
48 2 1 0.0828 0.0727 0.01480.462
summary(fit, times = seq(0, 48, 6))
Call: survfit(formula = Surv(time, status), data = aml)
time n.risk n.event survival std.err lower 95% CI upper 95% CI
0 23 0 1. 0. 1.1.000
6 21 2 0.9130 0.0588 0.80491.000
12 18 4 0.7391 0.0916 0.57980.942
18 14 2 0.6460 0.1011 0.47530.878
24 11 2 0.5466 0.1073 0.37210.803
30 9 2 0.4417 0.1095 0.27170.718
36 5 3 0.2761 0.1020 0.13380.569
42 5 0 0.2761 0.1020 0.13380.569
48 2 3 0.0828 0.0727 0.01480.462
HTH,
Marc Schwartz
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Re: [R] Spectral Analysis of Time Series in R
You can do (1) and (2) [with some additional coding] using mvspec.R, which you can download from http://www.stat.pitt.edu/stoffer/tsa2/chap7.htm ... scroll down to the Spectral Envelope section and you'll find it there. You can look at the top part of the examples to get an idea of how to use mvspec.R ... once you have the spectral matrix estimate, you can code up the extraction of the partial coherence and so on. Alexander Schnebel wrote: Dear R Community, I am currently student at the Vienna University of Technology writing my Diploma thesis on causality in time series and doing some analyses of time series in R. I have the following questions: (1) Is there a function in R to estimate the PARTIAL spectral coherence of a multivariate time series? If yes, how does this work? Is there an test in R if the partial spectral coherence between two variables is zero? The functions I know (spectrum, etc.) only work to estimate the spectral coherence. (2) For some causality analysis I need an estimate of the inverse of the spectral density matrix of a multivariate time series. Is there any possibility in R to get this? Actually, I would be happy if I could at least get a functional estimate of the spectral density matrix. I guess this should work because R can plot the kernel density estimator of the spectral density, so it should be possible to extract the underlying function estimate. (3) Is there any possibility to do Granger Causality in R? That means fitting an VAR model and testing if some coefficients are zero. Thank you very much in advance! Best Regards, Alexander T __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - The power of accurate observation is commonly called cynicism by those who have not got it. George Bernard Shaw -- View this message in context: http://www.nabble.com/Spectral-Analysis-of-Time-Series-in-R-tp20814256p20866634.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.
