Re: [R] The error in R while using bugs.R function
On Mon, 16 May 2005, Li, Jia wrote: I followed the instuctions on Dr. Gelman's web to install all of documents that bugs.R needs, but when I try to run the school example that the web posted in R, I got an error: couldn't find function bugs, what's wrong? I suggest you ask Dr Gelman. There is a function bugs in package R2WinBUGS: perhaps you don't have that installed or that package loaded? -- 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
RE: [R] NA erase your data trick
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Anders Schwartz Corr
Sent: Monday, May 16, 2005 10:38 PM
To: r-help@stat.math.ethz.ch
Subject: [R] NA erase your data trick
Oops,
I just erased all my data using this gizmo that I thought would replace -9
with NA.
A) Can I get my tcn5 back?
I don't think there is any going back.
B) How do I do it right next time, I learned my lesson, I'll never do it
again, I promise!
How about something like
x[x == -9] - NA
Dan Nordlund
Bothell, WA
Anders Corr
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
+
+ }
+ }
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Re: [R] The error in R while using bugs.R function
Li, Jia wrote: Dear R users, I followed the instuctions on Dr. Gelman's web to install all of documents that bugs.R needs, but when I try to run the school example that the web posted in R, I got an error: couldn't find function bugs, what's wrong? It sounds as though you missed an instruction, or he did. I'm guessing you didn't run library() to load the package. Generally when a contributed package doesn't work, you should ask the maintainer for help. Duncan Murdoch __ 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
Re: [R] NA erase your data trick
Anders Schwartz Corr wrote:
Oops,
I just erased all my data using this gizmo that I thought would replace -9
with NA.
A) Can I get my tcn5 back?
As you got it the first time. There is nothing like undo.
B) How do I do it right next time, I learned my lesson, I'll never do it
again, I promise!
By vectorization:
tcn5[tcn5 == -9] - NA
Uwe Ligges
Anders Corr
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
+
+ }
+ }
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Re: [R] NA erase your data trick
Anders Schwartz Corr wrote:
Oops,
I just erased all my data using this gizmo that I thought would replace -9
with NA.
A) Can I get my tcn5 back?
Not if you don't have it backed up somewhere else.
I wouldn't recommend keeping your only copy of anything in an R
workspace. It's too easy to accidentally delete or overwrite it. Keep
the original in a file.
B) How do I do it right next time, I learned my lesson, I'll never do it
again, I promise!
Anders Corr
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
For some values of i and n, this last line simplifies to
tcn5[TRUE] - NA
which is why you lost your data.
You want to (a) think in vectors, or (b) use an if statement:
(a) Replace your whole series of statements with
tcn5[is.na(tcn5) | tcn5 == -9] - NA
or
(b) Replace just the last line above with
if (is.na(tcn5[n,i]) | tcn5[n,i] == -9) tcn5[n,i] - NA
I'd choose (a); it's a lot cleaner and will run faster.
Duncan Murdoch
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Re: [R] The error in R while using bugs.R function
Li, Jia wrote: Dear R users, I followed the instuctions on Dr. Gelman's web to install all of documents that bugs.R needs, but when I try to run the school example that the web posted in R, I got an error: couldn't find function bugs, what's wrong? Have you forgot to source() Andrew's bugs.R file? Anyway, you might want to give the CRAN package R2WinBUGS a try, which is based on Andrew's code known as bugs.R. There is also a developer version BRugs (an interface to OpenBUGS) available at http://www.statistik.uni-dortmund.de/~ligges/BRugs/ (will move to CRAN very soon now). Uwe Ligges Thanks, Jia __ 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 __ 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
Re: [R] NA erase your data trick
Hi
Maybe
tcn5[tcn5 == -9] - NA
if tcn5 is matrix
mat-matrix(rnorm(100),10,10)
mat[5,6:7]- -9
mat[mat == -9]-NA
Read some intro on data manipulation, it helps you to avoid
thinking in loops
Cheers
Petr
On 17 May 2005 at 1:37, Anders Schwartz Corr wrote:
Oops,
I just erased all my data using this gizmo that I thought would
replace -9 with NA.
A) Can I get my tcn5 back?
B) How do I do it right next time, I learned my lesson, I'll never do
it again, I promise!
Anders Corr
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
+
+ }
+ }
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Petr Pikal
[EMAIL PROTECTED]
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Re: [R] A question about bugs.R: functions for running WinBUGs from R
Li, Jia wrote: Dear R users, I've found bugs.R : the functions for running WinBUGs from R that is writen by Dr. Andrew Gelman who is a professor from Columbia University. The bugs.R would be very useful for me, and I think many of you know it as well. I followed the instuctions on Dr. Gelman's web to install all of documents that bugs.R needs, but when I try to run the school example the web posted in R, I got an error. Would you please help me out? I am stuck on this for a while and really frustrated now. The program and the error as follow. Thanks a lot in advance! Why do you post twice? See my former message, you forgot to source() bugs.R. BTW: Since this code is from the website from Andrew Gelman, why do you think posting to R-help is the appropriate way to get help? R has a standardized mechanism for distributing code - called package. And the appropriate package R2WinBUGS you are looking for has been made available - thanks to Andrew's effort in writing the original code and thanks to the efforts of two others to generalize the code and package it for you. Uwe Ligges Jiaa _ # R code for entering the data and fitting the Bugs model for 8 schools # analysis from Section 5.5 of Bayesian Data Analysis. # To run, the Bugs model must be in the file schools.txt in your working # directory and you must load in the functions in the bugs.R file (see # http://www.stat.columbia.edu/~gelman/bugsR/). J - 8 y - c(28,8,-3,7,-1,1,18,12) sigma.y - c(15,10,16,11,9,11,10,18) schools.data - list (J, y, sigma.y) schools.inits - function() + list (theta=rnorm(J,0,1), mu.theta=rnorm(1,0,100), + sigma.theta=runif(1,0,100)) schools.parameters - c(theta, mu.theta, sigma.theta) #run in winbugs14 schools.sim - bugs (schools.data, schools.inits, schools.parameters, schools.bug, n.chains=3, n.iter=1000, version=1.4) Error: couldn't find function bugs --- [[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 __ 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
Re: [R] NA erase your data trick
On Tue, 17 May 2005 08:33:00 +0200
Uwe Ligges [EMAIL PROTECTED] wrote:
Anders Schwartz Corr wrote:
Oops,
I just erased all my data using this gizmo that I thought would replace -9
with NA.
A) Can I get my tcn5 back?
As you got it the first time. There is nothing like undo.
If you´re lucky it still lives inside an old, not overwritten when leaving the
depressing R session .RData.
Detlef
B) How do I do it right next time, I learned my lesson, I'll never do it
again, I promise!
By vectorization:
tcn5[tcn5 == -9] - NA
Uwe Ligges
Anders Corr
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
+
+ }
+ }
__
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PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
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Re: [R] NA erase your data trick
On Tue, 17 May 2005, Uwe Ligges wrote:
Anders Schwartz Corr wrote:
Oops,
I just erased all my data using this gizmo that I thought would replace -9
with NA.
A) Can I get my tcn5 back?
As you got it the first time. There is nothing like undo.
B) How do I do it right next time, I learned my lesson, I'll never do it
again, I promise!
By vectorization:
tcn5[tcn5 == -9] - NA
That will work if tcn5 contains NAs, but only because NA indices on the
lhs are now ignored for matrices (if tcn5 is a matrix, which seems
unstated) -- this used not to be the case. I would prefer
tcn5[tcn %in% -9] - NA
Using %in% rather than == in computed indices is a good habit to acquire:
it also makes things like
tcn5[tcn %in% c(-9, -99)] - NA
work as expected.
If tcn is a data frame, you have to do this column-by-column, as in
tcn5[] - lapply(tcn5, function(x) x[x %in% -9] - NA)
or by a logical index matrix, which is harder to construct.
for(i in 1:dim(tcn5)[2]){ ##for the number of columns
+ for(n in 1:dim(tcn5)[1]){ ##for the number of rows
+ tcn5[is.na(tcn5[n,i]) | tcn5[n,i] == -9] - NA
+
+ }
+ }
--
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] parsing speed
BertG == Berton Gunter [EMAIL PROTECTED]
on Mon, 16 May 2005 15:20:01 -0700 writes:
BertG (just my additional $.02) ... and as a general rule
BertG (subject to numerous exceptions, caveats, etc.)
BertG 1) it is programming and debugging time that most
BertG impacts overall program execution time; 2) this is
BertG most strongly impacted by code readability and size
BertG (the smaller the better); 3) both of which are
BertG enhanced by modular construction and reuseability,
BertG which argues for avoiding inline code and using
BertG separate functions.
BertG These days, i would argue that most of the time it is
BertG program clarity and correctness (they are related)
BertG that is the important issue, not execution speed.
BertG ... again, subject to exceptions and caveats, etc.
Yes indeed; very good points very well put!
Just to say it again:
We strongly recommend not to inline your code, but rather
program modularly, i.e. call small `utility' functions.
If execution time ever becomes crucial for your problem
(not often), the chances are considerable that the time spent is
not there [[ but you have to measure! - use Rprof() ! ]]
and if it *was* there, then you have your bottleneck in one
simple function that you could start optimizing... even a good
reason for not inlining that code..
Martin Maechler, ETH Zurich
BertG -- Bert Gunter Genentech Non-Clinical Statistics
BertG South San Francisco, CA
BertG The business of the statistician is to catalyze the
BertG scientific learning process. - George E. P. Box
-Original Message- From:
[EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Duncan Murdoch Sent: Monday, May 16, 2005 3:09 PM To:
[EMAIL PROTECTED] Cc: r-help Subject: Re: [R]
parsing speed
Federico Calboli wrote: Hi everyone,
I have a question on parsing speed.
I have two functions:
F1 F2
As things are now, F2 calls F1 internally:
F2 = function(x){ if (something == 1){ y = F1(x)
} if (something ==2){ do whatever } }
*Assuming there could be some difference*, is is faster
to use the code as written above or should I actually
write the statements of F1 to make the parsing faster?
The parsing only happens once when you define the
functions, and is (almost always) a negligible part of
total execution time. I think you're really worried
about execution time. You'll probably get more execution
time with a separate function because function calls take
time.
However, my guess is that putting F1 inline won't make
enough difference to notice.
Duncan
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[R] Vuong test
Hi, I have two questions. First, I'd like to compare a ZINB model to a negativ binomial model with the Vuong test, but I can't find how to performe it from the zicount package. Does a programm exist to do it ? Second, I'd like to know in which cases we have to use a double hurdle model instead of a zero inflated model. Many thanks, Stéphanie Payet REES France Réseau d'Evaluation en Economie de la Santé 28, rue d'Assas 75006 PARIS Tél. +33 (0)1 44 39 16 90 Fax +33 (0)1 44 39 16 92 Mèl. [EMAIL PROTECTED] Site Internet : http://www.rees-france.com __ 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
Re: [R] cluster results using fanny
Barbara Diaz wrote: Hi, I am using fanny and I have estrange results. I am wondering if someone out there can help me understand why this happens. First of all in most of my tries, it gives me a result in which each object has equal membership in all clusters. I have read that that means the clustering is entirely fuzzy. Looking at the graphics it is really difficult to understand how objects with so different scores for the variables have the same membership for all the clusters. Hi Barbara, I think, there is a problem with fanny, when you have standardised data. For example: library(mvoutlier) library(cluster) data(chorizon) a - fanny(chorizon[,101:110],4) b - fanny(scale(chorizon[,101:110]),4) a$mem # is ok, but b$mem # have same memberships Better to use function cmeans in package e1071, which gives correct memberships! Best, Matthias __ 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 __ 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
Re: [R] Vuong test
Hi Stéphanie, The Vuong test can be done in Stata (http://www.stata.com/support/faqs/stat/vuong.html), but I am also looking for its code in R. In addition to zicounts, Dr. Simon Jackman (http://pscl.stanford.edu/) has provided the code for fitting the zero-inflated (http://pscl.stanford.edu/zeroinfl.r) and hurdle (http://pscl.stanford.edu/hurdle.r) count models. Reza On 5/17/05, Stéphanie PAYET [EMAIL PROTECTED] wrote: Hi, I have two questions. First, I'd like to compare a ZINB model to a negativ binomial model with the Vuong test, but I can't find how to performe it from the zicount package. Does a programm exist to do it ? Second, I'd like to know in which cases we have to use a double hurdle model instead of a zero inflated model. Many thanks, Stéphanie Payet REES France Réseau d'Evaluation en Economie de la Santé 28, rue d'Assas 75006 PARIS Tél. +33 (0)1 44 39 16 90 Fax +33 (0)1 44 39 16 92 Mèl. [EMAIL PROTECTED] Site Internet : http://www.rees-france.com __ 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 __ 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
Re: [R] a problem sourcing a file using chdir=TRUE
This and some related problems should be fixed in tomorrow's R-patched
snapshot.
On Mon, 16 May 2005, Luca Scrucca wrote:
Dear R-users,
I used to give commands such as:
source(file=~/path/to/file.R, chdir=TRUE)
but with the latest v. 2.1.0 it does not seem to work anymore.
I tried to figure out what it was going on and it seems that the string
for which
class(file)
[1] character
is changed to
class(file)
[1] file connection
when the connection is open by
file - file(file, r, encoding = encoding)
But this force the following if statement
if (chdir is.character(file) (path - dirname(file)) != .)
{ owd - getwd()
on.exit(setwd(owd))
setwd(path)
}
to be FALSE and then non changing of current directory is done.
Is this the desired behavior or some bug fix is required?
--
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] parsing speed
On Tue, May 17, 2005 at 09:50:20AM +0200, Martin Maechler wrote: BertG == Berton Gunter [EMAIL PROTECTED] on Mon, 16 May 2005 15:20:01 -0700 writes: BertG (just my additional $.02) ... and as a general rule BertG (subject to numerous exceptions, caveats, etc.) BertG 1) it is programming and debugging time that most BertG impacts overall program execution time; 2) this is BertG most strongly impacted by code readability and size BertG (the smaller the better); 3) both of which are BertG enhanced by modular construction and reuseability, BertG which argues for avoiding inline code and using BertG separate functions. BertG These days, i would argue that most of the time it is BertG program clarity and correctness (they are related) BertG that is the important issue, not execution speed. BertG ... again, subject to exceptions and caveats, etc. Yes indeed; very good points very well put! Just to say it again: We strongly recommend not to inline your code, but rather program modularly, i.e. call small `utility' functions. Generally, I fully agree -- modular coding is good, not only in R. However, with regard to execution time, modularisation that involves passing of large amounts of data (100 x 1000 data frames etc.) can cause problems. Best regards, Jan -- +- Jan T. Kim ---+ |*NEW*email: [EMAIL PROTECTED] | |*NEW*WWW: http://www.cmp.uea.ac.uk/people/jtk | *-= hierarchical systems are for files, not for humans =-* __ 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
Re: [R] adjusted p-values with TukeyHSD?
Dear Christoph You can use the multcomp package. Please have a look at the following example: library(multcomp) The first two lines were already proposed by Erin Hodgess: summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks)) TukeyHSD(fm1, tension, ordered = TRUE) Tukey multiple comparisons of means 95% family-wise confidence level factor levels have been ordered Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks) $tension difflwr upr M-H 4.72 -4.6311985 14.07564 L-H 14.72 5.3688015 24.07564 L-M 10.00 0.6465793 19.35342 By using the functions simtest or simint you can get the p-values, too: summary(simtest(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous tests: Tukey contrasts Call: simtest.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj tensionH-tensionL -14.722 -3.8023.872 0.000 0.001 0.001 tensionM-tensionL -10.000 -2.5823.872 0.013 0.026 0.024 tensionH-tensionM -4.722 -1.2193.872 0.228 0.228 0.228 or if you prefer to get the confidence intervals, too, you can use: summary(simint(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous 95% confidence intervals: Tukey contrasts Call: simint.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 95 % quantile: 2.415 Coefficients: Estimate 2.5 % 97.5 % t value Std.Err. p raw p Bonf p adj tensionM-tensionL -10.000 -19.352 -0.648 -2.5823.872 0.013 0.038 0.034 tensionH-tensionL -14.722 -24.074 -5.370 -3.8023.872 0.000 0.001 0.001 tensionH-tensionM -4.722 -14.074 4.630 -1.2193.872 0.228 0.685 0.447 - Please be careful: The resulting confidence intervals in simint are not associated with the p-values from 'simtest' as it is described in the help page of the two functions. - I had not the time to check the differences in the function or read the references given on the help page. If you are interested in the function you can check those to find out which one you prefer. Best regards, Christoph Buser -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND phone: x-41-44-632-4673 fax: 632-1228 http://stat.ethz.ch/~buser/ -- Christoph Strehblow writes: hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for holm, hochberg, bonferroni, but not Tukey. Is it not possbile to get adjusted p-values after Tukey-correction? sorry, if this is an often-answered-question, but i didn´t find it on the list archive... thx a lot, list, Chris Christoph Strehblow, MD Department of Rheumatology, Diabetes and Endocrinology Wilhelminenspital, Vienna, Austria [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 __ 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
Re: [R] parsing speed
Jan T. Kim wrote: Generally, I fully agree -- modular coding is good, not only in R. However, with regard to execution time, modularisation that involves passing of large amounts of data (100 x 1000 data frames etc.) can cause problems. I've just tried a few simple examples of throwing biggish (3000x3000) matrices around and haven't encountered any pathological behaviour yet. I tried modifying the matrices within the functions, tried looping a few thousand times to estimate the matrix passing overhead, and in most cases the modular version run pretty much as fast as - or occasionally faster than - the inline version. There was some variability in CPU time taken, probably due to garbage collection. Does anyone have a simple example where passing large data sets causes a huge increase in CPU time? I think R is pretty smart with its parameter passing these days - anyone who thinks its still like Splus version 2.3 should update their brains to the 21st Century. Baz __ 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
[R] setting value arg of pdSymm() in nlme
Dear All, I wish to model random effects that have known between-group covariance structure using the lme() function from library nlme. However, I have yet to get even a simple example to work. No doubt this is because I am confusing my syntax, but I would appreciate any guidance as to how. I have studied Pinheiro Bates carefully (though it's always possible I've missed something), the few posts mentioning pdSymm (some of which suggest lme is suboptimal here anyway) and ?pdSymm (which has only a trivial example, see later) but have not yet found a successful example of syntax for this particular problem. I am using the pdSymm class to specify a positive definite matrix corresponding to the covariance structure of a random batch effect, and passing this to lme() through the random= argument. To do this, I must set the value= argument of pdSymm. Consider the following simple and self-contained example: library(nlme) # make response and batch data batch.names - c(A, B, C) data.df - data.frame( response = rnorm(100), batch = factor(sample(batch.names, 100, replace=T)) ) # make covariance matrix for batch batch.mat - matrix(c(1,.5,.2, .5, 1, .3, .2, .3, 1), ncol=3) colnames(batch.mat) - batch.names rownames(batch.mat) - batch.names # fit batch as a simple random intercept lme(response ~ 1, data=data.df, random=~1|batch) # ...works fine # do the same using pdSymm notation lme(response ~ 1, data=data.df, random=list( batch=pdSymm(form=~1) ) ) # ...works fine also # specify cov structure using value arg lme(response ~ 1, data=data.df, random=list( batch=pdSymm( value=batch.mat, form=~1, nam=batch.names) ) ) # throws error below ---snip--- Error in Names-.pdMat(`*tmp*`, value = (Intercept)) : Length of names should be 3 traceback() 7: stop(paste(Length of names should be, length(dn))) 6: Names-.pdMat(`*tmp*`, value = (Intercept)) 5: Names-(`*tmp*`, value = (Intercept)) 4: Names-.reStruct(`*tmp*`, value = list(batch = (Intercept))) 3: Names-(`*tmp*`, value = list(batch = (Intercept))) 2: lme.formula(response ~ 1, data = data.df, random = list(batch = pdSymm(value = batch.mat, form = ~1, nam = batch.names))) 1: lme(response ~ 1, data = data.df, random = list(batch = pdSymm(value = batch.mat, form = ~1, nam = batch.names))) ---snip--- The length of batch.names is 3, so I find this error enigmatic. Note that I had to specify all three of value, form and nam otherwise I got missing args errors. Also note that doing pdSymm(value=batch.mat, form=~1, nam=batch.names) on the command line, like the similar invocation described on ?pdSymm, works fine also. It's just lme() that doesn't like it. Can anybody show me what I should be doing instead? Some successful code will greatly clarify the issue. (My version details are below). Also, I notice the pdMat scheme is absent from lme() in lme4. Is this functionality deprecated in lme4 and excluded from lmer? Many thanks, William Version details: running R 2.1.0 on windows XP, using nlme 3.1-57. =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Dr William Valdar ++44 (0)1865 287 717 Wellcome Trust Centre [EMAIL PROTECTED] for Human Genetics, Oxford www.well.ox.ac.uk/~valdar __ 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
Re: [R] adjusted p-values with TukeyHSD?
Hi! Thanks a lot, works as advertised. If i used Tukey, it even gives raw, Bonferroni- and Tukey-corrected p-values! Thx for the help, Christoph Strehblow, MD Department of Rheumatology, Diabetes and Endocrinology Wilhelminenspital, Vienna, Austria [EMAIL PROTECTED] Am 17.05.2005 um 13:23 schrieb Christoph Buser: Dear Christoph You can use the multcomp package. Please have a look at the following example: library(multcomp) The first two lines were already proposed by Erin Hodgess: summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks)) TukeyHSD(fm1, tension, ordered = TRUE) Tukey multiple comparisons of means 95% family-wise confidence level factor levels have been ordered Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks) $tension difflwr upr M-H 4.72 -4.6311985 14.07564 L-H 14.72 5.3688015 24.07564 L-M 10.00 0.6465793 19.35342 By using the functions simtest or simint you can get the p-values, too: summary(simtest(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous tests: Tukey contrasts Call: simtest.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj tensionH-tensionL -14.722 -3.8023.872 0.000 0.001 0.001 tensionM-tensionL -10.000 -2.5823.872 0.013 0.026 0.024 tensionH-tensionM -4.722 -1.2193.872 0.228 0.228 0.228 or if you prefer to get the confidence intervals, too, you can use: summary(simint(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous 95% confidence intervals: Tukey contrasts Call: simint.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 95 % quantile: 2.415 Coefficients: Estimate 2.5 % 97.5 % t value Std.Err. p raw p Bonf p adj tensionM-tensionL -10.000 -19.352 -0.648 -2.5823.872 0.013 0.038 0.034 tensionH-tensionL -14.722 -24.074 -5.370 -3.8023.872 0.000 0.001 0.001 tensionH-tensionM -4.722 -14.074 4.630 -1.2193.872 0.228 0.685 0.447 - Please be careful: The resulting confidence intervals in simint are not associated with the p-values from 'simtest' as it is described in the help page of the two functions. - I had not the time to check the differences in the function or read the references given on the help page. If you are interested in the function you can check those to find out which one you prefer. Best regards, Christoph Buser -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH (Federal Inst. Technology)8092 Zurich SWITZERLAND phone: x-41-44-632-4673fax: 632-1228 http://stat.ethz.ch/~buser/ -- Christoph Strehblow writes: hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for holm, hochberg, bonferroni, but not Tukey. Is it not possbile to get adjusted p-values after Tukey-correction? sorry, if this is an often-answered-question, but i didn´t find it on the list archive... thx a lot, list, Chris Christoph Strehblow, MD Department of Rheumatology, Diabetes and Endocrinology Wilhelminenspital, Vienna, Austria [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 __ 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
RE: [R] install.packages parameters
Caveat: I know next to nothing about Mac... That said, my guess is that you installed R from binary, rather than building from source. In that case the compilers and flags, etc., are configured to the machine that the binary is built on. You can look in $RHOME/etc/Makeconf to see the settings, and see if changing them helps. Andy From: [EMAIL PROTECTED] Hello. R is having some trouble installing a package because it passed arguments to gcc which were non-existent directories and files. It also didn't find g77, although it's in a directory in my $PATH; I tricked it by making a sym link in /usr/bin. What file does R get these parameters from? I've looked for the parameters in the package source, the install.packages help pages, and the R preferences menu, all to no avail. I am running R 2.1.0 on Mac 10.3.8, and three days ago I installed a different package from source where installation involved gcc without any problems, and nothing has changed since then. The packages I'm trying to install are Joe Schafer's mix, norm, and cat. Thanks, Janet __ 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 __ 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
Re: [R] adjusted p-values with TukeyHSD?
Hi Chris and Chris, I was keeping my eye on this thread as I have also been discovering multiple comparisons recently. Your instructions are very clear! Thanks. Now I would love to see an R boffin write a nifty function to produce a graphical representation of the multiple comparison, like this one: http://www.theses.ulaval.ca/2003/21026/21026024.jpg Should not be too difficult.[any one up for the challenge?] I came across more multiple comparison info here; http://www.agr.kuleuven.ac.be/vakken/statisticsbyR/ANOVAbyRr/multiplecomp.htm Cheers, Sander. Christoph Buser wrote: Dear Christoph You can use the multcomp package. Please have a look at the following example: library(multcomp) The first two lines were already proposed by Erin Hodgess: summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks)) TukeyHSD(fm1, tension, ordered = TRUE) Tukey multiple comparisons of means 95% family-wise confidence level factor levels have been ordered Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks) $tension difflwr upr M-H 4.72 -4.6311985 14.07564 L-H 14.72 5.3688015 24.07564 L-M 10.00 0.6465793 19.35342 By using the functions simtest or simint you can get the p-values, too: summary(simtest(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous tests: Tukey contrasts Call: simtest.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 Coefficients: Estimate t value Std.Err. p raw p Bonf p adj tensionH-tensionL -14.722 -3.8023.872 0.000 0.001 0.001 tensionM-tensionL -10.000 -2.5823.872 0.013 0.026 0.024 tensionH-tensionM -4.722 -1.2193.872 0.228 0.228 0.228 or if you prefer to get the confidence intervals, too, you can use: summary(simint(breaks ~ wool + tension, data = warpbreaks, whichf=tension, type = Tukey)) Simultaneous 95% confidence intervals: Tukey contrasts Call: simint.formula(formula = breaks ~ wool + tension, data = warpbreaks, whichf = tension, type = Tukey) Tukey contrasts for factor tension, covariable: wool Contrast matrix: tensionL tensionM tensionH tensionM-tensionL 0 0 -110 tensionH-tensionL 0 0 -101 tensionH-tensionM 0 00 -11 Absolute Error Tolerance: 0.001 95 % quantile: 2.415 Coefficients: Estimate 2.5 % 97.5 % t value Std.Err. p raw p Bonf p adj tensionM-tensionL -10.000 -19.352 -0.648 -2.5823.872 0.013 0.038 0.034 tensionH-tensionL -14.722 -24.074 -5.370 -3.8023.872 0.000 0.001 0.001 tensionH-tensionM -4.722 -14.074 4.630 -1.2193.872 0.228 0.685 0.447 - Please be careful: The resulting confidence intervals in simint are not associated with the p-values from 'simtest' as it is described in the help page of the two functions. - I had not the time to check the differences in the function or read the references given on the help page. If you are interested in the function you can check those to find out which one you prefer. Best regards, Christoph Buser -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND phone: x-41-44-632-4673 fax: 632-1228 http://stat.ethz.ch/~buser/ -- Christoph Strehblow writes: hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for holm, hochberg, bonferroni, but not Tukey. Is it not possbile to get adjusted p-values after Tukey-correction? sorry, if this is an often-answered-question, but i didn´t find it on the list archive... thx a lot, list, Chris Christoph Strehblow, MD Department of Rheumatology, Diabetes and Endocrinology Wilhelminenspital, Vienna, Austria [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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read
[R] Finding the right number of clusters
SAS has something called the cubic criterion cutoff for finding the most appropriate number of clusters. Does R have anything that would replicate that? I've been searching the lists and can't seem to find anything that would point me in the right direction. Thank in advance, Philip Bermingham __ 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
Re: [R] Finding the right number of clusters
Le 17.05.2005 14:42, Philip Bermingham a écrit : SAS has something called the cubic criterion cutoff for finding the most appropriate number of clusters. Does R have anything that would replicate that? I've been searching the lists and can't seem to find anything that would point me in the right direction. Thank in advance, Philip Bermingham Hello, Package fpc has a function cluster.stats with a lot of criterion like G2, G3, etc ... Romain -- ~ ~~ Romain FRANCOIS - http://addictedtor.free.fr ~~ Etudiant ISUP - CS3 - Industrie et Services ~~http://www.isup.cicrp.jussieu.fr/ ~~ Stagiaire INRIA Futurs - Equipe SELECT ~~ http://www.inria.fr/recherche/equipes/select.fr.html~~ ~ __ 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
RE: [R] parsing speed
From: Barry Rowlingson Jan T. Kim wrote: Generally, I fully agree -- modular coding is good, not only in R. However, with regard to execution time, modularisation that involves passing of large amounts of data (100 x 1000 data frames etc.) can cause problems. I've just tried a few simple examples of throwing biggish (3000x3000) matrices around and haven't encountered any pathological behaviour yet. I tried modifying the matrices within the functions, tried looping a few thousand times to estimate the matrix passing overhead, and in most cases the modular version run pretty much as fast as - or occasionally faster than - the inline version. There was some variability in CPU time taken, probably due to garbage collection. Does anyone have a simple example where passing large data sets causes a huge increase in CPU time? I think R is pretty smart with its parameter passing these days - anyone who thinks its still like Splus version 2.3 should update their brains to the 21st Century. I think one example of this is using the formula interface to fit models on large data sets, especially those with tons of variables. Some model fitting functions have the default interface f(x, y, ...), along with a formula method f(formula, data, ...). If x has lots of variables (say over 1000), using the formula interface can take several times longer than calling the raw interface directly. Andy Baz __ 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 __ 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
RE: [R] Survey of moving window statistical functions - still l ooking for fast mad function
On 9 Oct 2004, Brian D. Ripley wrote: On Fri, 8 Oct 2004, Tuszynski, Jaroslaw W. wrote: Finally a question: I still need to get moving windows mad function faster my runmad function is not that much faster than apply/embed combo, and that I used before, and this is where my code spends most of its time. I need something like runmed but for a mad function. Any suggestions? Write your own C-level implementation, as runmed and most of the other fast functions you cite are. I did as suggested and just released runmean, runmax, runmin, runmad and runquantile functions as part of caMassClass package. Thanks for the suggestion, now my codes run 20 min. instead of overnight. Jarek \=== Jarek Tuszynski, PhD. o / \ Science Applications International Corporation \__,| (703) 676-4192 \ [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
[R] installing R on Irix
Hello veeryone, I nedd some help here. The problem is I was trying to install R on my Irix system, with little success: I got the following ugly error messages: watch out: begin installing recommended package mgcv Cannot create directory : No such file or directory * Installing *source* package 'mgcv' ... ** libs gmake[3]: Entering directory `/tmp/R.INSTALL.13709658/mgcv/src' gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c gcv.c -o gcv.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c magic.c -o magic.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c mat.c -o mat.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c matrix.c -o matrix.o as: Error: /var/tmp/ccAomyDd.s, line 23679: register expected dmtc1 244($sp),$f0 as: Error: /var/tmp/ccAomyDd.s, line 23679: Undefined symbol: 244 gmake[3]: *** [matrix.o] Error 1 gmake[3]: Leaving directory `/tmp/R.INSTALL.13709658/mgcv/src' ERROR: compilation failed for package 'mgcv' ** Removing '/usr/home/csmatyi/programs/R/R-2.1.0/library/mgcv' gmake[2]: *** [mgcv.ts] Error 1 gmake[2]: Leaving directory `/usr/home/csmatyi/programs/R/R- 2.1.0/src/library/Recommended' gmake[1]: *** [recommended-packages] Error 2 gmake[1]: Leaving directory `/usr/home/csmatyi/programs/R/R- 2.1.0/src/library/Recommended' gmake: *** [stamp-recommended] Error 2 What could the problem be here? Thanks, Matthew C. __ 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
Re: [R] adjusted p-values with TukeyHSD?
Shame I can not get hold of Hsu, J. C. and M. Peruggia (1994) just now.
I am quite curious to see what their graphs look like. Would you be able
to give an example in R.? ;-)
The graph I put forward is typically used by ecologists to summarize
data. It comes down to a simple means plot with error bars. Significant
differences of multiple comparisons are then added using the letters a,
b, c etc. If two bars have the same letter, they are not significantly
different. It can become quite complicated when mean one is different
from mean three but not from mean two and mean two is different from
mean three but not mean one. You then get: a, ab, c for mean one, two
and three respectively.
Of course what is often used does not constitute the best way of doing it.
Sander.
Liaw, Andy wrote:
From: Sander Oom
Hi Chris and Chris,
I was keeping my eye on this thread as I have also been discovering
multiple comparisons recently. Your instructions are very
clear! Thanks.
One thing to note, though: Multcomp does not do Dunnett's or
Tukey's multiple comparisons per se. Those names in multcomp
refer to the contrasts being used (comparison to a control for
Dunnett and all pairwise comparison for Tukey). The actual
methods used are as described in the references of the help
pages.
Now I would love to see an R boffin write a nifty function to
produce a
graphical representation of the multiple comparison, like this one:
http://www.theses.ulaval.ca/2003/21026/21026024.jpg
Should not be too difficult.[any one up for the challenge?]
I beg to differ: That's probably as bad a way as one can use to
graphically show multiple comparison. The shaded bars serve no
purpose.
Two alternatives that I'm aware of are
- Multiple comparison circles, due to John Sall, and not
surprisingly, implemented in JMP and SAS/Insight. See:
http://support.sas.com/documentation/onlinedoc/v7/whatsnew/insight/sect4.htm
- The mean-mean display proposed by Hsu and Peruggia:
Hsu, J. C. and M. Peruggia (1994).
Graphical representations of Tukey's multiple comparison method.
Journal of Computational and Graphical Statistics 3, 143{161
Andy
I came across more multiple comparison info here;
http://www.agr.kuleuven.ac.be/vakken/statisticsbyR/ANOVAbyRr/m
ultiplecomp.htm
Cheers,
Sander.
Christoph Buser wrote:
Dear Christoph
You can use the multcomp package. Please have a look at the
following example:
library(multcomp)
The first two lines were already proposed by Erin Hodgess:
summary(fm1 - aov(breaks ~ wool + tension, data = warpbreaks))
TukeyHSD(fm1, tension, ordered = TRUE)
Tukey multiple comparisons of means
95% family-wise confidence level
factor levels have been ordered
Fit: aov(formula = breaks ~ wool + tension, data = warpbreaks)
$tension
difflwr upr
M-H 4.72 -4.6311985 14.07564
L-H 14.72 5.3688015 24.07564
L-M 10.00 0.6465793 19.35342
By using the functions simtest or simint you can get the
p-values, too:
summary(simtest(breaks ~ wool + tension, data = warpbreaks,
whichf=tension,
type = Tukey))
Simultaneous tests: Tukey contrasts
Call:
simtest.formula(formula = breaks ~ wool + tension, data =
warpbreaks,
whichf = tension, type = Tukey)
Tukey contrasts for factor tension, covariable: wool
Contrast matrix:
tensionL tensionM tensionH
tensionM-tensionL 0 0 -110
tensionH-tensionL 0 0 -101
tensionH-tensionM 0 00 -11
Absolute Error Tolerance: 0.001
Coefficients:
Estimate t value Std.Err. p raw p Bonf p adj
tensionH-tensionL -14.722 -3.8023.872 0.000 0.001 0.001
tensionM-tensionL -10.000 -2.5823.872 0.013 0.026 0.024
tensionH-tensionM -4.722 -1.2193.872 0.228 0.228 0.228
or if you prefer to get the confidence intervals, too, you can
use:
summary(simint(breaks ~ wool + tension, data = warpbreaks,
whichf=tension,
type = Tukey))
Simultaneous 95% confidence intervals: Tukey contrasts
Call:
simint.formula(formula = breaks ~ wool + tension, data =
warpbreaks,
whichf = tension, type = Tukey)
Tukey contrasts for factor tension, covariable: wool
Contrast matrix:
tensionL tensionM tensionH
tensionM-tensionL 0 0 -110
tensionH-tensionL 0 0 -101
tensionH-tensionM 0 00 -11
Absolute Error Tolerance: 0.001
95 % quantile: 2.415
Coefficients:
Estimate 2.5 % 97.5 % t value Std.Err.
p raw p Bonf p adj
tensionM-tensionL -10.000 -19.352 -0.648 -2.5823.872
0.013 0.038 0.034
tensionH-tensionL -14.722 -24.074 -5.370 -3.8023.872
0.000 0.001 0.001
tensionH-tensionM -4.722 -14.074 4.630 -1.2193.872
0.228 0.685 0.447
-
Please be careful: The resulting confidence intervals in
simint are not associated with the p-values
Re: [R] installing R on Irix
Cserháti Mátyás [EMAIL PROTECTED] writes: Hello veeryone, I nedd some help here. The problem is I was trying to install R on my Irix system, with little success: I got the following ugly error messages: watch out: begin installing recommended package mgcv Cannot create directory : No such file or directory * Installing *source* package 'mgcv' ... ** libs gmake[3]: Entering directory `/tmp/R.INSTALL.13709658/mgcv/src' gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c gcv.c -o gcv.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c magic.c -o magic.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c mat.c -o mat.o gcc -I/usr/home/csmatyi/programs/R/R-2.1.0/include - I/usr/local/include -g -O2 -c matrix.c -o matrix.o as: Error: /var/tmp/ccAomyDd.s, line 23679: register expected dmtc1 244($sp),$f0 as: Error: /var/tmp/ccAomyDd.s, line 23679: Undefined symbol: 244 gmake[3]: *** [matrix.o] Error 1 gmake[3]: Leaving directory `/tmp/R.INSTALL.13709658/mgcv/src' ERROR: compilation failed for package 'mgcv' ** Removing '/usr/home/csmatyi/programs/R/R-2.1.0/library/mgcv' gmake[2]: *** [mgcv.ts] Error 1 gmake[2]: Leaving directory `/usr/home/csmatyi/programs/R/R- 2.1.0/src/library/Recommended' gmake[1]: *** [recommended-packages] Error 2 gmake[1]: Leaving directory `/usr/home/csmatyi/programs/R/R- 2.1.0/src/library/Recommended' gmake: *** [stamp-recommended] Error 2 What could the problem be here? Offhand: You seem to be using the system assembler as on the output of gcc. Sometimes one needs the GNU assembler (gas) in which case you likely need to install the GNU binutils. If that is the case, it is quite surprising that you got that far, but stranger things have happened... -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- 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
Re: [R] setting value arg of pdSymm() in nlme
Hi, I'm afraid that I don't understand what you are trying to do. With a formula of ~ 1 the pdSymm generator creates a 1x1 variance-covariance matrix, which you are initializing to a 3x3 matrix. Oh... I had a feeling I was doing something wrong there. What is batch.mat supposed to represent? I would like to use batch.mat to specify a correlation structure for the batches A, B and C. Specifically, I wish to work out the contribution to the variance of the batch random effect, given that I know some pairs of batches (eg, A and B) are going to be more similar than other pairs (eg, A and C) and how similar they are likely to be. My (mis?)reading of PB p165 suggested this may be possible turning some of the pdIdents into pdSymms. I was hoping to use this test example as a prelude to using genetic relationship data to impose a correlation structure on a subject-level or family-level effect. I understand from an earlier post (Jarrod Hadfield, 2003) that lme is not really optimized for this, but I would nontheless like to evaluate to what extent it can do it anyway. The example used in that post was extreme: each case was effectively a different realization of a random effect where the correlation between levels is known. My needs are simpler: in the example I gave above, batches A, B and C might represent three families related by different degrees. Yes, although I think you mean lmer in lme4. Because the lmer function allows multiple nested or non-nested grouping factors, the need for the pdMat classes is eliminated (or greatly reduced) and the code can be simplified considerably. There is an article in the 2005/1 issue of R News describing the use of lmer. Thanks for pointing this out as I had missed this article. I know lmer is under development and am very interested in what it can do. Having found lme/lmer useful for more standard problems I would like to use it for genetic-type analysis where possible rather than resort to a different language (eg SAS) or specialized proprietary software (eg, ASREML). However, I understand that may not be possible. William =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Dr William Valdar ++44 (0)1865 287 717 Wellcome Trust Centre [EMAIL PROTECTED] for Human Genetics, Oxford www.well.ox.ac.uk/~valdar __ 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
[R] Combinations with two part column
Dear R-helpers, I am a beginner using R. This is the first question in this list. My question, Is there possible to make combinations with two part column? If I have a number 1,2,3,4,5,6,7,8. I need the result something like below: 1,2,3,4,5 6,7,8 1,2,3,4,7 5,6,8 2,3,4,5,6 1,7,8 1,2,3,6,7 4,5,8 1,2,3,4,8 5,6,7 3,4,6,7,8 1,2,5 I would be very happy if anyone could help me. Best regards, Sofyan __ 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
[R] Problem in lme longitudinal model
Hi R-masters! I trying model Heart disease mortality in my country with a lme model like this: m1.lme-lme(log(rdeath)~age*year,random=~age|year,data=dados) where: rdeath is rate of mortality per 10 person per age and year age: age of death (22 27 32 37 42 47 52 57 62 67 72 77 82) year: year of death (1980:2002) I don´t have problem to fit the model, but in residual analysis I have one problem. If i type acf(m1.lme$residuals) the graph show 4 plot with a sin curve (a example in attach) and I get this variogram: Variogram(m1.lme) variog dist n.pairs 1 0.39390651 276 2 0.64864522 253 3 0.96798703 230 4 1.27650944 207 5 1.41581475 184 6 1.42646856 161 7 1.40486087 138 8 1.19026138 115 9 0.96193309 92 10 0.7037427 10 69 11 0.8591257 11 46 12 1.2215657 12 23 Well I need help for solution this problem Somebody can help me? Thanks in advance Bernardo Rangel Tura, MD, MSc National Institute of Cardiology Laranjeiras Rio de Janeiro Brazil No virus found in this outgoing message. Checked by AVG Anti-Virus. __ 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
RE: [R] Omitting NAs in aggregate.ts()
I have a time series vector (not necessarily ts class) that has NAs
in it.
How can I omit the NAs when using aggregate.ts() to compute a
function
on each window? If there is at least one non-NA value in each
window,
I'd like to proceed with evaluating the function; otherwise, I would
like NA returned. I'm not wedded to aggregate.ts and don't need ts
class if there is another fast, vectorized way to handle this. Here
is what I am trying to do, with the invalid na.rm thrown in:
as.vector(aggregate.ts(x, ndeltat=24, FUN=min, na.rm=F))
I don't know is the short answer, but if I had the data I
might have tried mymin - function(x) min(x, na.rm = TRUE)
as.vector(aggregate.ts(x, ndeltat=24, FUN=mymin))
Thanks for the suggestions, and my apologies for not providing an
example and error messages. Using a custom function for FUN in
aggregate.ts() is the way to go. Here is a complete solution to my
problem:
e - c(2.3, 4.5, 6.2, 1.8)
f - c(2.3, NA, NA, NA)
mymin - function(x) {
+ if(length(x[!is.na(x)]) 0) return(min(x, na.rm = TRUE))
+ else return(NA)
+ }
as.vector(aggregate.ts(e, ndeltat=2, FUN=mymin))
[1] 2.3 1.8
as.vector(aggregate.ts(f, ndeltat=2, FUN=mymin))
[1] 2.3 NA
Scott Waichler
Pacific Northwest National Laboratory
[EMAIL PROTECTED]
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[R] simple question, i hope
How do you output a list to a text file without the extra line numbers and stuff? I have a list b, and tried zz-textConnection(captest.txt,w) sink(zz) b sink() close(zz) but that isnt what i want, because i get [[1]] [1] a etc. Is there a simple way to do the R equivalent of this perl code? open(OUT,out.txt); print OUT @b; close OUT Thank you for your help. I tried pouring over teh documentation for this, but couldnt find what I was lookign for. ~Erithid __ 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
Re: [R] Combinations with two part column
On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Dear R-helpers, I am a beginner using R. This is the first question in this list. My question, Is there possible to make combinations with two part column? If I have a number 1,2,3,4,5,6,7,8. I need the result something like below: 1,2,3,4,5 6,7,8 1,2,3,4,7 5,6,8 2,3,4,5,6 1,7,8 1,2,3,6,7 4,5,8 1,2,3,4,8 5,6,7 3,4,6,7,8 1,2,5 Try this: library(gtools) t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x __ 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
[R] predict() question
Hi, there: Following yesterday's question ( i had a new level for a categorical variable occurred in validation dataset and predict() complains about it: i made some python code to solve the problem), but here, I am just curious about some details about the mechanism: I believed rpart follows CART and for a categorical variable, the splitting criteria should be like, is it A or not? --yes, go to left branch --no, go to right So, when you predict, if you have a new level C,for example, the predict() should not complain about the occurrence of C (of course, if there are many C's in validation, it should complain). Maybe for robustness, predict() has to check first if there is new level or not. I am not sure if my understanding is right or not, please be advised! Thanks, -- Weiwei Shi, Ph.D 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
[R] Sweave and paths
Is there some way to encourage \SweaveInput{foo} to find foo in a
subdirectory of a file tree? Something along the lines of the
behavior of list.files(stuff, recursive=TRUE). This would be very
helpful at calling small modular files, such as solution sets and the
like.
I couldn't see anything in the documentation, and I looked in the
source code, but it seems that SweaveReadFiledoc() wants to look only
in the directory which contains foo.
Still, I'm guessing that I'm missing something. Any tips would be
much appreciated.
Bill
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[R] Error message glmmPQL
Hi, I'm fitting a model for two-nested binay data in glmmPQL function but I have this error message: Error in solve.default(estimates[dimE[1] - (p:1), dimE [2] - (p:1), drop = FALSE]) : system is computationally singular: reciprocal condition number = 1.14416e-018. How do I can solve this problem? Or the problem are the data? Thanks, Pedro A. Torres S. Graduate Student - Statistics Department of Mathematics University of Puerto Rico at Mayaguez (1 787) 832-4040 X 2537 __ 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
Re: [R] Combinations with two part column
Thanks for you quick answer. Could I extend my question? How to make the result for each rows with comma ,; library(gtools) comb8.5 - t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x comb8.5[,1:5] [,1] [,2] [,3] [,4] [,5] [1,]12345 [2,]12346 [3,]12347 [4,]12348 [5,]12356 [6,]12357 ... I mean like: 1,2,3,4,5 1,2,3,4,6 1,2,3,4,7 1,2,3,4,8 1,2,3,5,6 1,2,3,5,7 comb8.5[,6:8] [,1] [,2] [,3] [1,]678 [2,]578 [3,]568 [4,]567 [5,]478 [6,]468 ... this below like: 6,7,8 5,7,8 5,6,8 5,6,7 4,7,8 4,6,8 Best, Sofyan On 5/17/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Dear R-helpers, I am a beginner using R. This is the first question in this list. My question, Is there possible to make combinations with two part column? If I have a number 1,2,3,4,5,6,7,8. I need the result something like below: 1,2,3,4,5 6,7,8 1,2,3,4,7 5,6,8 2,3,4,5,6 1,7,8 1,2,3,6,7 4,5,8 1,2,3,4,8 5,6,7 3,4,6,7,8 1,2,5 Try this: library(gtools) t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x __ 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
Re: [R] Combinations with two part column
Try write.table: write.table(comb8.5[,1:5], sep = ,, row.names = FALSE, col.names = FALSE) write.table(comb8.5[,6:8], sep = ,, row.names = FALSE, col.names = FALSE) On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Thanks for you quick answer. Could I extend my question? How to make the result for each rows with comma ,; library(gtools) comb8.5 - t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x comb8.5[,1:5] [,1] [,2] [,3] [,4] [,5] [1,]12345 [2,]12346 [3,]12347 [4,]12348 [5,]12356 [6,]12357 ... I mean like: 1,2,3,4,5 1,2,3,4,6 1,2,3,4,7 1,2,3,4,8 1,2,3,5,6 1,2,3,5,7 comb8.5[,6:8] [,1] [,2] [,3] [1,]678 [2,]578 [3,]568 [4,]567 [5,]478 [6,]468 ... this below like: 6,7,8 5,7,8 5,6,8 5,6,7 4,7,8 4,6,8 Best, Sofyan On 5/17/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Dear R-helpers, I am a beginner using R. This is the first question in this list. My question, Is there possible to make combinations with two part column? If I have a number 1,2,3,4,5,6,7,8. I need the result something like below: 1,2,3,4,5 6,7,8 1,2,3,4,7 5,6,8 2,3,4,5,6 1,7,8 1,2,3,6,7 4,5,8 1,2,3,4,8 5,6,7 3,4,6,7,8 1,2,5 Try this: library(gtools) t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x __ 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 __ 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
[R] cumsum on chron objects
Hi, Is there some alternative to cumsum for chron objects? I have data frames that contain some chron objects that look like this: DateTime 13/10/03 12:30:35 NA NA NA 15/10/03 16:30:05 NA NA ... and I've been trying to replace the NA's so that a date/time sequence is created starting with the preceding available value. Because the number of rows with NA's following each available date/time is unknown, I've split the data frame using: splitdf - split(df, as.factor(df$DateTime)) so that I can later use lapply to work on each block of data. I thought I could use cumsum and set the NA's to the desired interval to create the date/time sequence starting with the first row. However, this function is not defined for chron objects. Does anybody know of alternatives to create such a sequence? Thanks in advance, -- Sebastian P. Luque __ 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
[R] Hmisc/Design and problem with cgroup
Hello,
I am trying to use the following to output a table to latex:
cohortbyagesummary - by(data.frame(age,ethnicity), cohort, summary)
w - latex.default(cohortbyagesummary,
caption=Five Number Age Summaries by Cohort,
label=agesummarybycohort,
cgroup=c('hello','goodbye','hello'),
colheads=c(Age,Ethnicity),
extracolheads=c('hello','goodbye'), # demonstration of subheadings
greek=TRUE,
ctable=TRUE)
I am not able to get the major column headings of cgroup to work. I
receive the error:
Object cline not found
I do not see in the examples or documentation that you must specify
cline or anything.
Any suggestions? Thanks very much,
aric
R 2.0.1
Design 2.0-11
Hmisc 3.0-5
OS X 10.3.9
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Re: [R] cumsum on chron objects
On 5/17/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 5/17/05, Sebastian Luque [EMAIL PROTECTED] wrote: Hi, Is there some alternative to cumsum for chron objects? I have data frames that contain some chron objects that look like this: DateTime 13/10/03 12:30:35 NA NA NA 15/10/03 16:30:05 NA NA ... and I've been trying to replace the NA's so that a date/time sequence is created starting with the preceding available value. Because the number of rows with NA's following each available date/time is unknown, I've split the data frame using: splitdf - split(df, as.factor(df$DateTime)) so that I can later use lapply to work on each block of data. I thought I could use cumsum and set the NA's to the desired interval to create the date/time sequence starting with the first row. However, this function is not defined for chron objects. Does anybody know of alternatives to create such a sequence? The 'zoo' package has na.locf which stands for Last Occurrence Carried Forward, which is what I believe you want. First let us create some test data, x: library(chron); library(zoo) x - chron(c(1.5, 2, NA, NA, 4, NA)) x [1] (01/02/70 12:00:00) (01/03/70 00:00:00) (NA NA) [4] (NA NA) (01/05/70 00:00:00) (NA NA) # na.locf is intended for zoo objects but we can convert # the chron object to zoo, apply na.locf and convert back: chron(as.vector(na.locf(zoo(as.vector(x) [1] (01/02/70 12:00:00) (01/03/70 00:00:00) (01/03/70 00:00:00) [4] (01/03/70 00:00:00) (01/05/70 00:00:00) (01/05/70 00:00:00) Just to reply to my own post, it can actually be done even more simply: chron(na.locf(as.vector(x))) Also in re-reading my post, I think the O in locf stands for observation rather than occurrence. __ 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
Re: [R] Combinations with two part column
Dear to Gabor Grothendieck and James Holtman, Thank you for giving me so much of your time to solve my problem. Many thanks and best regards, Sofyan On 5/17/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try write.table: write.table(comb8.5[,1:5], sep = ,, row.names = FALSE, col.names = FALSE) write.table(comb8.5[,6:8], sep = ,, row.names = FALSE, col.names = FALSE) On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Thanks for you quick answer. Could I extend my question? How to make the result for each rows with comma ,; library(gtools) comb8.5 - t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x comb8.5[,1:5] [,1] [,2] [,3] [,4] [,5] [1,]12345 [2,]12346 [3,]12347 [4,]12348 [5,]12356 [6,]12357 ... I mean like: 1,2,3,4,5 1,2,3,4,6 1,2,3,4,7 1,2,3,4,8 1,2,3,5,6 1,2,3,5,7 comb8.5[,6:8] [,1] [,2] [,3] [1,]678 [2,]578 [3,]568 [4,]567 [5,]478 [6,]468 ... this below like: 6,7,8 5,7,8 5,6,8 5,6,7 4,7,8 4,6,8 Best, Sofyan On 5/17/05, Gabor Grothendieck [EMAIL PROTECTED] wrote: On 5/17/05, Sofyan Iyan [EMAIL PROTECTED] wrote: Dear R-helpers, I am a beginner using R. This is the first question in this list. My question, Is there possible to make combinations with two part column? If I have a number 1,2,3,4,5,6,7,8. I need the result something like below: 1,2,3,4,5 6,7,8 1,2,3,4,7 5,6,8 2,3,4,5,6 1,7,8 1,2,3,6,7 4,5,8 1,2,3,4,8 5,6,7 3,4,6,7,8 1,2,5 Try this: library(gtools) t(apply(combinations(8,5), 1, function(x) c(x,setdiff(1:8, x __ 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 __ 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
Re: [R] cumsum on chron objects
On 5/17/05, Sebastian Luque [EMAIL PROTECTED] wrote: Hi, Is there some alternative to cumsum for chron objects? I have data frames that contain some chron objects that look like this: DateTime 13/10/03 12:30:35 NA NA NA 15/10/03 16:30:05 NA NA ... and I've been trying to replace the NA's so that a date/time sequence is created starting with the preceding available value. Because the number of rows with NA's following each available date/time is unknown, I've split the data frame using: splitdf - split(df, as.factor(df$DateTime)) so that I can later use lapply to work on each block of data. I thought I could use cumsum and set the NA's to the desired interval to create the date/time sequence starting with the first row. However, this function is not defined for chron objects. Does anybody know of alternatives to create such a sequence? The 'zoo' package has na.locf which stands for Last Occurrence Carried Forward, which is what I believe you want. First let us create some test data, x: library(chron); library(zoo) x - chron(c(1.5, 2, NA, NA, 4, NA)) x [1] (01/02/70 12:00:00) (01/03/70 00:00:00) (NA NA) [4] (NA NA) (01/05/70 00:00:00) (NA NA) # na.locf is intended for zoo objects but we can convert # the chron object to zoo, apply na.locf and convert back: chron(as.vector(na.locf(zoo(as.vector(x) [1] (01/02/70 12:00:00) (01/03/70 00:00:00) (01/03/70 00:00:00) [4] (01/03/70 00:00:00) (01/05/70 00:00:00) (01/05/70 00:00:00) __ 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
Re: [R] Centered overall title with layout()
Dear Pierre, On May 15, 2005, at 6:36 PM, Lapointe, Pierre wrote: Hello, I would like to have a centered overall title for a graphics page using the layout() function. Example, using this function: z - layout(matrix(c(1:6), 3,2, byrow = TRUE)) layout.show(6) I'd like to get this: Centered Overall Title | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | I really want to use layout(), not par(mfrow()) Thanks Pierre Lapointe Does mtext give you what you want? E.g., par(oma = c(0, 0, 3, 0)) z - layout(matrix(c(1:6), 3,2, byrow = TRUE)) layout.show(6) mtext(Centered Overall Title, side = 3, line = 1, outer = TRUE) Hope this helps, Stephen __ 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
Re: [R] cumsum on chron objects
Hello Gabor, Thanks for your reply. na.locf would replace the NA's with the most recent non-NA, so it wouldn't create a sequence of chron dates/times (via as.vector, as in your example). To expand my original example: On 5/17/05, Sebastian Luque [EMAIL PROTECTED] wrote: [...] DateTime 13/10/03 12:30:35 NA NA NA 15/10/03 16:30:05 NA NA ... I thought one could replace the NA's by the desired interval, say 1 day, so if the above chron object was named nachron, one could do: nachron[is.na(nachron)] - 1 and, for simplicity, applying on each block separately: cumsum(nachron) would give: DateTime 13/10/03 12:30:35 14/10/03 12:30:35 15/10/03 12:30:35 16/10/03 12:30:35 for the first block, and: DateTime 15/10/03 16:30:05 16/10/03 16:30:05 17/10/03 16:30:05 ... for the second one. Since there are not too many blocks I may end up doing it in Excel, but it would be nice to know how to do it in R! Cheers and thank you, -- Sebastian P. Luque __ 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
