[R] Segmentation fault/buffer overflow with fix() in Fedora Core 5 from Extras repository
The Fedora Extras update of R found its way onto my systems today and I noted that fix() and edit() no longer work. There is a program crash that closes up R, but it does not leave a core file. I've tested by turning off SELinux, it had no effect. Do you see it too? What do you think? It happens on both systems I've tested. As far as I know, both of these systems are up-to-date. I restarted with R -d gdb to try to get a backtrace, but gdb says the debugging symbols have been removed and I don't see the debuginfo package on the Extras archive. I'm attaching the gdb info later, but I don't think it helps much without line numbers.. I think my next step will be to re-build R on these systems and see if the problem disappears. Right? If it still crashes, I'll make sure I have debugging symbols and give you a full backtrace. If it does not crash, I'll let you know as well Here's the session that crashes library(car) data(Chile) edit(Chile) *** buffer overflow detected ***: /usr/lib/R/bin/exec/R terminated === Backtrace: = /lib/libc.so.6(__chk_fail+0x29)[0xa8079d] /lib/libc.so.6[0xa8195d] /usr/lib/R/modules//R_X11.so[0x7c094a] /usr/lib/R/modules//R_X11.so[0x7c20dd] /usr/lib/R/modules//R_X11.so[0x7c3428] /usr/lib/R/modules//R_X11.so(RX11_dataentry+0xa25)[0x7c4b15] /usr/lib/R/lib/libR.so[0x2bf4c5] /usr/lib/R/lib/libR.so[0x1dfd26] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so[0x1b4d28] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so[0x1b1887] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so(Rf_applyClosure+0x2a7)[0x1b2f67] /usr/lib/R/lib/libR.so[0x1e146f] /usr/lib/R/lib/libR.so(Rf_usemethod+0x609)[0x1e28d9] /usr/lib/R/lib/libR.so[0x1e30ae] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so(Rf_applyClosure+0x2a7)[0x1b2f67] /usr/lib/R/lib/libR.so(Rf_eval+0x2f4)[0x1b07e4] /usr/lib/R/lib/libR.so(Rf_ReplIteration+0x311)[0x1d01b1] /usr/lib/R/lib/libR.so[0x1d03c1] /usr/lib/R/lib/libR.so(run_Rmainloop+0x60)[0x1d0710] /usr/lib/R/lib/libR.so(Rf_mainloop+0x1c)[0x1d073c] /usr/lib/R/bin/exec/R(main+0x46)[0x8048696] /lib/libc.so.6(__libc_start_main+0xc6)[0x9c41fe] /usr/lib/R/bin/exec/R[0x8048591] === Memory map: 0011-00329000 r-xp 08:05 553625 /usr/lib/R/lib/libR.so 00329000-00336000 rwxp 00219000 08:05 553625 /usr/lib/R/lib/libR.so 00336000-003cd000 rwxp 00336000 00:00 0 003cd000-003d5000 r-xp 08:05 683486 /lib/libnss_files-2.4.90.so 003d5000-003d6000 r-xp 7000 08:05 683486 /lib/libnss_files-2.4.90.so 003d7000-003f5000 r-xp 08:05 1045723 /usr/lib/R/library/grDevices/libs/grDevices.so 003f5000-003f6000 rwxp 0001d000 08:05 1045723 /usr/lib/R/library/grDevices/libs/grDevices.so 003f6000-003fc000 r-xp 08:05 1046746 /usr/lib/R/library/methods/libs/methods.so 003fc000-003fd000 rwxp 5000 08:05 1046746 /usr/lib/R/library/methods/libs/methods.so 003fd000-0040 r-xp 08:05 1050384 /usr/lib/R/library/tools/libs/tools.so 0040-00401000 rwxp 2000 08:05 1050384 /usr/lib/R/library/tools/libs/tools.so 00413000-0043d000 r-xp 08:05 553410 /usr/lib/R/lib/libRblas.so 0043d000-0043e000 rwxp 00029000 08:05 553410 /usr/lib/R/lib/libRblas.so 0043e000-004b9000 r-xp 08:05 2868184/usr/lib/libgfortran.so.1.0.0 004b9000-004ba000 rwxp 0007b000 08:05 2868184/usr/lib/libgfortran.so.1.0.0 004ba000-0050b000 r-xp 08:05 1049782 /usr/lib/R/library/stats/libs/stats.so 0050b000-0050d000 rwxp 0005 08:05 1049782 /usr/lib/R/library/stats/libs/stats.so 0051-00511000 r-xp 0051 00:00 0 [vdso] 00511000-0060a000 r-xp 08:05 2868912/usr/lib/libX11.so.6.2.0 0060a000-0060e000 rwxp 000f9000 08:05 2868912/usr/lib/libX11.so.6.2.0 00664000-0067b000 r-xp 08:05 683622 /lib/libpcre.so.0.0.1 0067b000-00692000 rwxp 00017000 08:05 683622 /lib/libpcre.so.0.0.1 007bb000-007d4000 r-xp 08:05 1050764/usr/lib/R/modules/R_X11.so 007d4000-007d5000 rwxp 00018000 08:05 1050764/usr/lib/R/modules/R_X11.so 007d5000-007e1000 rwxp 007d5000 00:00 0 00896000-008eb000 r-xp 08:05 2876525/usr/lib/libXt.so.6.0.0 008eb000-008ef000 rwxp 00054000 08:05 2876525/usr/lib/libXt.so.6.0.0 0099-009a7000 r-xp 08:05 683431 /lib/ld-2.4.90.so 009a7000-009a8000 r-xp 00017000 08:05 683431 /lib/ld-2.4.90.so 009a8000-009a9000 rwxp 00018000 08:05 683431 /lib/ld-2.4.90.so 009ab000-00acf000 r-xp 08:05 683432 /lib/libc-2.4.90.so 00acf000-00ad1000 r-xp 00124000 08:05 683432 /lib/libc-2.4.90.so 00ad1000-00ad2000 rwxp 00126000 08:05 683432 /lib/libc-2.4.90.so 00ad2000-00ad5000 rwxp 00ad2000 00:00 0 00ad7000-00afc000 r-xp 08:05 683433 /lib/libm-2.4.90.so 00afc000-00afd000 r-xp 00024000 08:05 683433 /lib/libm-2.4.90.so 00afd000-00afe000 rwxp 00025000 08:05 683433 /lib/libm-2.4.90.so 00b0-00b02000 r-xp 08:05 683435
Re: [R] Segmentation fault/buffer overflow with fix() in Fedora Core 5 from Extras repository
Is this in a UTF-8 locale? If so, this is covered by Ei-ji Nakama's posting to both R-help and R-devel yesterday: see https://stat.ethz.ch/pipermail/r-devel/2006-October/039792.html You have three choices: 1) Use a single-byte locale. 2) Compile with the standard CFLAGS and not the extra flags used by FC. 3) Use R-patched, which has this fixed. As my dept still sets Linux boxes up in en_GB and not en_GB.utf8, I am using workaround 1 and so took a while to work out what the problem might be. What is happening is that FC sets CFLAGS to something other than the R default. This enables extra checks on buffer overflow and stack-smashing, but unfortunately removes the flag -std=gnu99 that is needed to allow C99 features to be used. Those extra checks are triggered by a few places in the MBCS code that Mr Nakama contributed, and some of those were patched prior to the release of 2.4.0. AFAIK the problems are not new but the detection has got more efficient. It is very helpful to include a concise description of your environment. You only mentioned the OS in the subject line, never the architecture, exact version of R (let alone the exact RPM) nor the locale. sessionInfo() provides such information in a compact form. On Thu, 19 Oct 2006, Paul Johnson wrote: The Fedora Extras update of R found its way onto my systems today and I noted that fix() and edit() no longer work. There is a program crash that closes up R, but it does not leave a core file. I've tested by turning off SELinux, it had no effect. Do you see it too? What do you think? It happens on both systems I've tested. As far as I know, both of these systems are up-to-date. I restarted with R -d gdb to try to get a backtrace, but gdb says the debugging symbols have been removed and I don't see the debuginfo package on the Extras archive. I'm attaching the gdb info later, but I don't think it helps much without line numbers.. I think my next step will be to re-build R on these systems and see if the problem disappears. Right? If it still crashes, I'll make sure I have debugging symbols and give you a full backtrace. If it does not crash, I'll let you know as well Here's the session that crashes library(car) data(Chile) edit(Chile) *** buffer overflow detected ***: /usr/lib/R/bin/exec/R terminated === Backtrace: = /lib/libc.so.6(__chk_fail+0x29)[0xa8079d] /lib/libc.so.6[0xa8195d] /usr/lib/R/modules//R_X11.so[0x7c094a] /usr/lib/R/modules//R_X11.so[0x7c20dd] /usr/lib/R/modules//R_X11.so[0x7c3428] /usr/lib/R/modules//R_X11.so(RX11_dataentry+0xa25)[0x7c4b15] /usr/lib/R/lib/libR.so[0x2bf4c5] /usr/lib/R/lib/libR.so[0x1dfd26] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so[0x1b4d28] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so[0x1b1887] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so(Rf_applyClosure+0x2a7)[0x1b2f67] /usr/lib/R/lib/libR.so[0x1e146f] /usr/lib/R/lib/libR.so(Rf_usemethod+0x609)[0x1e28d9] /usr/lib/R/lib/libR.so[0x1e30ae] /usr/lib/R/lib/libR.so(Rf_eval+0x483)[0x1b0973] /usr/lib/R/lib/libR.so(Rf_applyClosure+0x2a7)[0x1b2f67] /usr/lib/R/lib/libR.so(Rf_eval+0x2f4)[0x1b07e4] /usr/lib/R/lib/libR.so(Rf_ReplIteration+0x311)[0x1d01b1] /usr/lib/R/lib/libR.so[0x1d03c1] /usr/lib/R/lib/libR.so(run_Rmainloop+0x60)[0x1d0710] /usr/lib/R/lib/libR.so(Rf_mainloop+0x1c)[0x1d073c] /usr/lib/R/bin/exec/R(main+0x46)[0x8048696] /lib/libc.so.6(__libc_start_main+0xc6)[0x9c41fe] /usr/lib/R/bin/exec/R[0x8048591] === Memory map: 0011-00329000 r-xp 08:05 553625 /usr/lib/R/lib/libR.so 00329000-00336000 rwxp 00219000 08:05 553625 /usr/lib/R/lib/libR.so 00336000-003cd000 rwxp 00336000 00:00 0 003cd000-003d5000 r-xp 08:05 683486 /lib/libnss_files-2.4.90.so 003d5000-003d6000 r-xp 7000 08:05 683486 /lib/libnss_files-2.4.90.so 003d7000-003f5000 r-xp 08:05 1045723 /usr/lib/R/library/grDevices/libs/grDevices.so 003f5000-003f6000 rwxp 0001d000 08:05 1045723 /usr/lib/R/library/grDevices/libs/grDevices.so 003f6000-003fc000 r-xp 08:05 1046746 /usr/lib/R/library/methods/libs/methods.so 003fc000-003fd000 rwxp 5000 08:05 1046746 /usr/lib/R/library/methods/libs/methods.so 003fd000-0040 r-xp 08:05 1050384 /usr/lib/R/library/tools/libs/tools.so 0040-00401000 rwxp 2000 08:05 1050384 /usr/lib/R/library/tools/libs/tools.so 00413000-0043d000 r-xp 08:05 553410 /usr/lib/R/lib/libRblas.so 0043d000-0043e000 rwxp 00029000 08:05 553410 /usr/lib/R/lib/libRblas.so 0043e000-004b9000 r-xp 08:05 2868184/usr/lib/libgfortran.so.1.0.0 004b9000-004ba000 rwxp 0007b000 08:05 2868184/usr/lib/libgfortran.so.1.0.0 004ba000-0050b000 r-xp 08:05 1049782 /usr/lib/R/library/stats/libs/stats.so 0050b000-0050d000 rwxp 0005 08:05 1049782 /usr/lib/R/library/stats/libs/stats.so 0051-00511000
[R] Re : CI
Get the the package fortunes in the CRAN and try
fortune(Harrell) for experience of and advance R-user. For the second
question, the binomial distribution converge to normal distribution
asymptomaticaly. You can not expect get the same result by using the two
approximations on a finite sample. Watching a football macth in a TV screen is
not the same thing that see the game in a staduim.
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Ethan Johnsons [EMAIL PROTECTED]
À : Liaw, Andy [EMAIL PROTECTED]
Cc : r-help@stat.math.ethz.ch
Envoyé le : Jeudi, 19 Octobre 2006, 5h43mn 36s
Objet : Re: [R] CI
Thx so much.
I just got into R world for my small research.
I thought that R is free so doesn't have many features, but it seems I
was wrong.
Why do these two return different values?
0.2666456 0.6133544
0.2698531 0.6213784
I think the diff is ignorable, but would ask.
ej
On 10/19/06, Liaw, Andy [EMAIL PROTECTED] wrote:
You did ask for CI of mean, so that's what you got. If you want CI for
proportion, here are two (non-bootstrap) ways:
R confint(lm(I(x == 1) ~ 1), level=.9)
5 % 95 %
(Intercept) 0.2666456 0.6133544
R binom.test(sum(x == 1), length(x), conf.level=.9)
Exact binomial test
data: sum(x == 1) and length(x)
number of successes = 11, number of trials = 25, p-value = 0.69
alternative hypothesis: true probability of success is not equal to 0.5
90 percent confidence interval:
0.2698531 0.6213784
sample estimates:
probability of success
0.44
I hope these are not HW problems?
Andy
From: Ethan Johnsons
Thank you so much for the feedback.
The random numbers are working great. I have tried
non-random numbers, and the outcome is not correct with confint.
Is there a way to compute i.e. a 90% confidence interval for
percent of 1?
i.e. where 1 = apple; 2 = orange
x
[1] 2 2 2 2 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 1 1 2 2 2
table (x)
x
1 2
11 14
x =11
confint(lm(x~1), level=0.90)
5 % 95 %
(Intercept) NaN NaN
ej
On 10/18/06, Liaw, Andy [EMAIL PROTECTED] wrote:
Here's one way:
R x - c(6,11,5,14,30,11,17,3,9,3,8,8) confint(lm(x~1), level=.9)
5 %95 %
(Intercept) 6.546834 14.2865
Andy
From: Ethan Johnsons
I have a quick question, please.
Does R have function to compute i.e. a 90% confidence
interval for
the mean for these numbers?
mean (6,11,5,14,30,11,17,3,9,3,8,8)
[1] 6
I thought pt or qt would give me the interval, but it seems not.
thx much.
ej
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Re: [R] creating bins for a plot
Jeffrey Stratford stratja at auburn.edu writes: I'm trying to plot the ratio of used versus unused bird houses (coded 1 or 0) versus a continuous environmental gradient (proportion of urban cover [purban2]) that I would like to convert into bins (0 - 0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck figuring this out. I ran a logistic regression and purban2 ends up driving the probability of a box being occupied so it would be nice to show this relationship. I'm also plotting the fitted values vs. purban2 but that's done. Check the example under predict.glm. It does not use binning, though. --- Code below added because gmane complains about too much quoted text. ## example from Venables and Ripley (2002, pp. 190-2.) ldose - rep(0:5, 2) numdead - c(1, 4, 9, 13, 18, 20, 0, 2, 6, 10, 12, 16) sex - factor(rep(c(M, F), c(6, 6))) SF - cbind(numdead, numalive=20-numdead) budworm.lg - glm(SF ~ sex*ldose, family=binomial) summary(budworm.lg) Dieter __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Nested source() errors
Pierce, Ken ken.pierce at oregonstate.edu writes: Does anyone know of any issues with nesting source() calls within multiple scripts? I have at least one script which always finds errors when I source it but runs fine when run on its own. It containd source() calls to other scripts and it seems to fail during the first nested source() command. This could happen when the last line in one of the files has no linefeed appended. But probably better take Andy's advice. Dieter __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Getting group size in a data frame
Hi all, I have a data frame with some measured values of some animals. Sometimes the measurement failed, resulting in a NA for a measurement and sometimes the animal died, resulting in NA for all measurements. I have several groups of animals. How do I find the size of each group with only alive animals? And how do I find the size of the groups for each measurement? An example: l1 - list(factor=c(24,24,24), val1=c(2, 3, NA), val2=c(4, NA, NA)) df - as.data.frame(l1) df$factor - factor(df$factor) The size of factors should be 2 and not 3. The number of measurement in val1 should be 2 and the number of measurements in val2 should be 1 Thanks in advance for any help and suggestions Ulrik [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating bins for a plot
Jeffrey,
not sure what you mean by showing the relationship between the prob and
purban2, but something like this could work:
library(gtools)
#inverse logit
fl - function(x,beta){inv.logit(beta[1]+beta[2]*x)}
# for beta fill in your intercept and the coefficient of purban2
flb-function(x,beta=c(-0.09,3.65)){fl(x,beta)}
# plot over the range of purban2 by:
plot(flb,0,1)
hth, Ingmar
On 10/19/06 12:24 AM, Jeffrey Stratford [EMAIL PROTECTED] wrote:
Hi. I'm trying to plot the ratio of used versus unused bird houses
(coded 1 or 0) versus a continuous environmental gradient (proportion of
urban cover [purban2]) that I would like to convert into bins (0 -
0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck
figuring this out. I ran a logistic regression and purban2 ends up
driving the probability of a box being occupied so it would be nice to
show this relationship. I'm also plotting the fitted values vs. purban2
but that's done. Any suggestions would be appreciated.
Many thanks,
Jeff
Data sample:
box use purbank purban2
1 1 0.003813435 0.02684564
2 1 0.04429451 0.1610738
3 1 0.04458785 0.06040268
4 1 0.06072162 0.2080537
5 0 0.6080962 0.6979866
6 1 0.6060428 0.6107383
7 1 0.3807568 0.4362416
8 0 0.3649164 0.3154362
9 0 0.3505427 0.2483221
10 0 0.3476093 0.1409396
11 0 0.3719566 0.3020134
12 1 0.09238011 0.1342282
13 0 0.08616111 0.1073826
14 0 0.07388724 0.04026845
15 1 0.07046477 0.03355705
.
.
.
Jeffrey A. Stratford, Ph.D.
Postdoctoral Associate
331 Funchess Hall
Department of Biological Sciences
Auburn University
Auburn, AL 36849
334-329-9198
FAX 334-844-9234
http://www.auburn.edu/~stratja
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Ingmar Visser
Department of Psychology, University of Amsterdam
Roetersstraat 15, 1018 WB Amsterdam
The Netherlands
http://users.fmg.uva.nl/ivisser/
tel: +31-20-5256735
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Re: [R] Adding locfit confidence intervals in trelis xyplot
Deepayan Sarkar deepayan.sarkar at gmail.com writes: On 10/18/06, juan f poyatos jpoyatos at cnio.es wrote: Dear all, I am trying to include confidence intervals in a xyplot. ... Well, panel.locfit doesn't have any options to draw confidence bands, so you'll have to write a panel function that does. Shouldn't be hard to extend panel.locfit if you know how to extract that information from a locfit object. plot.locfit has bands. I think Juan mixed trellis and standard plot documentation. x - rnorm(100) y - dnorm(x) + rnorm(100) / 5 plot(locfit(y~x), band=global) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 locfit confidence intervals in trelis xyplot
But it is *within* a trellis plot when I cannot plot the bands!
Building up on Dieter's example:
x - rnorm(100)
y - dnorm(x) + rnorm(100) / 5
plot(locfit(y~x), band=global)
Plots two nice confidence bands.
But try this
z - dnorm(y) + rnorm(100) / 3
Z - equal.count(z, number = 4, overlap = .1)
xyplot(z ~ y|Z, alpha = 1,band = global,panel = panel.locfit)
and you will not see the bands,
indeed band is completely ignored as
xyplot(z ~ y|Z, alpha = 1,band = rubbish,panel = panel.locfit)
works fine but without the bands!!
- juan
On Thu, 2006-10-19 at 07:49 +, Dieter Menne wrote:
Deepayan Sarkar deepayan.sarkar at gmail.com writes:
On 10/18/06, juan f poyatos jpoyatos at cnio.es wrote:
Dear all,
I am trying to include confidence intervals in a xyplot.
...
Well, panel.locfit doesn't have any options to draw confidence bands,
so you'll have to write a panel function that does. Shouldn't be hard
to extend panel.locfit if you know how to extract that information
from a locfit object.
plot.locfit has bands. I think Juan mixed trellis and standard plot
documentation.
x - rnorm(100)
y - dnorm(x) + rnorm(100) / 5
plot(locfit(y~x), band=global)
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Juan F. Poyatos
Structural and Computational Biology Programme
Spanish National Cancer Centre (CNIO)
Melchor Fernandez Almagro, 3/E-28029 Madrid SPAIN
Phone:+34 912 246 900/Fax: +34 912 246 980
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Re: [R] Redhat compilers and lme4 with R-2.4.0
Prof Brian Ripley schrieb: Not surprising: you have not compiled R with -std=gnu99 to allow C99 features, as recommended in the R-admin manual, and selected by default. (Comment to Doug Bates: it is probably better still to write in C90 if you can.) This seems to be a problem with the flags used to build the FC5 RPMs. I think you can circumvent it by adding PKG_CFLAGS=-std=gnu99 to lme4/src/Makevars. I am having the same problem on Fedora Core 5 with the 2.4.0 rpm from CRAN. Update from lme4 0.9975-3 to lme4 0.9975-6 failed due to exactly the same error message. I would have expected the cran rpm is compiled according to the R-admin guide...? message (the german version ;-) ): * Installing *source* package 'lme4' ... ** libs gcc -I/usr/lib/R/include -I/usr/lib/R/include -I/usr/local/include -I/usr/lib/R/library/Matrix/include -fpic -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m32 -march=i386 -mtune=generic -fasynchronous-unwind-tables -c glmer.c -o glmer.o In file included from lmer.h:4, from glmer.h:4, from glmer.c:1: lme4_utils.h: In Funktion »internal_symmetrize«: lme4_utils.h:82: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet lme4_utils.h:83: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet lme4_utils.h: In Funktion »internal_make_named«: lme4_utils.h:105: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet lme4_utils.h: In Funktion »internal_getElement«: lme4_utils.h:124: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet glmer.c: In Funktion »random_effects_deviance«: glmer.c:433: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet glmer.c: In Funktion »glmer_init«: glmer.c:583: Fehler: Anfangsdeklaration in »for«-Schleife außerhalb C99-Modus verwendet make: *** [glmer.o] Fehler 1 ERROR: compilation failed for package 'lme4' __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extract arguments from a list
Hi, I would like to know how to extract the arguments from a list : For example, I have a list of charchacter x x- c(Bentazone,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine,Atrazine desethyl,Epoxiconazol,Atrazine desethyl,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine) and I am searching for a (basic) function which would return arguments of the list x, aphabeticaly reordered like this : (Atrazine,Atrazine desethyl,Bentazone,Epoxiconazol,Fenpropimorph,Metolachlor,Simazine). Can anyone help me ? Thanks by advance Jessica [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Re : extract arguments from a list
If you want to sort see ?sort If you want to get subsample see ?subset Justin BEM Elève Ingénieur Statisticien Economiste BP 294 Yaoundé. Tél (00237)9597295. - Message d'origine De : [EMAIL PROTECTED] [EMAIL PROTECTED] À : r-help@stat.math.ethz.ch Envoyé le : Jeudi, 19 Octobre 2006, 10h13mn 46s Objet : [R] extract arguments from a list Hi, I would like to know how to extract the arguments from a list : For example, I have a list of charchacter x x- c(Bentazone,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine,Atrazine desethyl,Epoxiconazol,Atrazine desethyl,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine) and I am searching for a (basic) function which would return arguments of the list x, aphabeticaly reordered like this : (Atrazine,Atrazine desethyl,Bentazone,Epoxiconazol,Fenpropimorph,Metolachlor,Simazine). Can anyone help me ? Thanks by advance Jessica [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ___ Découvrez une nouvelle façon d'obtenir des réponses à toutes vos questions ! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] extract arguments from a list
First, that isn't what is called a list in R, it is a character vector. Second, to get what you want, try sort(unique(x)) [1] Atrazine Atrazine desethyl Bentazone [4] Epoxiconazol Fenpropimorph Metolachlor [7] Simazine On 19/10/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hi, I would like to know how to extract the arguments from a list : For example, I have a list of charchacter x x- c(Bentazone,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine,Atrazine desethyl,Epoxiconazol,Atrazine desethyl,Atrazine,Epoxiconazol,Metolachlor,Epoxiconazol,Atrazine desethyl,Fenpropimorph,Epoxiconazol,Metolachlor,Simazine) and I am searching for a (basic) function which would return arguments of the list x, aphabeticaly reordered like this : (Atrazine,Atrazine desethyl,Bentazone,Epoxiconazol,Fenpropimorph,Metolachlor,Simazine). Can anyone help me ? Thanks by advance Jessica [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Rug-like density plots on margins of figure
Hi everyone,
I'm trying to plot density curves on the axes of a plot, in a
similar way to the 'rug' function.
I have had a look at a few approaches and libraries, including:
* layout http://addictedtor.free.fr/graphiques/RGraphGallery.php?
graph=78
* fancygraph http://addictedtor.free.fr/graphiques/RGraphGallery.php?
graph=81
* ggplot - gave up since it needs R 2.3.0 and today I need to run on
less than that
* grid - probably powerful, good approach, but too much effort for a
midnight hack
* rug - since it does something similar
* lattice - it has multiple graphs, but I think they may need to be
homogeneous
* plot + lines
* plot + plot
And here is the solution I am using at the moment, based upon layout.
Can anyone suggest a better way to do this?
-Alex Brown
# some data
x1 = rnorm(100)
y1 = rnorm(100)
# setup the layout
opar = par(no.readonly=TRUE)
omar = par(mar)
l = layout(matrix(c(2,0,1,3),2),c(1,4),c(4,1))
par(cex=1,bty=n)
# first plot : main plot
par(mar=omar * c(0,0,1,1)) # just top and right margins
plot(x1, y1, axes=FALSE, main=Density Margins,
xlab=, ylab=
)
# vars
usepoly - TRUE
plott = ifelse(usepoly,n,l)
par(col=blue)
# second plot : y axis density
par(mar=omar * c(0,1,1,0))
par(xpd=NA)
yd = density(y1,from=min(y1),to=max(y1))
ydd= data.frame(x=-yd$y, y=yd$x)
plot(ydd, xaxt=n,xlim=c(0,min(ydd$x)),type= plott,bty=n,ylab=y
(density),xlab=)
if(usepoly) {
ydd=rbind(ydd, c(0,max(ydd$y)), c(0,min(ydd$y)))
polygon(ydd,col=grey,border=NA)
}
# third plot : x axis density
par(mar=omar * c(1,0,0,1))
par(xpd=NA)
xd = density(x1,from=min(y1),to=max(y1))
xdd= data.frame(x=xd$x, y=-xd$y)
plot(xdd, yaxt=n,ylim=c(0,min(xdd$y)),type= plott,xlab=x
(density),ylab=)
if(usepoly) {
xdd=rbind(xdd,c(max(xdd$x),0), c(min(xdd$x),0))
polygon(xdd,col=grey,border=NA)
}
par(opar)
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Re: [R] Redhat compilers and lme4 with R-2.4.0
On Wed, 18-Oct-2006 at 09:35AM +0100, Prof Brian Ripley wrote:
| There are no known problems with 'RedHat compilers' and lme4.
|
| You do need to have installed an up-to-date Matrix prior to lme4, and the
| command-line you show indicates that you have not done so (or that it is
| not in a library known to R_LIBS).
I did Matrix immediately before trying lme4 but because I wasn't
putting it in the default library, Dealing with that issue is
different on the two machines, but they are essentially the same
problem.
Thank you for adding some wisdom to the information errors. Fairly
obvious once you know, of course.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Labels for Points- 2 character labels?
Is this what you want?
plot(e,n)
text(e,n,labels=e_order,pos=4)
On 18/10/06, ableape [EMAIL PROTECTED] wrote:
I would like to be able to label each point in a scatter plot with
the numeric order of that point. for example, I create the following plot:
plot(e,n)
#
# now I go back and create my labels
#
for(i in 1:length(e)) {# lets say e 10
pc - as.character(e_order[i]) # e_order has an
integer array 1,4,3,2... which is the order of e
points(e15[i], n15[i],pch=pc)# this will plot
0-9 as data labels
}
#
The above works for single characters. Now if the length of e[] is
greater than 9, the character string converted by as.character will
yield more than one digit. At this point, my labeling trick fails:-(
1) is there a way to create two character labels for a plot?
2) Or even better, is there a way to do what I am trying to
do. Label each point by its numeric order?
Thanks for your help.
Renaldo
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--
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
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Re: [R] Redhat compilers and lme4 with R-2.4.0
On Thu, 19 Oct 2006, Stefan Grosse wrote: Prof Brian Ripley schrieb: Not surprising: you have not compiled R with -std=gnu99 to allow C99 features, as recommended in the R-admin manual, and selected by default. (Comment to Doug Bates: it is probably better still to write in C90 if you can.) This seems to be a problem with the flags used to build the FC5 RPMs. I think you can circumvent it by adding PKG_CFLAGS=-std=gnu99 to lme4/src/Makevars. I am having the same problem on Fedora Core 5 with the 2.4.0 rpm from CRAN. Update from lme4 0.9975-3 to lme4 0.9975-6 failed due to exactly the same error message. I would have expected the cran rpm is compiled according to the R-admin guide...? But it is not, as you can check by looking at R_HOME/etc/Makeconf. (That seems like basic homework to be done _before_ posting: I can only know that because I happen to have that RPM installed on my own machine.) My understanding is that rpmbuild is overridding the R defaults, and needs to be overridden in turn. I've copied in Martyn Plummer, who will know for sure what happens and what needs to be done about this. From offline correspondence, I gathered that the next lme4 revision will revert to C90, which would resolve this. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to get multiple Correlation Coefficients
Hi
I have used a polycor package for categorical correlation coefficients.
I run the following script. But there were no results.
Could you tell me how to correct the script?
Thanks in advance,
vars - names(sdi)
for (i in 1:length(vars)) {
for (j in 1:length(vars)) {
paste(vars[i], and , vars[j])
polychor(vars[i], vars[j])
# corr
}
}
--
Kum-Hoe Hwang, Ph.D.Phone : 82-31-250-3516Email : [EMAIL PROTECTED]
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Re: [R] spectral analysis of time series
Thomas Hoffmann [EMAIL PROTECTED] writes: Dear List-Members, I would like to draw the amplitudes of different frequencies from a time series as shown in the attached figure. Does anybody has an idea how to do it? Have a look at par(ask=T) ; example(spectrum) help(spectrum) -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] CTRL-C behaviour with RODBC on Solaris2.8
This is nothing to do with RODBC, and not reproducible on any of my systems. Investigate your unstated ODBC driver manager and driver for possible causes. Something other than R has uninstalled the signal handler for SIGINT, and a rogue shared library is the likely cause. And please learn to give credit where it is due, instead of blame where it is not. (RODBC attracts far too many careless attributions of blame, this being the second just today. Yet words of thanks are very, very thin on the ground.) On Tue, 17 Oct 2006, [EMAIL PROTECTED] wrote: After loading the RODBC package version 1.1-7, Ctrl-C changes its behaviour and is quitting R and returning to the (unix-)command prompt on the solaris2.8 platform here. Here's what happened before and after loading RODBC for (i in 1:10^5) rnorm(10) ^C library(RODBC) for (i in 1:10^5) rnorm(10) ^C bash-3.00$ platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) This version of R was built with gcc-3.3 (too old?) and ODBC_INCLUDE, ODBC_LIBS pointing to non-standard locations /quant/temp/jagat/usr/local/include, /quant/temp/jagat/usr/local/lib, respectively. Will be glad to provide further details. Any ideas on how to correct this would be greatly appreciated. Thanks -- Jagat K. Sheth Prepayment Modeling and Economics Wells Fargo Home Mortgage 7911 Forsyth Boulevard Suite 500, M5001-061 Clayton, MO 63105 Tel: (314)-726-4496 Fax: (314)-726-4483 [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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 multiple Correlation Coefficients
Oh, and of course you need to use an explicit print inside a for loop.
So, the line would be:
print( polychor(sdi[,i], sdi[,j]))
On 19/10/06, David Barron [EMAIL PROTECTED] wrote:
The problem is that in the expression polychor(vars[i], vars[j]),
vars[i] and vars[j] refer to the names of the variables, not the
variables themselves. So, use sdi[,i] and sdi[,j] instead.
On 19/10/06, Kum-Hoe Hwang [EMAIL PROTECTED] wrote:
Hi
I have used a polycor package for categorical correlation coefficients.
I run the following script. But there were no results.
Could you tell me how to correct the script?
Thanks in advance,
vars - names(sdi)
for (i in 1:length(vars)) {
for (j in 1:length(vars)) {
paste(vars[i], and , vars[j])
polychor(vars[i], vars[j])
# corr
}
}
--
Kum-Hoe Hwang, Ph.D.Phone : 82-31-250-3516Email : [EMAIL PROTECTED]
__
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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.
--
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
--
=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
__
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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] cluster in R
On Wed, 18 Oct 2006, Weiwei Shi wrote: Dear Chris: I tried to use cor+1 but it still gives me sil width 0 for average. Well, then it seems that the clustering is not that good. I don't know your data and there is no theoretical reason why it has to be positive. You should read the Kaufman and Rousseeuw book to understand the average silhouette width better. Best wishes, Christian set.seed(1000) t9 - cor(t(x), method=pearson)+1 # here i add 1 t8 - as.dist(t9) t7 - cutree(hclust(t8), 4) cluster.stats(t8, t7)$avg.silwidth [1] -0.008750826 set.seed(1000) t9 - cor(t(x), method=pearson) # here I did not add 1 t8 - as.dist(t9) t7 - cutree(hclust(t8), 4) cluster.stats(t8, t7)$avg.silwidth [1] -0.09543089 On 10/18/06, Weiwei Shi [EMAIL PROTECTED] wrote: Dear Chris: thanks for the prompt reply! You are right, dist from pearson has negatives there, which I should use cor+1 in my case (since negatively correlated genes should be considered farthest). Thanks. as to the ?cluster.stats, I double-checked it and I found I need to restart my JGR, until then the help page function starts to accept newly loaded package, like fpc for this case. sorry for the confusion, weiwei On 10/18/06, Christian Hennig [EMAIL PROTECTED] wrote: Dear Weiwei, btw, ?cluster.stats does not work on my Mac machine. version _ platform i386-apple-darwin8.6.1 arch i386 os darwin8.6.1 system i386, darwin8.6.1 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) Because I don't have access to a Mac, I can't tell you anything about this, unfortunately. I always thought that my package should work on all platforms if it passes all the standard tests for packages? (Is there anyone else who could comment on this please?) I have a sample like this dim(dd.df) [1] 142 28 and I want to cluster rows; first of all, I followed the examples for cluster.stats() by d.dd - dist(dd.df) # use Euclidean d.4 - cutree(hclust(d.dd), 4) # 4 clusters I tried cluster.stats(d.dd, d.4) # gives me some results like this: $cluster.size [1] 133 5 2 2 $avg.silwidth [1] 0.9857916 but when I tried to use pearson dist here, by visualization, i think 4 or 5 clusters are good for pearson dist, but it gave me a very bad avg.siqlwidth d.dd - as.dist(cor(t(x),method=pearson)) # is This correct? $cluster.size [1] 86 31 6 19 $avg.silwidth [1] -0.09543089 cor can give negative values, which doesn't fit the usual definition of a distance. I don't know what as.dist does in this case, but I think that, depending on your application, you should rather use the absolute value of the correlation, or 1+cor. btw, what's $seperation? where can I find the detailed explanation on the output from cluster.stats? This is documented on the cluster.stats help page: separation: vector of clusterwise minimum distances of a point in the cluster to a point of another cluster. Best regards, Christian *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 [EMAIL PROTECTED], www.homepages.ucl.ac.uk/~ucakche -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 [EMAIL PROTECTED], www.homepages.ucl.ac.uk/~ucakche __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Getting group size in a data frame
I can't quite tell what you are looking for, but try the following: measurecols = c(val1,val2) df2 - df[!apply(is.na(df[1:nrow(df),measuredcols]),1,all),] to remove rows which have no measurements in. a simple count of the rows (nrow) will then give you the number of animals that didn't die, and then table(df2$factor) will tell you how many per group didn't die table(df2$factor[!is.na(df2$val1)] and names(measurecols) = measurecols lapply(measurecols, function(x)table(df2$factor[!is.na(df2[,x])])) will tell you for each measurement, how many of each group you got. -Alex On 19 Oct 2006, at 08:37, Ulrik Stervbo wrote: Hi all, I have a data frame with some measured values of some animals. Sometimes the measurement failed, resulting in a NA for a measurement and sometimes the animal died, resulting in NA for all measurements. I have several groups of animals. How do I find the size of each group with only alive animals? And how do I find the size of the groups for each measurement? An example: l1 - list(factor=c(24,24,24), val1=c(2, 3, NA), val2=c(4, NA, NA)) df - as.data.frame(l1) df$factor - factor(df$factor) The size of factors should be 2 and not 3. The number of measurement in val1 should be 2 and the number of measurements in val2 should be 1 Thanks in advance for any help and suggestions Ulrik [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] creating bins for a plot
I suggest you look at the functions cut and tapply. for instance: breaks = 0:40 / 40 bucket - cut(purban2, breaks) used.c = tapply(used, bucket, sum) unused.c = tapply(1 - used, bucket, sum) used.c[is.na(used.c)] = 0 unused.c[is.na(unused.c)] = 0 plot(breaks[-length(breaks)], used.c / unused.c) -Alex Brown On 18 Oct 2006, at 23:24, Jeffrey Stratford wrote: Hi. I'm trying to plot the ratio of used versus unused bird houses (coded 1 or 0) versus a continuous environmental gradient (proportion of urban cover [purban2]) that I would like to convert into bins (0 - 0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck figuring this out. I ran a logistic regression and purban2 ends up driving the probability of a box being occupied so it would be nice to show this relationship. I'm also plotting the fitted values vs. purban2 but that's done. Any suggestions would be appreciated. Many thanks, Jeff Data sample: box use purbank purban2 1 1 0.003813435 0.02684564 2 1 0.04429451 0.1610738 3 1 0.04458785 0.06040268 4 1 0.06072162 0.2080537 5 0 0.6080962 0.6979866 6 1 0.6060428 0.6107383 7 1 0.3807568 0.4362416 8 0 0.3649164 0.3154362 9 0 0.3505427 0.2483221 100 0.3476093 0.1409396 110 0.3719566 0.3020134 121 0.09238011 0.1342282 130 0.08616111 0.1073826 140 0.07388724 0.04026845 151 0.07046477 0.03355705 . . . Jeffrey A. Stratford, Ph.D. Postdoctoral Associate 331 Funchess Hall Department of Biological Sciences Auburn University Auburn, AL 36849 334-329-9198 FAX 334-844-9234 http://www.auburn.edu/~stratja __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Using the ROracle package with Oracle 10 ?
Hi, The ROracle package works very well reasonable easy R-code, the documention for ROracle is very usefull. Thank you for this contribution to R ! A tip for complex SQL: I did however find difficulties to handle complex sql statements in an Oracle script (something with handling of spaces and line feeds) This problem was solved by instead creating views in oracle to do the complex sql and then only using simple SQL statement in R e.g. select * from viewname Window binary for ROracle 10 ? Is there anyone who could help with a version of ROracle for windows which is compatible with Oracle 10 or more general just uses the default-home The available (not officially released) version only supports Oracle 9 I work in a clinical microbiology laboratory. We try to develop the Roracle functionality to retrieve current data from the laboratory information system to generate graphics for epidemiologic surveillance and controlcharts for ELISA robots. Greetings Ram Dessau, Naestved Denmark Ram Dessau Overlæge Klinisk Mikrobiologisk afdeling Storstrømmens sygehus, Næstved Ringstedgade 61 4700 Næstved tel. 5572 9000 tone 4148 el. 4141 fax. 5572 6024 BEGIN:VCARD VERSION:2.1 X-GWTYPE:USER FN:Dessau, Ram TEL;WORK:4148 ORG:;Klinisk Mikrobiologisk afd. EMAIL;WORK;PREF;NGW:[EMAIL PROTECTED] N:Dessau;Ram TITLE:Overlæge END:VCARD __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 multiple Correlation Coefficients
The problem is that in the expression polychor(vars[i], vars[j]),
vars[i] and vars[j] refer to the names of the variables, not the
variables themselves. So, use sdi[,i] and sdi[,j] instead.
On 19/10/06, Kum-Hoe Hwang [EMAIL PROTECTED] wrote:
Hi
I have used a polycor package for categorical correlation coefficients.
I run the following script. But there were no results.
Could you tell me how to correct the script?
Thanks in advance,
vars - names(sdi)
for (i in 1:length(vars)) {
for (j in 1:length(vars)) {
paste(vars[i], and , vars[j])
polychor(vars[i], vars[j])
# corr
}
}
--
Kum-Hoe Hwang, Ph.D.Phone : 82-31-250-3516Email : [EMAIL PROTECTED]
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=
David Barron
Said Business School
University of Oxford
Park End Street
Oxford OX1 1HP
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Re: [R] spectral analysis of time series
On Thu, 2006-10-19 at 11:59 +0200, Thomas Hoffmann wrote:
Dear List-Members,
I would like to draw the amplitudes of different frequencies from a time
series as shown in the attached figure. Does anybody has an idea how to
do it?
Best wishes
Thomas
Thomas,
If your data are really like those (Foram derived \delta^{18}O
measurements?) palaeoceanographic data and are irregular in time, then
the nuSpectral package may be useful. It isn't on CRAN, but the files
and a paper in JSS about the package are available here:
http://www.jstatsoft.org/index.php?vol=11
HTH
G
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Gavin Simpson [t] +44 (0)20 7679 0522
ECRC ENSIS, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
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[R] Writing a script for multiplying a progressively larger lists of items
Hi, I would like to get guidance as to how to write code in R that can deal with the following situation: v1 is a vector containing a sequence of numbers (Ex. 1:10) I want to store the sequence of numbers resulting from multiplying the first and the second element in the vector, then the product of the 1st, 2nd and 3rd, then 1st,2nd,3rd and 4th and so forth until all the elements in the vector are multiplied. Many thanks, Fabio Fabio Sánchez, MD, MSc, PhD Molecular Dermatology Department of Medicine Karolinska Institute SE-17176 Karolinska University Hospital Tel: +46 8 51772158 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] GHK Simulator for Ordered Probit
Hi, I am wondering if anybody has already written a GHK Simulator function for an ordered probit model in R? We have a system of three equations where the variables can take ordered categorial values. Thanks a million, Pensi __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Writing a script for multiplying a progressively larger listsof items
probably you're looking for ?cumprod(), e.g., cumprod(1:10) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm - Original Message - From: Fabio Sanchez [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Thursday, October 19, 2006 1:49 PM Subject: [R] Writing a script for multiplying a progressively larger listsof items Hi, I would like to get guidance as to how to write code in R that can deal with the following situation: v1 is a vector containing a sequence of numbers (Ex. 1:10) I want to store the sequence of numbers resulting from multiplying the first and the second element in the vector, then the product of the 1st, 2nd and 3rd, then 1st,2nd,3rd and 4th and so forth until all the elements in the vector are multiplied. Many thanks, Fabio Fabio Sánchez, MD, MSc, PhD Molecular Dermatology Department of Medicine Karolinska Institute SE-17176 Karolinska University Hospital Tel: +46 8 51772158 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Getting group size in a data frame [Broadcast]
Is this sort of what you want?
R aggregate(df[2:3], df[1], function(x) sum(!is.na(x)))
factor val1 val2
1 2421
Andy
From: Ulrik Stervbo
Hi all,
I have a data frame with some measured values of some
animals. Sometimes the
measurement failed, resulting in a NA for a measurement and
sometimes the
animal died, resulting in NA for all measurements.
I have several groups of animals. How do I find the size of
each group with
only alive animals? And how do I find the size of the groups for each
measurement?
An example:
l1 - list(factor=c(24,24,24), val1=c(2, 3, NA), val2=c(4, NA, NA))
df - as.data.frame(l1)
df$factor - factor(df$factor)
The size of factors should be 2 and not 3. The number of
measurement in
val1 should be 2 and the number of measurements in val2 should be 1
Thanks in advance for any help and suggestions
Ulrik
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Re: [R] CTRL-C behaviour with RODBC on Solaris2.8
The ODBC driver manager is unixODBC-2.2.11. I will continue investigating other causes and am extremely grateful for the time you took to see if the problem was reproducible on any of your systems. The lack of reproducibility is valuable information to me. Thanks very much again for the time you took to investigate and reply. Thank you very much also to the R project and all the great things it has done. I wouldn't be here without you! Best wishes, Jagat -Original Message- From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Sent: Thursday, October 19, 2006 5:23 AM To: Sheth, Jagat K Cc: [EMAIL PROTECTED] Subject: Re: CTRL-C behaviour with RODBC on Solaris2.8 This is nothing to do with RODBC, and not reproducible on any of my systems. Investigate your unstated ODBC driver manager and driver for possible causes. Something other than R has uninstalled the signal handler for SIGINT, and a rogue shared library is the likely cause. And please learn to give credit where it is due, instead of blame where it is not. (RODBC attracts far too many careless attributions of blame, this being the second just today. Yet words of thanks are very, very thin on the ground.) On Tue, 17 Oct 2006, [EMAIL PROTECTED] wrote: After loading the RODBC package version 1.1-7, Ctrl-C changes its behaviour and is quitting R and returning to the (unix-)command prompt on the solaris2.8 platform here. Here's what happened before and after loading RODBC for (i in 1:10^5) rnorm(10) ^C library(RODBC) for (i in 1:10^5) rnorm(10) ^C bash-3.00$ platform sparc-sun-solaris2.8 arch sparc os solaris2.8 system sparc, solaris2.8 status major 2 minor 3.1 year 2006 month 06 day01 svn rev38247 language R version.string Version 2.3.1 (2006-06-01) This version of R was built with gcc-3.3 (too old?) and ODBC_INCLUDE, ODBC_LIBS pointing to non-standard locations /quant/temp/jagat/usr/local/include, /quant/temp/jagat/usr/local/lib, respectively. Will be glad to provide further details. Any ideas on how to correct this would be greatly appreciated. Thanks -- Jagat K. Sheth Prepayment Modeling and Economics Wells Fargo Home Mortgage 7911 Forsyth Boulevard Suite 500, M5001-061 Clayton, MO 63105 Tel: (314)-726-4496 Fax: (314)-726-4483 [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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem Reading from .txt
I apologize that I've asked a similar question before, but being new to R I don't think I did a very good job of formating the question. I've included a text file since the date set is somewhat large. What I have is a huge string of numbers in a text file. The numbers are all separated by comma's and the groups are separated by a semicolon. What I would like to do is read each group in as a single column. I had hoped that I could just do something like a-cbind(c(big string of numbers)) However I receive several syntax errors when I do that. Included in the text document are two lines that represent the first point that error occurs in the included data set. If I run line one it works. Line two is identical I just selected one more number from the data set and it will not work. If anyone could tell me why this is happening, and possibly some way to read these large strings into a single column I would be grateful. Bill Wyatt Associate Instructor Ergonomics Graduate Student Department of Kinesiology School of Health, Physical Education, and Recreation Indiana University O:(812)856-5924 [EMAIL PROTECTED] //working line b1x-cbind(c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-132.5,-132.5,-132.5,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-133.6,-133.6,-133.05,-133.05,-131.35,-131.35,-129.7,-129.7,-126.9,-126.9,-124.7,-124.7,-121.35,-121.35,-118,-118,-114.7,-114.7,-110.8,-110.8,-106.9,-106.9,-102.45,-102.45,-98,-98,-92.95,-92.95,-87.95,-87.95,-82.95,-82.95,-77.95,-77.95,-72.35,-67.3 5,-67.35,-61.8,-61.8,-55.65)) //non-working line b1x-cbind(c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-132.5,-132.5,-132.5,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-133.6,-133.6,-133.05,-133.05,-131.35,-131.35,-129.7,-129.7,-126.9,-126.9,-124.7,-124.7,-121.35,-121.35,-118,-118,-114.7,-114.7,-110.8,-110.8,-106.9,-106.9,-102.45,-102.45,-98,-98,-92.95,-92.95,-87.95,-87.95,-82.95,-82.95,-77.95,-77.95,-72.35,-67.3 5,-67.35,-61.8,-61.8,-55.65,-55.65)) //output from non-working line b1x-cbind(c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-132.5,-132.5,-132.5,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-134.15,-133.6,-133.6,-133.05,-133.05,-131.35,-131.35,-129.7,-129.7,-126.9,-126.9,-124.7,-124.7,-121.35,-121.35,-118,-118,-114.7,-114.7,-110.8,-110.8,-106.9,-106.9,-102.45,-102.45,-98,-98,-92.95,-92.95,-87.95,-87.95,-82.95,-82.95,-77.95,-77.95,-72.35,-67 .35,-67.35,-61.8,-61.8,-55.65,-55. + 5)) Error: syntax error //one group in orginal data set //I would like to do this
Re: [R] How to get multiple Correlation Coefficients
Dear David and Kum-Hoe Hwang,
David has pointed out to me that polychor() will incorrectly return a
polychoric correlation when its arguments are length-one character vectors:
polychor(a, b)
[1] 0.1055909
Actually, the problem is more general, since polychor() will erroneously
compute a polychoric correlation in any event when the contingency table for
the data contains only one cell:
polychor(1, 2)
[1] 0.1055909
polychor(rep(a, 100), rep(b, 100))
[1] 0.1055909
I've fixed this bug so that polychor() and polyserial() report an error when
a categorical variable has just one level. I'll upload the new version of
the package to CRAN shortly.
Regards,
John
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Barron
Sent: Thursday, October 19, 2006 5:54 AM
To: Kum-Hoe Hwang; r-help
Subject: Re: [R] How to get multiple Correlation Coefficients
The problem is that in the expression polychor(vars[i],
vars[j]), vars[i] and vars[j] refer to the names of the
variables, not the variables themselves. So, use sdi[,i] and
sdi[,j] instead.
On 19/10/06, Kum-Hoe Hwang [EMAIL PROTECTED] wrote:
Hi
I have used a polycor package for categorical correlation
coefficients.
I run the following script. But there were no results.
Could you tell me how to correct the script?
Thanks in advance,
vars - names(sdi)
for (i in 1:length(vars)) {
for (j in 1:length(vars)) {
paste(vars[i], and , vars[j])
polychor(vars[i], vars[j])
# corr
}
}
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[R] Re : Writing a script for multiplying a progressively larger lists of items
multi-function(x0) {
result-rep(0,length(x0))
for ( i in 1:length(x0)) result[i]-prod(x0[1:i])
result
}
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Fabio Sanchez [EMAIL PROTECTED]
À : r-help@stat.math.ethz.ch
Envoyé le : Jeudi, 19 Octobre 2006, 12h49mn 12s
Objet : [R] Writing a script for multiplying a progressively larger lists of
items
Hi, I would like to get guidance as to how to write code in R that can
deal with the following situation:
v1 is a vector containing a sequence of numbers (Ex. 1:10)
I want to store the sequence of numbers resulting from multiplying the
first and the second element in the vector, then the product of the
1st, 2nd and 3rd, then 1st,2nd,3rd and 4th and so forth until all the
elements in the vector are multiplied.
Many thanks,
Fabio
Fabio Sánchez, MD, MSc, PhD
Molecular Dermatology
Department of Medicine
Karolinska Institute
SE-17176 Karolinska University Hospital
Tel: +46 8 51772158
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[R] out.format for chron
Dear R users, Do you know of a way to precise an out.format for a chron object that would use numbers for months and yet 4 digits for the year? I have tried out.format=c(d-m-year) (note the m instead of either mon or month) but got 27-Feb-1992. Also, the help for chron tells us how to define an out.format when we create a chron object, but how can you change the out.format of an existing chron object? Thanks in advance, Denis __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a loop-created list
Hello,
I create a certain number (n*m) matrices, that I would like to store. These
matrices have different dimensions. As far as I know, for matrices of different
dimensions, there's no alternative to a list.
Here's how I store them on the fly :
my_list - NULL
for (i in 1:n)
{for (j in (i+1):m)
{my_list - list(my_list,i_create_a_matrix(i,j)))
}
}
But the indexation of the result is horrible ! (see below a simplified version,
with no matrices created but i*10+j added to the list instead). Is there a
cleaner way to get the same result ?
Any help will be very welcomed.
Nicolas
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
[[1]][[1]][[1]][[1]]
[[1]][[1]][[1]][[1]][[1]]
[[1]][[1]][[1]][[1]][[1]][[1]]
NULL
[[1]][[1]][[1]][[1]][[1]][[2]]
[1] 12
[[1]][[1]][[1]][[1]][[2]]
[1] 13
[[1]][[1]][[1]][[2]]
[1] 14
[[1]][[1]][[2]]
[1] 23
[[1]][[2]]
[1] 24
[[2]]
[1] 34
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Re: [R] Re : CI
Great analogy!
Thanks much for the explanantion.
ej
On 10/19/06, justin bem [EMAIL PROTECTED] wrote:
Get the the package fortunes in the CRAN and try
fortune(Harrell) for experience of and advance R-user. For the second
question, the binomial distribution converge to normal distribution
asymptomaticaly. You can not expect get the same result by using the two
approximations on a finite sample. Watching a football macth in a TV screen
is not the same thing that see the game in a staduim.
Justin BEM
Elève Ingénieur Statisticien Economiste
BP 294 Yaoundé.
Tél (00237)9597295.
- Message d'origine
De : Ethan Johnsons [EMAIL PROTECTED]
À : Liaw, Andy [EMAIL PROTECTED]
Cc : r-help@stat.math.ethz.ch
Envoyé le : Jeudi, 19 Octobre 2006, 5h43mn 36s
Objet : Re: [R] CI
Thx so much.
I just got into R world for my small research.
I thought that R is free so doesn't have many features, but it seems I
was wrong.
Why do these two return different values?
0.2666456 0.6133544
0.2698531 0.6213784
I think the diff is ignorable, but would ask.
ej
On 10/19/06, Liaw, Andy [EMAIL PROTECTED] wrote:
You did ask for CI of mean, so that's what you got. If you want CI for
proportion, here are two (non-bootstrap) ways:
R confint(lm(I(x == 1) ~ 1), level=.9)
5 % 95 %
(Intercept) 0.2666456 0.6133544
R binom.test(sum(x == 1), length(x), conf.level=.9)
Exact binomial test
data: sum(x == 1) and length(x)
number of successes = 11, number of trials = 25, p-value = 0.69
alternative hypothesis: true probability of success is not equal to 0.5
90 percent confidence interval:
0.2698531 0.6213784
sample estimates:
probability of success
0.44
I hope these are not HW problems?
Andy
From: Ethan Johnsons
Thank you so much for the feedback.
The random numbers are working great. I have tried
non-random numbers, and the outcome is not correct with confint.
Is there a way to compute i.e. a 90% confidence interval for
percent of 1?
i.e. where 1 = apple; 2 = orange
x
[1] 2 2 2 2 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 1 1 2 2 2
table (x)
x
1 2
11 14
x =11
confint(lm(x~1), level=0.90)
5 % 95 %
(Intercept) NaN NaN
ej
On 10/18/06, Liaw, Andy [EMAIL PROTECTED] wrote:
Here's one way:
R x - c(6,11,5,14,30,11,17,3,9,3,8,8) confint(lm(x~1), level=.9)
5 %95 %
(Intercept) 6.546834 14.2865
Andy
From: Ethan Johnsons
I have a quick question, please.
Does R have function to compute i.e. a 90% confidence
interval for
the mean for these numbers?
mean (6,11,5,14,30,11,17,3,9,3,8,8)
[1] 6
I thought pt or qt would give me the interval, but it seems not.
thx much.
ej
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[R] Tukey-Kramer test
Hello All, I found the TukeyHSD() function. Is there a Tukey-Kramer test for unbalanced data? Andrew __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] write data to pdf
Hello! Is there a possibility in R to save data in pdf-format? I do not want to save a plot but some lines of simple text. Regards, Franco Mendolia __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] out.format for chron
As discussed the Help Desk article in R News 4/1, the 2 vs 4 year length is controlled by the chron.year.abb option, e.g. options(chron.year.abb = FALSE) chron(20) however, as also discussed there its not really recommended that you use this option so try this instead: ddmm - function( x ) with( month.day.year( x ), sprintf( %02.f-%02.f-%04.f, day, month, year) ) chron( 20, out.format = ddmm ) On 10/19/06, Denis Chabot [EMAIL PROTECTED] wrote: Dear R users, Do you know of a way to precise an out.format for a chron object that would use numbers for months and yet 4 digits for the year? I have tried out.format=c(d-m-year) (note the m instead of either mon or month) but got 27-Feb-1992. Also, the help for chron tells us how to define an out.format when we create a chron object, but how can you change the out.format of an existing chron object? Thanks in advance, Denis __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] out.format for chron
For your second question: x - chron(1) x - chron(x, out.format = ddmm) using the ddmm from below. On 10/19/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: As discussed the Help Desk article in R News 4/1, the 2 vs 4 year length is controlled by the chron.year.abb option, e.g. options(chron.year.abb = FALSE) chron(20) however, as also discussed there its not really recommended that you use this option so try this instead: ddmm - function( x ) with( month.day.year( x ), sprintf( %02.f-%02.f-%04.f, day, month, year) ) chron( 20, out.format = ddmm ) On 10/19/06, Denis Chabot [EMAIL PROTECTED] wrote: Dear R users, Do you know of a way to precise an out.format for a chron object that would use numbers for months and yet 4 digits for the year? I have tried out.format=c(d-m-year) (note the m instead of either mon or month) but got 27-Feb-1992. Also, the help for chron tells us how to define an out.format when we create a chron object, but how can you change the out.format of an existing chron object? Thanks in advance, Denis __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] write data to pdf
Give a look at Sweave library(tools) ?Sweave or RNews 2002 vol 2 number 3 and 2003 vol 3 number 2 Stefano On Thu, Oct 19, 2006 at 04:09:56PM +0200, Franco Mendolia wrote: FrancoHello! Franco FrancoIs there a possibility in R to save data in pdf-format? FrancoI do not want to save a plot but some lines of simple text. Franco FrancoRegards, Franco FrancoFranco Mendolia Franco Franco__ FrancoR-help@stat.math.ethz.ch mailing list Francohttps://stat.ethz.ch/mailman/listinfo/r-help FrancoPLEASE do read the posting guide http://www.R-project.org/posting-guide.html Francoand provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a loop-created list
__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] predict.Arima question
Hi,
I am trying to forecast a model using predict.Arima
I found arima model for a data set: x={x1,x2,x3,...,x(t)}
arima_model = arima(x,order=c(1,0,1))
I am forecasting the next N lags using predict:
arima_pred = predict(arima_model,n.ahead = N, se.fit=T)
If I have one more point in my series, let's say x(t+1). I do not want to
recalibrate themodel, I just want to forecast the next N-1 lags using the
same model for x={x1,x2,...x(t)} but without recalibrate arima.
How to do it using arima + predict.Arima ?
My problem is that I am trying to fit arima models by brute force ( trying
lots of combinations for p and q and chosing the best model by AIC and BIC )
I have a big time series and I am running calibration for some sub-sequence
and I trying to forecast some points. I repeat this process for the next
contiguous subsequence and try to forecast again, until the big series end.
Thanks
Felipe
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Re: [R] write data to pdf
One possible option is the sweave package, available on CRAN. This package allows you to generate reports including static text, inlined and pretty printed data frames, plots, etc. -Alex On 19 Oct 2006, at 15:57, Franco Mendolia wrote: Hi Alex! I'll try to explain. I am writing a tool to simulate the lifetime of special machines. When finisched with simulating I display the results in a GUI written with the tcltk-package. It looks somehow like this: Lifetime mean : 7 years variance:1 standard deviation: 1 xyz: abc .. .. .. I now would like to save the results in a pdf-file in the same format as above and add one or two plots. In R the results are written in a data frame. Franco Mendolia Original-Nachricht Von: Alex Brown [EMAIL PROTECTED] An: Franco Mendolia [EMAIL PROTECTED] Betreff: Re:[R] write data to pdf Datum: 19.10.2006 16:38 Hi Franco There are several possible answers to this. To get the best answer, it would help if you could describe what you actually want, and why. Even better, give an example of what you would like to write in R, and the output you expect. -Alex Brown On 19 Oct 2006, at 15:09, Franco Mendolia wrote: Hello! Is there a possibility in R to save data in pdf-format? I do not want to save a plot but some lines of simple text. Regards, Franco Mendolia __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a loop-created list
hi,
try this:
my_list - NULL
for (i in 1:n)
{for (j in (i+1):m)
{my_list - c(my_list,list(i_create_a_matrix(i,j)))
}
}
Best,
Romain
Romain Lorrillière
UMR 8079 Laboratoire Ecologie, Systématique et Evolution
Bât. 362
Université Paris-Sud
91405 Orsay cedex
France
tel : 01 69 15 56 85
fax : 01 69 15 56 96
mobile : 06 81 70 90 70
email : [EMAIL PROTECTED]
Nicolas Prune a écrit :
Hello,
I create a certain number (n*m) matrices, that I would like to store. These
matrices have different dimensions. As far as I know, for matrices of
different
dimensions, there's no alternative to a list.
Here's how I store them on the fly :
my_list - NULL
for (i in 1:n)
{for (j in (i+1):m)
{my_list - list(my_list,i_create_a_matrix(i,j)))
}
}
But the indexation of the result is horrible ! (see below a simplified
version,
with no matrices created but i*10+j added to the list instead). Is there a
cleaner way to get the same result ?
Any help will be very welcomed.
Nicolas
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
[[1]][[1]][[1]][[1]]
[[1]][[1]][[1]][[1]][[1]]
[[1]][[1]][[1]][[1]][[1]][[1]]
NULL
[[1]][[1]][[1]][[1]][[1]][[2]]
[1] 12
[[1]][[1]][[1]][[1]][[2]]
[1] 13
[[1]][[1]][[1]][[2]]
[1] 14
[[1]][[1]][[2]]
[1] 23
[[1]][[2]]
[1] 24
[[2]]
[1] 34
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Re: [R] write data to pdf
Franco Mendolia franco.mendolia at gmx.de writes: Is there a possibility in R to save data in pdf-format? I do not want to save a plot but some lines of simple text. I remember that about 4 years ago there was a thread on the subject, but I could not find it any more. But, maybe, it's anyway better to think the other way round, asking How can I generate nice pdf tables with R. For that, you should have a look at Sweave (latex), or package odfWeave if you prefer Open Office output. Dieter __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] binom.test
R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 sample estimates: probability of success 0.24 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] out.format for chron
Got it Gabor, Thank you very much. Denis Le 06-10-19 à 10:38, Gabor Grothendieck a écrit : For your second question: x - chron(1) x - chron(x, out.format = ddmm) using the ddmm from below. On 10/19/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: As discussed the Help Desk article in R News 4/1, the 2 vs 4 year length is controlled by the chron.year.abb option, e.g. options(chron.year.abb = FALSE) chron(20) however, as also discussed there its not really recommended that you use this option so try this instead: ddmm - function( x ) with( month.day.year( x ), sprintf( %02.f-%02.f-%04.f, day, month, year) ) chron( 20, out.format = ddmm ) On 10/19/06, Denis Chabot [EMAIL PROTECTED] wrote: Dear R users, Do you know of a way to precise an out.format for a chron object that would use numbers for months and yet 4 digits for the year? I have tried out.format=c(d-m-year) (note the m instead of either mon or month) but got 27-Feb-1992. Also, the help for chron tells us how to define an out.format when we create a chron object, but how can you change the out.format of an existing chron object? Thanks in advance, Denis __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Time conversion from Win32 64bit FILETIME?
Windows-32 has a time structure called FILETIME, a 64-bit value representing the number of 100-nanosecond intervals since January 1, 1601 (UTC). That is not a typo, the year is 1601. Does anyone have a clue(or algorhithm)for how this is converted to something a little more POSIX-like ? Thank you, Derek -- Derek N. Eder Gothenburg University VINKLA - Vigilance and Neurocognition laboratory SU/Sahlgrenska Utvecklingslab 1, Med Gröna stråket 8 SE 413 45 Göteborg (Gothenburg) Sverige (Sweden) +46 (031)* 342 8261 (28261 inom Sahlgrenska) +46 0704 915 714 (mobile) +46 (031) 25 97 07 (home) * omit the 0 when calling from outside Sweden personal web page: www.derek-eder.org __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] write data to pdf
Franco Mendolia franco.mendolia at gmx.de writes: Hello! Is there a possibility in R to save data in pdf-format? I do not want to save a plot but some lines of simple text. Regards, Franco Mendolia You could also use pdf() and textplot() in the gplots package Mark Lyman __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] write data to pdf
Franco Mendolia wrote: Hello! Is there a possibility in R to save data in pdf-format? I do not want to save a plot but some lines of simple text. Howabout: R - RSPython[1] - ReportLab[2] - PDF [1] http://www.omegahat.org/RSPython/ [2] http://www.reportlab.org/ ReportLab is a very nice PDF generator library, closer integration with R would be useful... But no I dont have time to do it :( Barry __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] binom.test
Ethan Johnsons wrote: R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 sample estimates: probability of success 0.24 You might consider binconf() in the Hmisc package too: library(Hmisc) binconf(12, 50, method=all) PointEstLowerUpper Exact 0.24 0.130610 0.381691 Wilson 0.24 0.142974 0.374127 Asymptotic 0.24 0.121621 0.358379 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] binom.test
On 19-Oct-06 Ethan Johnsons wrote: R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? What do you mean by better? For a symmetrical 2-sided exact binomial confidence interval, binom.test gives the result quickly and, to within the precision of pbinom, correctly (as I've just verified by hand!). And you can get 1-sided CIs by setting the 'alternative' option, or asymmetrical CI's by finding the two 1-sided CIs (e.g. for conf.level = 0.03 and 0.07) that you want. What do you want to improve on? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 r-12 ; n-50 1-pbinom(r-1,n, 0.14471815) [1] 0.0499 1-pbinom(r-1,n, 0.14471816) [1] 0.0501 pbinom(r,n, 0.35965569) [1] 0.0501 pbinom(r,n, 0.35965570) [1] 0.05 pbinom(r,n, 0.35965571) [1] 0.0498 sample estimates: probability of success 0.24 thx much ej Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 19-Oct-06 Time: 16:53:19 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] CI with sd
Please let me ask you another quick question. I have results for e coli, and am trying to get 95% CI with the sd (1.783956). I got the result from another tool as (1.21, 3.42). But, I like to verify it with R. What function do you use for this? e.coli=c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) sd(e.coli, na.rm = FALSE) [1] 1.783956 Sorry for the newbie question. thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a loop-created list
m - 2 n - 3 matlist - matrix(vector(list), m, n) for (i in 1:m) for (j in 1:n) matlist[[i,j]] - matrix(sample(12), 3, 4) matlist[[1,2]] ## access with double[[ subscripts __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time conversion from Win32 64bit FILETIME?
One way might be to convert the number to POSIXct by scaling it based
that POSIXct is from 1970 and your number is from 1601. So if you
subtract the difference between 1601 and 1970 then you should have a
compliant number for R:
# read your number
x - 12345678901234567890 # big number (your 64-bit time)
x.sec - x / 10^9 # convert to seconds
xBase - unclass(as.POSIXct('1601-1-1')) # your time base, relative to 1970
x.sec - x.sec - abs(xBase) # scale to 1970
x.time - structure(x.sec, class=c(POSIXt, POSIXct)) # convert to POSIXct
On 10/19/06, Derek Eder [EMAIL PROTECTED] wrote:
Windows-32 has a time structure called FILETIME, a 64-bit value
representing the number of 100-nanosecond intervals since January 1,
1601 (UTC). That is not a typo, the year is 1601.
Does anyone have a clue(or algorhithm)for how this is converted to
something a little more POSIX-like ?
Thank you,
Derek
--
Derek N. Eder
Gothenburg University
VINKLA - Vigilance and Neurocognition laboratory
SU/Sahlgrenska
Utvecklingslab 1, Med
Gröna stråket 8
SE 413 45 Göteborg (Gothenburg)
Sverige (Sweden)
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+46 0704 915 714 (mobile)
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personal web page: www.derek-eder.org
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What is the problem you are trying to solve?
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Re: [R] Time conversion from Win32 64bit FILETIME?
On Thu, 19 Oct 2006, Derek Eder wrote: Windows-32 has a time structure called FILETIME, a 64-bit value representing the number of 100-nanosecond intervals since January 1, 1601 (UTC). That is not a typo, the year is 1601. Does anyone have a clue(or algorhithm)for how this is converted to something a little more POSIX-like ? ISOdatetime(1601, 1, 1, 0, 0, UTC) + x/1e7 looks about right, although you won't manage to get the full 64-bit time into R. However, you can also use Windows system calls such as FileTimeToSystemTime to do the conversion. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] binom.test
Thx much for the feedback. It is a big help. ej On 10/19/06, Ted Harding [EMAIL PROTECTED] wrote: On 19-Oct-06 Ethan Johnsons wrote: R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? What do you mean by better? For a symmetrical 2-sided exact binomial confidence interval, binom.test gives the result quickly and, to within the precision of pbinom, correctly (as I've just verified by hand!). And you can get 1-sided CIs by setting the 'alternative' option, or asymmetrical CI's by finding the two 1-sided CIs (e.g. for conf.level = 0.03 and 0.07) that you want. What do you want to improve on? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 r-12 ; n-50 1-pbinom(r-1,n, 0.14471815) [1] 0.0499 1-pbinom(r-1,n, 0.14471816) [1] 0.0501 pbinom(r,n, 0.35965569) [1] 0.0501 pbinom(r,n, 0.35965570) [1] 0.05 pbinom(r,n, 0.35965571) [1] 0.0498 sample estimates: probability of success 0.24 thx much ej Best wishes, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 19-Oct-06 Time: 16:53:19 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] A question regarding Wireframe in Package Lattice
Hello, The following code produces a quadrilateral: q-matrix(c(1,3,1,2,3,1,2,4,2,1,4,2),nrow=4,byrow=T) qc-xyz.coords(q) wireframe(z~y*x,qc) I have 2 questions 1) How can i remove the bounding box i.e the cube encompassing the quadrilateral? 2) Is there any function to get the 2D coordinates of the quadrilateral actually used in the final plot ? I could manually calculate the 2D co-ordinates of the projection of the quad if I knew the sequnce of 3D transformations 'wireframe' performed upto the final step before plotting. Thanks Saptarshi Saptarshi Guha | [EMAIL PROTECTED] | http://www.stat.purdue.edu/~sguha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] predict.Arima question
On Thu, 19 Oct 2006, Felipe Santos wrote:
Hi,
I am trying to forecast a model using predict.Arima
I found arima model for a data set: x={x1,x2,x3,...,x(t)}
arima_model = arima(x,order=c(1,0,1))
I am forecasting the next N lags using predict:
arima_pred = predict(arima_model,n.ahead = N, se.fit=T)
If I have one more point in my series, let's say x(t+1). I do not want to
recalibrate themodel, I just want to forecast the next N-1 lags using the
same model for x={x1,x2,...x(t)} but without recalibrate arima.
How to do it using arima + predict.Arima ?
The short answer is that you cannot. However, these are built on top of a
Kalman filter implementation, and you could use the underlying C code.
It would be easier to make use of a modification of predict.arima0,
though.
[...]
--
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
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Re: [R] binom.test
You can also plot an uncertainty distribution of p, using an uninformed prior (uniform(0,1)), using beta(s+1, n-s+1) i.e. x - seq(0.091, 0.469, length=100) plot(x, dbeta(.x, shape1=13, shape2=39), xlab=x, ylab=Density, main=Uncertainy distribution for p: beta(a = 12+1, b = 50-12+1), type=l) Cheers, Francisco Dr. Francisco J. Zagmutt College of Veterinary Medicine and Biomedical Sciences Colorado State University From: Chuck Cleland [EMAIL PROTECTED] To: Ethan Johnsons [EMAIL PROTECTED] CC: r-help@stat.math.ethz.ch Subject: Re: [R] binom.test Date: Thu, 19 Oct 2006 11:27:35 -0400 Ethan Johnsons wrote: R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 sample estimates: probability of success 0.24 You might consider binconf() in the Hmisc package too: library(Hmisc) binconf(12, 50, method=all) PointEstLowerUpper Exact 0.24 0.130610 0.381691 Wilson 0.24 0.142974 0.374127 Asymptotic 0.24 0.121621 0.358379 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. _ Stay in touch with old friends and meet new ones with Windows Live Spaces __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] CI with sd
Ethan Johnsons wrote: Please let me ask you another quick question. I have results for e coli, and am trying to get 95% CI with the sd (1.783956). I got the result from another tool as (1.21, 3.42). But, I like to verify it with R. What function do you use for this? e.coli=c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) sd(e.coli, na.rm = FALSE) [1] 1.783956 Sorry for the newbie question. e.coli - c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) library(nlme) mod1 - gls(e.coli ~ 1) intervals(mod1) Approximate 95% confidence intervals Coefficients: lower est.upper (Intercept) 24.16206 25.5 26.90460 attr(,label) [1] Coefficients: Residual standard error: lower est.upper 1.204986 1.783956 3.417651 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] binom.test
See also the binconf function in the Hmisc package. Ethan Johnsons wrote: R-experts: A quick question, please. From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative hypothesis: true probability of success is not equal to 0.24 90 percent confidence interval: 0.1447182 0.3596557 sample estimates: probability of success 0.24 thx much ej -- Kevin E. Thorpe Biostatistician/Trialist, Knowledge Translation Program Assistant Professor, Department of Public Health Sciences Faculty of Medicine, University of Toronto email: [EMAIL PROTECTED] Tel: 416.946.8081 Fax: 416.946.3297 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] CI with sd
Thx so much, Chuck. R is a reallt sweet tool to use.. I can try many things with i.e. intervals (gls(e.coli ~ 1), level=0.90). It is a great help. ej On 10/19/06, Chuck Cleland [EMAIL PROTECTED] wrote: Ethan Johnsons wrote: Please let me ask you another quick question. I have results for e coli, and am trying to get 95% CI with the sd (1.783956). I got the result from another tool as (1.21, 3.42). But, I like to verify it with R. What function do you use for this? e.coli=c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) sd(e.coli, na.rm = FALSE) [1] 1.783956 Sorry for the newbie question. e.coli - c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) library(nlme) mod1 - gls(e.coli ~ 1) intervals(mod1) Approximate 95% confidence intervals Coefficients: lower est.upper (Intercept) 24.16206 25.5 26.90460 attr(,label) [1] Coefficients: Residual standard error: lower est.upper 1.204986 1.783956 3.417651 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Question about random sampling in R
Hi, I looked up the help file on sample(), but didn't find the info I was looking for. When sample() is used to resample from a distribution, e.g., bootstrap, how does it do it? Does it use an uniform distribution, e.g., runif(), or something else? And, when the help file says:sample(x) generates a random permutation of the elements of x (or 1:x), would I be correct if I translate the statement as follows: it means that the order of sequence, which was generated from a uniform distribution, would look like a random normal distribution. Thanks, Tom [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Indexing a loop-created list
Romain Lorrilliere wrote:
try this:
my_list - NULL
for (i in 1:n)
{for (j in (i+1):m)
{my_list - c(my_list,list(i_create_a_matrix(i,j)))
}
}
Why it does not work as expected in this case?
my_list - NULL
for (i in 1:10)
my_list - c(my_list, function(x) x^i)
f - my_list[[2]]
f(10)
[1] 1e+10
Alberto Monteiro
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Re: [R] Problem Reading from .txt
Bill, It looks like you are having from some source a hard return in your string. When that hard return is in the middle of a number, R sees this as two numbers, not separated by a comma (,) and throws an error. Like this: c(13,12 + .3) Error: syntax error in: c(13,12 .3 When the hard return is next to a comma (,) it is valid R, so it passes. In your case it might be around 990th character that the hard occurs. However, the txt file you added had some additional hard returns so my observations might be wrong. Kees On Thursday 19 October 2006 15:06, Bill Wyatt wrote: I apologize that I've asked a similar question before, but being new to R I don't think I did a very good job of formating the question. I've included a text file since the date set is somewhat large. What I have is a huge string of numbers in a text file. The numbers are all separated by comma's and the groups are separated by a semicolon. What I would like to do is read each group in as a single column. I had hoped that I could just do something like a-cbind(c(big string of numbers)) However I receive several syntax errors when I do that. Included in the text document are two lines that represent the first point that error occurs in the included data set. If I run line one it works. Line two is identical I just selected one more number from the data set and it will not work. If anyone could tell me why this is happening, and possibly some way to read these large strings into a single column I would be grateful. Bill Wyatt Associate Instructor Ergonomics Graduate Student Department of Kinesiology School of Health, Physical Education, and Recreation Indiana University O:(812)856-5924 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] CI with sd
I thought of another interesting scenario, and I would just go ahead and ask this. I have a mean (i.e. 26.2) where the number of e coli samples is 250. The mean is already calculated, so it is a discrete value, not series. What fucntion can be used for the calc with CI i.e. 90%? thx much, ej On 10/19/06, Chuck Cleland [EMAIL PROTECTED] wrote: Ethan Johnsons wrote: Please let me ask you another quick question. I have results for e coli, and am trying to get 95% CI with the sd (1.783956). I got the result from another tool as (1.21, 3.42). But, I like to verify it with R. What function do you use for this? e.coli=c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) sd(e.coli, na.rm = FALSE) [1] 1.783956 Sorry for the newbie question. e.coli - c(27.5,24.6,25.3,28.7,23,26.8,24.7,24.3,24.9) library(nlme) mod1 - gls(e.coli ~ 1) intervals(mod1) Approximate 95% confidence intervals Coefficients: lower est.upper (Intercept) 24.16206 25.5 26.90460 attr(,label) [1] Coefficients: Residual standard error: lower est.upper 1.204986 1.783956 3.417651 thx much ej __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time conversion from Win32 64bit FILETIME?
Derek Eder wrote: Windows-32 has a time structure called FILETIME, a 64-bit value representing the number of 100-nanosecond intervals since January 1, 1601 (UTC). That is not a typo, the year is 1601. It could be worse. VAX/VMS used an internal time that was the number of seconds since 1858-11-17 multiplied by 10,000,000 - I don't know what they were smoking when they came to this number. Does anyone have a clue(or algorhithm)for how this is converted to something a little more POSIX-like ? I would try to get the meaningful FILETIME for some recent date [like 2001-01-01] and then do a simple arithmetic. The real problem is that there is no _unique_ definition of 1601-01-01, because the Gregorian reform of the calendar was established some years before that, but Redmond didn't change to Gregorian until some date after that. So, there's no way to know if 1601-01-01 is Gregorian or Julian calendar. If it's Gregorian, then a simple calculation gives that 2001-01-01 would correspond to FILETIME = ((365 * 400) + 100 - 3) * 24 * 60 * 60 * (1e9 / 100) where: 365 * 400 = number of normal (non-Feb-29) days 100 = number of _Julian_ calendar Feb-29 days -3 = take this off, because 1700, 1800 and 1900 were not leap years And now I think the problem is much simpler. Alberto Monteiro __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] bi-directional sockets
Hello R-helpers! I am new to R, but having a rough time with the socketConnection function. I cannot seem to get bi-directional communication to work. I have tried loads of possible ways, based on the manual's examples, but the result is always one or the other process hanging. Could anyone give me a working example of R code that: - creates a socket, - listens for data , - reads the data, - writes a reply. Without hanging or causing the sending process to hang. I would greatly appreciate any help. thanks, gf: http://gratefulfrog.net [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 locfit confidence intervals in trelis xyplot
On 10/19/06, juan f poyatos [EMAIL PROTECTED] wrote:
But it is *within* a trellis plot when I cannot plot the bands!
Building up on Dieter's example:
x - rnorm(100)
y - dnorm(x) + rnorm(100) / 5
plot(locfit(y~x), band=global)
Plots two nice confidence bands.
But try this
z - dnorm(y) + rnorm(100) / 3
Z - equal.count(z, number = 4, overlap = .1)
xyplot(z ~ y|Z, alpha = 1,band = global,panel = panel.locfit)
and you will not see the bands,
indeed band is completely ignored as
xyplot(z ~ y|Z, alpha = 1,band = rubbish,panel = panel.locfit)
works fine but without the bands!!
And why does that come as a surprise to you?
Since you seem to have missed the point of the previous replies, let
me spell it out for you. A function does what it is instructed to do,
not what you _want_ it to do. The most reliable way to figure out what
those instructions are is to look at the source code of that function.
Since this is not very user friendly, R also requires that all
functions be documented, and specifically, that the documentation for
any function, say foo, can be accessed by typing help(foo). Since you
are using the function panel.locfit, you are supposed to look at
help(panel.locfit), which does not claim to have any support for
bands.
In this case, the source of panel.locfit is actually simpler:
panel.locfit
function (x, y, subscripts, z, xyz.labs, xyz.axes, xyz.mid, xyz.minmax,
xyz.range, col.regions, at, drape, contour, region, groups,
...)
{
panel.xyplot(x, y, ...)
args - list(x = x, y = y, ...)
ok - names(formals(locfit.raw))
llines.locfit(do.call(locfit.raw, args[ok[ok %in% names(args)]]))
}
(removing the irrelevant part)
-Deepayan
- juan
On Thu, 2006-10-19 at 07:49 +, Dieter Menne wrote:
Deepayan Sarkar deepayan.sarkar at gmail.com writes:
On 10/18/06, juan f poyatos jpoyatos at cnio.es wrote:
Dear all,
I am trying to include confidence intervals in a xyplot.
...
Well, panel.locfit doesn't have any options to draw confidence bands,
so you'll have to write a panel function that does. Shouldn't be hard
to extend panel.locfit if you know how to extract that information
from a locfit object.
plot.locfit has bands. I think Juan mixed trellis and standard plot
documentation.
x - rnorm(100)
y - dnorm(x) + rnorm(100) / 5
plot(locfit(y~x), band=global)
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Juan F. Poyatos
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Spanish National Cancer Centre (CNIO)
Melchor Fernandez Almagro, 3/E-28029 Madrid SPAIN
Phone:+34 912 246 900/Fax: +34 912 246 980
http://bioinfo.cnio.es/~jpoyatos/
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Re: [R] Question about random sampling in R
When sampling with replacement (like ordinary bootstrap), each draw is
done independently, and in each draw every point has equal probability
of being drawn. When sampling without replacement (random permutation),
all possible sequences (permutations) have equal probability of
occurring. E.g., if the data is 1:2, then (1, 2) has the same
probability of occurring as (2, 1).
Andy
From: tom soyer
Hi,
I looked up the help file on sample(), but didn't find the
info I was looking for.
When sample() is used to resample from a distribution, e.g.,
bootstrap, how does it do it? Does it use an uniform
distribution, e.g., runif(), or something else? And, when the
help file says:sample(x) generates a random permutation of
the elements of x (or 1:x), would I be correct if I
translate the statement as follows: it means that the order
of sequence, which was generated from a uniform distribution,
would look like a random normal distribution.
Thanks,
Tom
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Re: [R] bi-directional sockets
Check out these packages: - Ryacas (on omegahat) -- sockets between R and yacas - mimR (on CRAN) -- sockets between R and mim (Windows only) On 10/19/06, Grateful Frog [EMAIL PROTECTED] wrote: Hello R-helpers! I am new to R, but having a rough time with the socketConnection function. I cannot seem to get bi-directional communication to work. I have tried loads of possible ways, based on the manual's examples, but the result is always one or the other process hanging. Could anyone give me a working example of R code that: - creates a socket, - listens for data , - reads the data, - writes a reply. Without hanging or causing the sending process to hang. I would greatly appreciate any help. thanks, gf: http://gratefulfrog.net [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 random sampling in R
On Thu, 2006-10-19 at 12:07 -0500, tom soyer wrote: Hi, I looked up the help file on sample(), but didn't find the info I was looking for. When sample() is used to resample from a distribution, e.g., bootstrap, how does it do it? Does it use an uniform distribution, e.g., runif(), or something else? And, when the help file says:sample(x) generates a random permutation of the elements of x (or 1:x), would I be correct if I translate the statement as follows: it means that the order of sequence, which was generated from a uniform distribution, would look like a random normal distribution. Thanks, Tom In the simplest case, where you have not specified a set of probability weights, sample() uses a uniform distribution, such that each element has an equal probability of being selected. In the case of sampling WITHOUT replacement (the default), each element in the vector has an equal probability of being selected. Once selected, that element is removed from the sampling space and the process is repeated with the remaining elements until all elements have been selected. So: sample(10) [1] 3 8 5 9 7 1 4 2 10 6 yields a random permutation of 1:10. In the case of 'replace = TRUE', which is sampling WITH replacement, after an element is selected it is retained in the sampling space, thus can be selected multiple times. So: sample(10, replace = TRUE) [1] 1 4 1 8 7 8 6 7 5 9 If you specify a set of probability weights from the sampling vector, then the probability for each element in being selected is affected accordingly. In the case of bootstrapping, sampling WITH replacement is used. You might find the following post helpful in this scenario: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/67421.html If you want to investigate further, you can review the C source code for the relevant R functions in random.c in the R source tarball. The file will be in ../src/main. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] hit return key before next gaph appears
i am looping and creating plots which are coming to the screen. i am in linux and remember ( in a previopus life ) i used to use a command so that the next graph n + 1 didn't appear on the screen until i hit the return key after graph n appeared. i thought i remeber using the command unix but when i type unix at the r prompt, it gices me system. it's probably a deprecated command then but i still don't remember what i did to make the next graph not appear until i hit the return key. or maybe it was any key for that matter. thanks a lot. mark This is not an offer (or solicitation of an offer) to buy/sell the securities/instruments mentioned or an official confirmation. Morgan Stanley may deal as principal in or own or act as market maker for securities/instruments mentioned or may advise the issuers. This is not research and is not from MS Research but it may refer to a research analyst/research report. Unless indicated, these views are the author's and may differ from those of Morgan Stanley research or others in the Firm. We do not represent this is accurate or complete and we may not update this. Past performance is not indicative of future returns. For additional information, research reports and important disclosures, contact me or see https://secure.ms.com/servlet/cls. You should not use e-mail to request, authorize or effect the purchase or sale of any security or instrument, to send transfer instructions, or to effect any other transactions. We cannot guarantee that any such requests received via ! e-mail will be processed in a timely manner. This communication is solely for the addressee(s) and may contain confidential information. We do not waive confidentiality by mistransmission. Contact me if you do not wish to receive these communications. In the UK, this communication is directed in the UK to those persons who are market counterparties or intermediate customers (as defined in the UK Financial Services Authority's rules). [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] hit return key before next gaph appears
Leeds, Mark (IED) wrote: i am looping and creating plots which are coming to the screen. i am in linux and remember ( in a previopus life ) i used to use a command so that the next graph n + 1 didn't appear on the screen until i hit the return key after graph n appeared. i thought i remeber using the command unix but when i type unix at the r prompt, it gices me system. it's probably a deprecated command then but i still don't remember what i did to make the next graph not appear until i hit the return key. or maybe it was any key for that matter. thanks a lot. par(ask=TRUE) plot(rnorm(10)) mark This is not an offer (or solicitation of an offer) to buy/sell the securities/instruments mentioned or an official confirmation. Morgan Stanley may deal as principal in or own or act as market maker for securities/instruments mentioned or may advise the issuers. This is not research and is not from MS Research but it may refer to a research analyst/research report. Unless indicated, these views are the author's and may differ from those of Morgan Stanley research or others in the Firm. We do not represent this is accurate or complete and we may not update this. Past performance is not indicative of future returns. For additional information, research reports and important disclosures, contact me or see https://secure.ms.com/servlet/cls. You should not use e-mail to request, authorize or effect the purchase or sale of any security or instrument, to send transfer instructions, or to effect any other transactions. We cannot guarantee that any such requests received vi a ! e-mail will be processed in a timely manner. This communication is solely for the addressee(s) and may contain confidential information. We do not waive confidentiality by mistransmission. Contact me if you do not wish to receive these communications. In the UK, this communication is directed in the UK to those persons who are market counterparties or intermediate customers (as defined in the UK Financial Services Authority's rules). [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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. 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] hit return key before next gaph appears
On Thu, 2006-10-19 at 14:23 -0400, Leeds, Mark (IED) wrote: i am looping and creating plots which are coming to the screen. i am in linux and remember ( in a previopus life ) i used to use a command so that the next graph n + 1 didn't appear on the screen until i hit the return key after graph n appeared. i thought i remeber using the command unix but when i type unix at the r prompt, it gices me system. it's probably a deprecated command then but i still don't remember what i did to make the next graph not appear until i hit the return key. or maybe it was any key for that matter. thanks a lot. mark Mark, It is par(ask). See ?par for more information. par(ask = TRUE) plot(1:10) par(ask = FALSE) HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 random sampling in R
On 19-Oct-06 tom soyer wrote:
Hi,
I looked up the help file on sample(), but didn't find the
info I was looking for.
When sample() is used to resample from a distribution, e.g.,
bootstrap, how does it do it? Does it use an uniform distribution,
e.g., runif(), or something else?
I don't know the details of the algorithm, but since sample()
has flexible options it may be helpful to describe the effect
of sample() in different cases.
1. sample(x,r) where x is a vector of length n
In effect, the index values (1:n) of x are sampled from
without replacement (default) with a uniform probability
distribution over the available elements at all stages.
Hence, i1 is sampled from (1:n) with probability 1/n for
each possibility. Then i2 is sampled from the remainder
with probability 1/(n-1) for each, and so on until r items
(all distinct) have been sampled. If the resulting indices
are {i1,i2,...,ir} then the result is x[i1],x[i2],...,x[ir].
Thus, if some of the values in x[1],...,x[n] are equal,
you can get 2 or more items in the sample which are equal
even though the sampling is done without replacement (since
it is the indices which are sampled).
[NB I'm describing the *effect* here, not saying that this
is how the algorithm operates]
2. sample(x, replace=TRUE)
Similar to [1], except that the sampled index is returned
to the pool and is available to be sampled again, so at each
stage the probability of any value being chosen is 1/n.
3. sample(x, replace=TRUE, prob=p) where p is a vector of
probability weights (which must not all be 0, and none
negative).
First, p is converted into a probability distribution
(summing to 1) (in effect by dividing by the sum).
Then an index i1 is sampled from (1:n) with probability
p[i] that i is chosen. This is repeated (with previously
sampled i's still available) until r index values have been
sampled -- i1,...,ir. The result is x[i1],...,x[ir].
4. sample(x, prob=p) [without replacement]
First p is scaled to sum to 1, then i1 is sampled as in [3].
The remaining p-values are rescaled so as to sum to 1,
and i2 is sampled from the remaining i's; and so on.
These are the essential variants of the use of sample().
runif() can be used to sample i1 from (1:n) with equal
probabilities by selecting i if runif() is = i and (i-1)
for i = 1:n.
Similarly runif() can be used to sample i1 from (1:n)
with probabilities p1,...,pn by selecting i if
p[1] + ... + p[i-1] runif() = p[1] + ... + p[i]
[LHS=0 if i=0], since the probability of this happening is p[i].
And, when the help file
says:sample(x) generates a random permutation of the elements
of x (or 1:x),
Since the default value of r (size of sample) is the length
of x, say n, sample(x) (see [1] above) will sample n elements
without replacement from the n elements of x with uniform
probabilities at each stage. In effect, n elements i1,i2,...,in
will be sampled without replacement from (1:n), giving a
random permutation of (1:n), so the result x[i1],...,x[in]
will be a random permutation of x[1],...,x[n] (though
different random permutations may look identical if there
are equal values in x[1],...,x[n]).
would I be correct if I translate the statement
as follows: it means that the order of sequence, which was
generated from a uniform distribution, would look like a
random normal distribution.
No. A normal distribution has nothing to do with it!
*Unless* the values x[1],...,x[n] already loooked like values
which had already been sampled from a normal distribution (but
were, say, in increasing order of size). Then sample(x) would
shuffle them into random order so the result could then look
like a real sample according ot eh order in which the data
came in.
Hoping this helps!
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 19-Oct-06 Time: 19:34:13
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Re: [R] Time conversion from Win32 64bit FILETIME?
Jim Holtman wrote:
One way might be to convert the number to POSIXct by scaling it based
that POSIXct is from 1970 and your number is from 1601. So if you
subtract the difference between 1601 and 1970 then you should have a
compliant number for R:
# read your number
x - 12345678901234567890 # big number (your 64-bit time)
x.sec - x / 10^9 # convert to seconds
No, it's not the number of ms, it's the number of 100ns, so
the way to convert is
x.sec - x / 1000 # 10 million - I hope I didn't place an extra 0
# BTW: is there a number format with thousand separators?
x.sec - x / 10e7 # better to write this way :-)
xBase - unclass(as.POSIXct('1601-1-1')) # your time base, relative
to 1970
x.sec - x.sec - abs(xBase) # scale to 1970
x.time - structure(x.sec, class=c(POSIXt, POSIXct)) # convert
to POSIXct
I would do it in a simpler way:
win.xp.filetime - (400 * 365 + 100 - 3) * 10e7 # 2001-01-01
x - as.Date(1601-01-01) + win.xp.filetime / 10e7
x
[1] 2001-01-01
This supposes, of course, that Windows is compliant with
the Gregorian Calendar [until a few years ago, Excel did
consider 1900 a leap year, so care must be taken here].
Alberto Monteiro
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Re: [R] Question about random sampling in R
To add to the nice explanation by Marc, you can access the source directly from the web at https://svn.r-project.org/R/trunk/src/main/random.c If you prefer to look directly in the source tarball, notice the file is called random.c Francisco Dr. Francisco J. Zagmutt College of Veterinary Medicine and Biomedical Sciences Colorado State University From: Marc Schwartz [EMAIL PROTECTED] Reply-To: [EMAIL PROTECTED] To: tom soyer [EMAIL PROTECTED] CC: r-help r-help@stat.math.ethz.ch Subject: Re: [R] Question about random sampling in R Date: Thu, 19 Oct 2006 13:10:20 -0500 On Thu, 2006-10-19 at 12:07 -0500, tom soyer wrote: Hi, I looked up the help file on sample(), but didn't find the info I was looking for. When sample() is used to resample from a distribution, e.g., bootstrap, how does it do it? Does it use an uniform distribution, e.g., runif(), or something else? And, when the help file says:sample(x) generates a random permutation of the elements of x (or 1:x), would I be correct if I translate the statement as follows: it means that the order of sequence, which was generated from a uniform distribution, would look like a random normal distribution. Thanks, Tom In the simplest case, where you have not specified a set of probability weights, sample() uses a uniform distribution, such that each element has an equal probability of being selected. In the case of sampling WITHOUT replacement (the default), each element in the vector has an equal probability of being selected. Once selected, that element is removed from the sampling space and the process is repeated with the remaining elements until all elements have been selected. So: sample(10) [1] 3 8 5 9 7 1 4 2 10 6 yields a random permutation of 1:10. In the case of 'replace = TRUE', which is sampling WITH replacement, after an element is selected it is retained in the sampling space, thus can be selected multiple times. So: sample(10, replace = TRUE) [1] 1 4 1 8 7 8 6 7 5 9 If you specify a set of probability weights from the sampling vector, then the probability for each element in being selected is affected accordingly. In the case of bootstrapping, sampling WITH replacement is used. You might find the following post helpful in this scenario: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/67421.html If you want to investigate further, you can review the C source code for the relevant R functions in random.c in the R source tarball. The file will be in ../src/main. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] A question regarding Wireframe in Package Lattice
On 10/19/06, Saptarshi Guha [EMAIL PROTECTED] wrote: Hello, The following code produces a quadrilateral: q-matrix(c(1,3,1,2,3,1,2,4,2,1,4,2),nrow=4,byrow=T) qc-xyz.coords(q) wireframe(z~y*x,qc) I have 2 questions 1) How can i remove the bounding box i.e the cube encompassing the quadrilateral? wireframe(z~y*x, qc, par.settings = list(box.3d = list(col = transparent)) Add scales = list(col = transparent) or scales = list(draw = FALSE) to remove the arrows/ticks as well. One warning though: I eventually plan to separate box.3d into two sets, one for the lines in front, one for the `hidden' ones. Hopefully, by that time there will be better documentation for settings. 2) Is there any function to get the 2D coordinates of the quadrilateral actually used in the final plot ? I could manually calculate the 2D co-ordinates of the projection of the quad if I knew the sequnce of 3D transformations 'wireframe' performed upto the final step before plotting. There are two things you need to know: (1) The range of the bounding box in the 3D space is not the range of the data. Usually, it is [-0.5, 0.5]^3, but that can change if aspect is not c(1, 1). The data are scaled to fit into this box (actually, xlim, ylim and zlim are scaled to fit into this box). This is mostly to ensure that the origin is in the middle of the box, which makes the projection calculations simpler. Inside 'panel.3d.wireframe' (see ?panel.3dwire), where you should be doing any additional calculations, these limits are available as the [xyz]lim.scaled arguments. (2) Modulo the above, what you want is described in ?utilities.3d. Extending your example, and making use of the fact that you are using the defaults for most arguments, we can do: sq - scale(q, center = apply(q, 2, function(x) mean(range(x))), scale = apply(q, 2, function(x) diff(range(x rot.mat - ltransform3dMatrix(screen = list(z = 40, x = -60)) ans - ltransform3dto3d(t(sq), rot.mat, dist = 0.2) rot.mat ans wireframe(z~x*y, qc, par.settings = list(box.3d = list(col = transparent)), scales = list(draw = FALSE), xlab = , ylab = , zlab = ) trellis.focus(panel, 1, 1) panel.points(ans[1, ], ans[2, ]) Note that in the wireframe call I've changed z~y*x to the more natural z~x*y. Otherwise some reordering of the columns of sq would be involved. -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] hit return key before next gaph appears
On Thu, Oct 19, 2006 at 02:28:01PM -0400, Chuck Cleland wrote:
Leeds, Mark (IED) wrote:
i am looping and creating plots which are coming to the screen. i am in
linux and remember ( in a previopus life ) i used to use a command so
that the next graph n + 1 didn't appear on the screen until
i hit the return key after graph n appeared. i thought i remeber using
the command unix but when i type unix at the r prompt, it gices me
system. it's probably a deprecated
command then but i still don't remember what i did to make the next
graph not appear until i hit the return key. or maybe it was any key for
that matter. thanks a lot.
par(ask=TRUE)
plot(rnorm(10))
As a somewhat more advanced variant, it's also possible to manually
engineer the waiting for the return key. I sometimes do this in loops,
as in
hitReturn - function(msg)
{
invisible(readline(sprintf(%s -- hit return, msg)));
}
plotAllColumns - function(dframe, waitFunc = hitReturn)
{
for (n in colnames(dframe))
{
plot(dframe[[n]]);
waitFunc(n);
}
}
# demo:
dframe - data.frame(x = runif(10), y = rnorm(10), z = rexp(10));
plotAllColumns(dframe);
I find this useful to keep track of what the current plot actually is
displaying. Furthermore, it's possible to run a simplistic kind of
animation, as in
plotAllColumns(dframe, function(msg) { print(msg); Sys.sleep(3); });
Best regards, Jan
--
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| email: [EMAIL PROTECTED] |
| WWW: http://www.cmp.uea.ac.uk/people/jtk |
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[R] [R-pkgs] New package CreditMetrics
Dear R useRs, A new package 'CreditMetrics' is now available on CRAN. It is mainly a set of functions for computing the CreditMetrics risk model. This is the first version of the package and it is also my first try to build a package for R. The canonical reference is: Glasserman, Paul, Monte Carlo Methods in Financial Engineering, Springer 2004 Suggestions, bug reports and other comments are very welcome. enjoy and best regards Andreas ___ R-packages mailing list R-packages@stat.math.ethz.ch https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] incorrect result for Matrix() %*% Diagonal()
I'm exploring pedigree() in lme4 and found some strange results trying
to multiply a dtCMatrix by a diag() matrix using %*%. Studying the
Matrix documentation, I stumbled upon the following error in the
example on the ddiMatrix-class {Matrix} help page. I wonder if my
problem could be related to this.
matrix(cbind(1, 2:4),3,2) %*% diag(c(10,1)) # OK
[,1] [,2]
[1,] 102
[2,] 103
[3,] 104
Matrix(cbind(1, 2:4)) %*% Diagonal(x=c(10,1)) # KO
3 x 2 Matrix of class dgeMatrix
[,1] [,2]
[1,] 102
[2,]1 30
[3,] 104
===
sessionInfo()
R version 2.4.0 (2006-10-03)
x86_64-pc-linux-gnu
attached base packages:
[1] methods stats graphics grDevices utils
datasets [7] base
other attached packages:
lme4 Matrixlattice
0.9975-4 0.9975-3 0.14-9
==
Thanks in advance
Gerald Jansen
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[R] how to remove the empty element in the vector
Hello, Would anyone kindly tell me how to remove the empty element in the vector object? For example, x [1] a c c c d unique(x) [1] a c d How could I get the output like: a,c,d? Thanks, Yanqin - Get your email and more, right on the new Yahoo.com [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] ROracle error in Windows. Memory could not be read.
I've seen from earlier posts that other people had problems installing ROracle under Windows. I run R-2.3.1. I got the Windows binaries for ROracle from http://stat.bell-labs.com/RS-DBI/download/index.html Here is my session: require(ROracle) Loading required package: ROracle Loading required package: DBI [1] TRUE drv - dbDriver(Oracle) drv OraDriver:(1648) con - dbConnect(drv, user=USER, password=PASS, dbname=TEST) A window pops up, with the message: The instruction at 0x6260c621 referenced memory at 0x4a280ae2. The memory could not be read. And R freezes. Something is working, because I get to establish an Oracle driver. Any ideas would be much appreciated. Thank you, Adrian __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] bi-directional sockets
Thanks for your answer, but I wasn't able to find anything that looked like an example of sockets in either of those packages. I guess I must be even dumber than I previously thought! Any help would still be welcome. On 10/19/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Check out these packages: - Ryacas (on omegahat) -- sockets between R and yacas - mimR (on CRAN) -- sockets between R and mim (Windows only) On 10/19/06, Grateful Frog [EMAIL PROTECTED] wrote: Hello R-helpers! I am new to R, but having a rough time with the socketConnection function. I cannot seem to get bi-directional communication to work. I have tried loads of possible ways, based on the manual's examples, but the result is always one or the other process hanging. Could anyone give me a working example of R code that: - creates a socket, - listens for data , - reads the data, - writes a reply. Without hanging or causing the sending process to hang. I would greatly appreciate any help. thanks, gf: http://gratefulfrog.net [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 remove the empty element in the vector
On Thu, 2006-10-19 at 12:23 -0700, Yanqin Yang wrote: Hello, Would anyone kindly tell me how to remove the empty element in the vector object? For example, x [1] a c c c d unique(x) [1] a c d How could I get the output like: a,c,d? Thanks, Yanqin It depends upon what you mean by removing the empty elements. If you want to just get the set of values that are not : x[x != ] [1] a c c c d If you want the output exactly as you have it above, which is eliminating the repeated values: unique(x[x != ]) [1] a c d See ?Extract, ?Comparison and ?Syntax for more information. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 remove the empty element in the vector
How about x[-c(2, 3)] -roger Yanqin Yang wrote: Hello, Would anyone kindly tell me how to remove the empty element in the vector object? For example, x [1] a c c c d unique(x) [1] a c d How could I get the output like: a,c,d? Thanks, Yanqin - Get your email and more, right on the new Yahoo.com [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Roger D. Peng | http://www.biostat.jhsph.edu/~rpeng/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] randomize a matrix
Hello everyone, If I have an incidence matrix of 0 and 1's P=[1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0] I want to create a new uniform random matrix [a] that is filled with 0's and 1's but constrained so that the row and column sums are the same as in [P]. Does anyone know how to accomplish this? Thanks in advance Cameron Guenther, Ph.D. Associate Research Scientist FWC/FWRI, Marine Fisheries Research 100 8th Avenue S.E. St. Petersburg, FL 33701 (727)896-8626 Ext. 4305 [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 random sampling in R
Tom Soyer wrote:
I looked up the help file on sample(), but didn't find the info I was
looking for.
When sample() is used to resample from a distribution, e.g.,
bootstrap, how does it do it? Does it use an uniform distribution,
e.g., runif(), or something else? And, when the help file
says:sample(x) generates a random permutation of the elements of x
(or 1:x), would I be correct if I translate the statement as
follows: it means that the order of sequence, which was generated
from a uniform distribution, would look like a random normal distribution.
I think it's clear that sample (without repetition) simulates
what you would get if you wrote every element in a card, shuffled
the card, and extracted a sample.
In other words, take some number n, another m = n,
let x - 1:n and then simulate y - sample(x, m). If you
do it many times, y[1] (or y[2], or y[m]) will have the
discrete distribution given by Probability(y[1] = 1) = 1/n,
Prob(y[1] = 2) = 1/n, ..., Prob(y[1] = n) = 1/n. The same,
of course, is valid for y[2], etc.
Ok, too much talking, let's run an example:
x - 1:10
y3.hist - NULL
for (i in 1:1) {
y - sample(x, 5)
y3.hist[i] - y[3]
}
hist(y3.hist)
Alberto Monteiro
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[R] Newbie: Selecting data
I've been working with R for all of about 8 hours, so anyone with more experience than this should be able to help me. General comments about my methods of work are also welcomed. I have a table that I've imported thusly: w - read.table(woodford.data, header=T) w start thermsgas KWHs elect temp days 1 10-Jun-98 9 16.84 613 63.80 75 40 2 20-Jul-98 6 15.29 721 74.21 76 29 3 18-Aug-98 7 15.73 597 62.22 76 29 4 16-Sep-98 42 35.81 460 43.98 70 33 5 19-Oct-98105 77.28 314 31.45 57 29 6 17-Nov-98106 77.01 342 33.86 48 30 snip [This is real data on my house.] 'days' is number of days in bill, 'temp' is average temperature in Fahrenheit. I'd like to see if there is a relationship between the gas burned (therms) and the number of heating degree days. I compute therms per day and heating degree days like this: thermsperday - w[,2]/w[,7] hdd - (w[,6] -65)*w[,7] However, I only want the data for the months in which the average temperature is less than 65 (otherwise, it's a cooling degree day). I tried ifelse, but couldn't get it to work. What simple technique am I overlooking? Thanks so much for your help and suggestions, especially for your patience with a newbie. -Kevin Kevin Zembower Internet Services Group manager Center for Communication Programs Bloomberg School of Public Health Johns Hopkins University 111 Market Place, Suite 310 Baltimore, Maryland 21202 410-659-6139 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] read.table
hi, how could I let the colname be the numbers instead of X plus numbers when I use read.table. Or there is an alternative way? thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] unique sets of factors
All: I have a matrix, X, with a LARGE number of rows. Consider the following three rows of that matrix: 1 1 1 1 2 2 3 3 1 1 1 1 3 3 2 2 3 3 2 2 1 1 1 1 I wish to fit many one-way ANOVAs to some response variable using each row as a set of factors. For example, for each row above I will do something like anova(lm(Y~as.factor(X[1,]))). My problem is that in the above example, I do not want to fit models for both rows 1 and 2 as they are essentially duplicates in terms of the ANOVA model. Clearly row 3, although it has the same number of 1's, 2's, and 3's, is a different model. Is there some computationally efficient way to remove such factor duplicates from my large matrix? I have been banging my head against the wall all morning. Thanks!! Tony -- ### Tony Long Ecology and Evolutionary Biology Steinhaus Hall University of California at Irvine Irvine, CA 92697-2525 Tel: (949) 824-2562 (office) Tel: (949) 824-5994 (lab) Fax: (949) 824-2181 email: [EMAIL PROTECTED] http://hjmuller.bio.uci.edu/~labhome/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] read.table
On Thu, 2006-10-19 at 16:10 -0400, Weiwei Shi wrote: hi, how could I let the colname be the numbers instead of X plus numbers when I use read.table. Or there is an alternative way? thanks Sounds like you have imported the data, perhaps using 'header = TRUE' either without an actual header row of column names or perhaps an incomplete or improper (not syntactically valid) header row. If you are getting X's (and not V's) pre-pended to the column names, that sounds like the result of make.names() on your data. See ?make.names If you don't have a header row in the imported data set, you can use the 'colnames' argument in read.table() to explicitly provide them. Just be sure to set 'header = FALSE'. If there is a header row, but it is problematic, set 'header = FALSE' AND 'skip = 1', which will skip over the first row in the incoming file. See ?read.table HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] read.table
thanks. i know how to go around it but i feel read.table should have something like that to disable the process of adding X to the header :) On 10/19/06, Marc Schwartz [EMAIL PROTECTED] wrote: On Thu, 2006-10-19 at 16:10 -0400, Weiwei Shi wrote: hi, how could I let the colname be the numbers instead of X plus numbers when I use read.table. Or there is an alternative way? thanks Sounds like you have imported the data, perhaps using 'header = TRUE' either without an actual header row of column names or perhaps an incomplete or improper (not syntactically valid) header row. If you are getting X's (and not V's) pre-pended to the column names, that sounds like the result of make.names() on your data. See ?make.names If you don't have a header row in the imported data set, you can use the 'colnames' argument in read.table() to explicitly provide them. Just be sure to set 'header = FALSE'. If there is a header row, but it is problematic, set 'header = FALSE' AND 'skip = 1', which will skip over the first row in the incoming file. See ?read.table HTH, Marc Schwartz -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] read.table
On Thu, 2006-10-19 at 17:06 -0400, Weiwei Shi wrote: thanks. i know how to go around it but i feel read.table should have something like that to disable the process of adding X to the header :) You could try setting 'check.names = FALSE' to see what you end up with in terms of column names. That will effectively disable the use of make.names() to validate/adjust the incoming column names. The risk here is unknown depending upon the root etiology of the problem. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] [Q] How to fit data to a straight line
Dear R users I have a question about how to fit data to a straight line. I tried nls to do it, but it didn't work. The reason I want to fit data to a straight line is that I need to compare AIC or BIC values of the two models (a straight line model vs a nonlinear curve model). Fitting data to a nonlinear curve is straightforward, but I could not figure out how to fit the data to a straight line and therefore, I could not get the AIC/BIC values of this simple model. If I can get the AIC/BIC value of the straight line fit without actually doing a fit, that is also fine. Thanks in advance. Young-Jin [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
