[R] Writing list object to a file
Hi all, I am wondering how to write a 'list' object to a file. I already gone through some threads like http://mail.python.org/pipermail/python-list/2001-April/080639.html, however could not trace out any reliable solution. I tried following : write.table(calc, file=c:/data.csv) Error in data.frame(200501 = c(-0.000387071806652095, -0.000387221689252648, : arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27, 19, 23, 20, 11 Anybody can help? Regards, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] find directories
Hello, Function dir() is very useful for finding files. However, is there a analog function for finding directories based eg on a pattern ? Apologies if I've missed something obvious. Thanks Vincent __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] problem with data for metaMDS analysis please help
On Tue, 2008-04-22 at 20:19 -0400, stephen sefick wrote:
am at my wit's end. I am not sure what is wrong with this data matrix. It
is sparse because it is a matrix of species, but I have looked at the row
totals and column totals and they are positive.
rmetaMDS(x.d)
Error in if (autotransform xam 50) { :
missing value where TRUE/FALSE needed
What is wrong? And in the future how in God's name do I easily diagnose
whatever problem there is with this data. I am in debt to anyone who
figures this one out.
[All this presumes that x.d is your data as in the file you attached.
You could have helped by including the few lines to read in the data and
do the metaMDS so I didn't have to work this out and guess what x.d
was.]
The problem is very simple; you don't have *all* positive marginal
totals with the dataset you attached. Two variables have missing values
in the R sense, and unless you ask for R to remove missing values (which
in this case would defeat the object of checking the marginal totals),
you can't have found all positive values using the common R tools for
marginal totals.
Take a closer look at your data (that you attached). There are two
columns with NA in them.
This is how I diagnosed the problem:
# read in the data
dat - read.delim(trans.txt)
# generate species sums these will be NA is any are missing
csum - colSums(dat)
# check if any are missing
any(is.na(csum))
[1] TRUE
# yes, some missing, so which ones?
which(is.na(csum))
X49 X68
49 67
# Check how many are missing
summary(dat[, c(X49,X68)])
X49 X68
Min. : 0.00 Min. :0.0
1st Qu.: 0.00 1st Qu.:0.0
Median : 1.00 Median :0.0
Mean : 9.13 Mean :0.02094
3rd Qu.: 6.00 3rd Qu.:0.0
Max. :257.00 Max. :2.0
NA's : 1.00 NA's :1.0
These are rudimentary data management and checking steps in R. It would
be worth brushing up on these sorts of things in one of the online
documents. Alternatively, you could have included what you did that lead
you to believe that you had all positive marginal totals so we could see
where you went wrong.
HTH
G
--
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ECRC, UCL Geography, [f] +44 (0)20 7679 0565
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Re: [R] Writing list object to a file
Previous question that I asked was originated from this problem : suppose I have following time series : library(zoo) date1 = seq(as.Date(01/01/01, format = %m/%d/%y), as.Date(12/31/08, format = %m/%d/%y), by = 1) len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=0.5), nrow = len1), date1) Now I group those time series observation month wise : data2 = split(as.data.frame(data1), format(index(data1), %Y%m) Now I want to perform Levene test ( http://en.wikipedia.org/wiki/Levene's_test) to test whether the variance for each group (here month) is significantly different or not. Can anyone suggest me any easiest way how to construct the test statistic, described above? Thanks On Wed, Apr 23, 2008 at 12:21 PM, Arun Kumar Saha [EMAIL PROTECTED] wrote: Hi all, I am wondering how to write a 'list' object to a file. I already gone through some threads like http://mail.python.org/pipermail/python-list/2001-April/080639.html, however could not trace out any reliable solution. I tried following : write.table(calc, file=c:/data.csv) Error in data.frame(200501 = c(-0.000387071806652095, -0.000387221689252648, : arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27, 19, 23, 20, 11 Anybody can help? Regards, -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] select rows from data based on a vector of char strings
Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 etc And I have a vector flist-c(fun,food) Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food 3 3 When I do 'data$label==flist[1]' I get 'F T F F', so it works for one item in the char vector flist. But, when I do 'data$label==flist' I get 'F F F F' while I expected 'F T F T'. It seems that I can't perform this action with a vector of charstrings? Is there an other way to do so? -- View this message in context: http://www.nabble.com/select-rows-from-data-based-on-a-vector-of-char-strings-tp16832735p16832735.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] select rows from data based on a vector of char strings
On 4/23/2008 3:13 AM, Dirkheld wrote: Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 etc And I have a vector flist-c(fun,food) Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food 3 3 When I do 'data$label==flist[1]' I get 'F T F F', so it works for one item in the char vector flist. But, when I do 'data$label==flist' I get 'F F F F' while I expected 'F T F T'. It seems that I can't perform this action with a vector of charstrings? Is there an other way to do so? newdata - subset(data, label %in% flist) ?subset ?is.element -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Documentation General Comments
Good morning, Firstly I'd like to say that I'm a huge fan of R and I think it's great system. Part of the problem in searching for information is knowing what buzzwords / keywords to use. I was recently caught out like this as I didn't see my problem as a cumulative sum (keyword=cumsum) only as referencing one line of a dataframe from another. Academic papers and certain webpages add special classification keywords to the text of a page to help. Searching is a general problem - not just within R - ask any archivist or librarian! A partial solution is to have disambiguation pages, e.g. http://en.wikipedia.org/wiki/Comma Is it reasonable to have help pages with no specific R / package item behind it only a See Also section? Does somebody have access to the most frequent RSiteSearch() terms? It would probably help to increase the number of See Also details - for example when I run into a problem the first thing I do is try to recreate it as a reproduceable toy problem which I could send to this list (which incidentally is actually a great way of figuring out solution to the problem without having to bother the list!!!). To do this I invariably want to generate some random numbers, and I can never remember the names runif() or rnorm() so I say help.search(random) which doesn't actually reference either runif() or rnorm() directly so I look at ?RNG which leads me to rnorm() - and already knowing that this is what I'm looking for I'm ok - but if somebody didn't already know this it is not obvious. I appreciate that there is always a difficult balance when writing documentation between having enough and too much. Just looking at the core documentation for R-2.6.2 (and ignoring the many many additional packages) The introduction to R is 100 pages of PDF and the reference manual runs to 1,576 pages of PDF. Adding more information as many of us want would make the reference manual even more unwieldy and far too big to print out to peruse, which gives rise to a market for books which take over where the introduction manual leaves off... Part of the difficulty that we encounter is that sometimes our difficulties are pure R, and other times the difficulty is statistical or mathematical - more often than not the problem is between the two... and frequently those of us asking the question don't actually know where on the spectrum it is... Q: Could there be ways other than submitting a bug / patch to help improve R? Q: Should this discussion be on r-devel or r-help? Best Regards, Sean The root of the problem is that R is a voluntary/cooperative project and those who develop and maintain R are (generously) contributing their time and probably have little-to-no time left over to devote to the improvement of the documentation. snip... This is why the documentation tends to be opaque in the first place. The people who build R are so clever and understand so much that they cannot put themselves in the shoes of those of us who are not so blessed with intelligence and erudition. So they (often) write terse cryptic instructions which (often) depend on background knowledge that many of us lack. That background knowledge can of course be found ***if you know where to look*** --- or even if you don't, given that you are prepared to put in sufficient time and effort searching ***and*** are clever at searching. It's that last requirement that leaves *me* out in the cold. snip... -- View this message in context: http://www.nabble.com/Documentation-General-Comments-tp16821085p16833353.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regression inclusion of variable, effect on coefficients
At 20:06 21/04/2008, Thiemo Fetzer wrote: Hello :) I am happy to hear that I am not necessarily asking stupid questions. The thing is, that I have data on x1 and x4 for the whole sample. However, theoretically, it is clear that the informational content of x1 is not as high as of x4. x4 provides more accurate information to the subjects participating in the game, as it has been experimentally and theoretically shown that the x1 is biased. It sounds as though you are thinking of some sort of path analysis or structural equation model. If you look at the sem package you should find some links to helpful material by John Fox and located on his web site (I think). Of course I could be quite wrong here and you may want something completely different. I am not advocating SEMs if they are not needed as they provide many temptations to dredge into your data but you do seem to have a theory to test. So the experimentators introduced x4 in response to the biased x1. Both prevail however together, so that the subjects have available information on x1 and x4. Theoretically, I argued that the relative importance of x1 on y will decrease in light that information x4 is available, as x4 is more accurate. With a simple regression, however, I do not find significant relationships. For x1 it has been empirically and theoretically shown that it has a positive effect on y. The same should hold for x4. There is no necessary theoretical argument as how x1 and x4 interact mathematically, as they both are a measure of the same thing. Yet, x4 is more accurate and contains even more information. It could be any kind of interaction. They are positively correlated, which is also reasonable. Could you suggest me a simple interaction model, with which I could try my luck? Thanks a lot Thiemo -Original Message- From: Uwe Ligges [mailto:[EMAIL PROTECTED] Sent: Montag, 21. April 2008 18:54 To: Thiemo Fetzer Cc: r-help@r-project.org Subject: Re: [R] Regression inclusion of variable, effect on coefficients This is not a dump question. This is a serious problem and it depends on what you know or assume about the relastionship between x1 and x4. If you assume linear interaction, you might want to introduce some interaction term to the model for example. Uwe Ligges Thiemo Fetzer wrote: Hello dear R users! I know this question is not strictly R-help, yet, maybe some of the guru's in statistics can help me out. I have a sample of data all from the same population. Say my regression equation is now this: m1 - lm(y ~ x1 + x2 + x3) I also regress on m2 - lm(y ~ x1 + x2 + x3 + x4) The thing is, that I want to study the effect of information x4. I would hypothesize, that the coefficient estimate for x1 goes down as I introduce x4, as x4 conveys some of the information conveyed by x1 (but not only). Of course x1 and x4 are correlated, however multicollinearity does not appear to be a problem, the variance inflation factors are rather low (around 1.5 or so). I want to basically study, how the interplay between x1 and x4 is, when introducing x4 into the regression equation and whether my hypothesis is correct; i.e. that given I consider the information x4, not so much of the variation is explained via x1 anymore. I observe that introducing x4 into the regression, the coefficient estimate for x1 goes down; also the associated p-value becomes bigger; i.e. x1 becomes comparatively less significant. However, x4 is not significant. Yet, the observation is in line with my theoretical argument. The question is now simple: how can I work this out? I know this is likely a dumb question, but I would really appreciate some links or help. Regards Thiemo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Michael Dewey http://www.aghmed.fsnet.co.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] find directories
On Wed, 23 Apr 2008, mel wrote:
Hello,
Function dir() is very useful for finding files.
However, is there a analog function for finding directories
based eg on a pattern ?
No.
Apologies if I've missed something obvious.
list.dirs - function(...)
{
x - dir(...)
x[file_test(-d, x)]
}
Thanks
Vincent
--
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] find directories
Prof Brian Ripley a écrit :
On Wed, 23 Apr 2008, mel wrote:
Hello,
Function dir() is very useful for finding files.
However, is there a analog function for finding directories
based eg on a pattern ?
No.
Apologies if I've missed something obvious.
list.dirs - function(...)
{
x - dir(...)
x[file_test(-d, x)]
}
Thanks for the fast answer Prof Ripley.
In fact, I was using (on a tree of directories of depth 4)
x - dir(..., recursive=TRUE)
and didn't notice that the results from
dir(...) is not included in dir(..., recursive=TRUE)
If I'm not wrong, x - dir(..., recursive=TRUE)
seems to return the findable *files only* in the tree.
While x - dir(...)
returns the *files and the directories* present at the top level
of the tree.
The directories I'm searching are unfortunately not at the top level,
But, in my special case, I'm lucky enough that a simple trick allows
to identify the directories.
Thanks
Vincent
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Re: [R] how to read in multiple files with unequal number of columns
Thank you John.
It was useful to know about this package.
I tried merge_all and I got this error:
Error in .subset2(x, i, exact = exact) : subscript out of bounds
It could be due to the way my data is and I will try the other
solutions suggested by the other kind souls on this list.
Best wishes,
tania
On 22 Apr 2008, at 19:29, John Kane wrote:
You might want to have a look at the merge_all
function in the reshape package.
--- Tania Oh [EMAIL PROTECTED] wrote:
Dear all,
I want to read in 1000 files which contain varying
number of columns.
For example:
file[1] contains 8 columns (mixture of characters
and numbers)
file[2] contains 16 columns etc
I'm reading everything into one big data frame and
when I try rbind, R
returns an error of
Error in rbind(deparse.level, ...) :
numbers of columns of arguments do not match
Below is my code:
all - NULL
all - as.data.frame(all)
##read in the contents of the files
for (f in 1:length(fnames)){
tmp - try(read.table(fnames[f], header=F,
fill=T, sep=\t),
TRUE)
if (class(tmp) == try-error) {
next ## skip this file if it's
empty/non-existent
}else{
## combine all the file contents into one
big data frame
all - rbind(all, tmp)
}
}
Here is some example of what the data in the files:
L3 - LETTERS[1:3]
(d - data.frame(cbind(x=1, y=1:10), fac=sample(L3,
10, replace=TRUE)))
str(d)
'data.frame':10 obs. of 3 variables:
$ x : num 1 1 1 1 1 1 1 1 1 1
$ y : num 1 2 3 4 5 6 7 8 9 10
$ fac: Factor w/ 3 levels A,B,C: 1 3 1 2 2 2
2 1 1 2
my.fake.data - data.frame(cbind(x=1, y=2))
str(my.fake.data)
'data.frame':1 obs. of 2 variables:
$ x: num 1
$ y: num 2
all - rbind(d, my.fake.data)
Error in rbind(deparse.level, ...) :
numbers of columns of arguments do not match
I've searched the R-site but couldn't find any
relevant solution.I
might have used the wrong keywords to search, so if
this question has
been answered already, I'd be very grateful if
someone could point me
to the post. Else any help/suggestions would be
greatly appreciated.
Many thanks in advance,
tania
D.Phil student
Department of Physiology, Anatomy and Genetics
University of Oxford
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Re: [R] optimization setup
The error comes from the way you specify the parameters:
At least not as elegant, but it works:
function4 - function(x){
theta1 - x[1]
theta2 - x[2]
theta3 - x[3]
function3(theta1,theta2,theta3)
}
fit-optim(par=c(1, 1.2, .2), fn=function4)
Bart
threshold wrote:
Hi, here comes my problem, say I have the following functions (example
case)
#
function1 - function (x, theta)
{a - theta[1] ( 1 - exp(-theta[2]) ) * theta[3] )
b - x * theta[1] / theta[3]^2
return( list( a = a, b = b )) }
#---
function2-function (x, theta)
{P - function1(x, theta)
c - P$a * x * exp(-theta[2])
d - P$b * exp(x)
q - theta[1] / theta[3]
res - c + d + q; res}
# Function to be optimized
function3 - function(theta1,theta2,theta3) {
n - length(data)
-sum( function2(x = data[2:n], theta = c(theta1, theta2, theta3) ))}
# 'data' is my input ts class object
#--
Then I want to maximize function3 with respect to theta(s) (given some
starting values)
fit-optim(par=c(theta1=1, theta2=1.2, theta3=.2), fn=function3)
I get the following:
Error in function1(x, theta) :
argument theta2 is missing, with no default
Where I made a mistake? I will appreciate any help ...
r
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[R] Problems with Windows RGUI R-2.7.0 when PA TH includes 'æ'
I have a subdirectory named 'Fælles Filer' in my PATH. When I start the RGUI, I receive the following error/warning messages: Error in Sys.setenv(PATH = PATH) : invalid input in wtransChar and Warning message: package methods in options(defaultPackages) was not found The problems disappear, when I change the the subdirectory name to its short version, 'FLLES~1' in the PATH specification in the Windows setup. regards Erik Jørgensen Faculty of Agricultural Sciences University of Aarhus, Denmark platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 7.0 year 2008 month 04 day22 svn rev45424 language R version.string R version 2.7.0 (2008-04-22) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Time arithmetic
I am a bit worried I am reinventing the wheel. Isn't there a calendar
system in R?
I have written a function to add months to a date and return the number of
days resulting. I am newish to R so I am hoping there is a package that
can do this sort of date arithmetic for me...
Worik
DaysInMonths - function(s,d){
## Days in d months from s
sdate - MSTD(s)
## Get day, month and the year
sd - as.double(days(sdate))
sm - as.double(months(sdate))
sy - as.double.difftime(years(sdate))
sm - sm+d
while(sm 12){
sm - sm-12
sy - sy+1
}
sdate2 - MSTD(paste(sd,sm,sy,sep=/))
return(as.double(difftime(sdate2, sdate, units=days)))
}
MSTD - function(a){
if(class(a)[1] == dates) return (a)
return(dates(a, format=d/m/y))
}
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[R] R appears not to reallocate memory on linux
Hi, I have a question concerning the memory management of R 2.6.1 on a 32-bit linux with 3.2 GB RAM. Especially, I found the current memory size reported from gc() to be different from the amount reported by e.g. top. The 'max used' number of gc() seems more to match the actually used memory. so is it the case that R keeps the allocated memory for its own memory manager and does not return it to the OS? this is what the situation looks like, although what i read in the newsgroups sounded like the opposite (e.g. R gives all unneeded memory back) this is especially important for me since i am processing huge data and evaulate different learning methods, and i frequently ran into 'cannot allocate vector of (say) 70M' any hint would be appreciated thanks a lot lars __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] new package multipol
Hello List please find a new package, multipol, recently uploaded to CRAN. This package generalizes the polynom package (which handles univariate polynomials) to the multivariate case. A short article discussing the package will appear in the next issue of Rnews, Insha'Allah enjoy -- Robin Hankin Uncertainty Analyst and Neutral Theorist, National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743 ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] g(x,y) = f(x,y) - e(x)- e(y)?
Thanks Phipp very much for your help. I had meant, given that I'd
computed the matrix f[x,y] and the vector e[x], how to take the
difference. What is confusing is how to subtract a vector from a
matrix. I don't want the recycling rule.
Cheers
Bill
On Tue, Apr 22, 2008 at 9:53 AM, Philipp Pagel [EMAIL PROTECTED] wrote:
g(x,y) = f(x,y) - e(x)- e(y)
These are continuous functions. I am not sure how to do this with the
discrete equivalents in R.
Is this what you are looking for?
g - function(x, y) {
f(x,y) - e(x) - e(y)
}
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://mips.gsf.de/staff/pagel
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[R] Density estimation
Hi, I am analysing a dataset containing genetic distances within and between species. I want to show a overlap of the distributions of the intra- and interspecific values; on a second graph I use a cut-off value to determine these boundaries. As the dataset contains 30 000 values, I would like to do this with a simple line rather than superimposed histograms. Hence, density plots. With the standard settings of plot(density(x)), I receive the desirable result, except that the function extends slightly in negative x values (which is impossible, distance values are always positive). Furthermore, in the second figure I supplied the cut-off value a priori, so overlap between the two classes is zero by definition. Is there a way to visualise this information in another way, or to readjust the density parameters so intraspecific values below zero have zero probability in the first case, and there is no overlap in the second case? Thanks, gunther __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems with Windows RGUI R-2.7.0 when P ATH includes 'æ'
Your email was not properly encoded by the time it reached me (see the ? in crucial places), so I may be missing something. I am guessing you are in a Danish locale with codepage CP1252, and fortunately the subject line appears to have worked. I think I have a reproduction test, in which case this will be fixed in R-patched shortly (but please test that for us). No one reported this in the alpha/beta/RC test period, and we do know that the Japanese did some testing of the new features. I guess that real testing started yesterday Thanks for the report -- if you find further misfeatures it would be helpful to have the locale and an email in a marked encoding. On Wed, 23 Apr 2008, Erik Jørgensen wrote: I have a subdirectory named 'F?lles Filer' in my PATH. When I start the RGUI, I receive the following error/warning messages: Error in Sys.setenv(PATH = PATH) : invalid input in wtransChar and Warning message: package methods in options(defaultPackages) was not found The problems disappear, when I change the the subdirectory name to its short version, 'FLLES~1' in the PATH specification in the Windows setup. regards Erik J?rgensen Faculty of Agricultural Sciences University of Aarhus, Denmark platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 7.0 year 2008 month 04 day22 svn rev45424 language R version.string R version 2.7.0 (2008-04-22) [[alternative HTML version deleted]] -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Time arithmetic
library(zoo)
?yearmon
library(chron)
?month.day.year
?strptime
See R News 4/1.
On Wed, Apr 23, 2008 at 6:13 AM, Worik R [EMAIL PROTECTED] wrote:
I am a bit worried I am reinventing the wheel. Isn't there a calendar
system in R?
I have written a function to add months to a date and return the number of
days resulting. I am newish to R so I am hoping there is a package that
can do this sort of date arithmetic for me...
Worik
DaysInMonths - function(s,d){
## Days in d months from s
sdate - MSTD(s)
## Get day, month and the year
sd - as.double(days(sdate))
sm - as.double(months(sdate))
sy - as.double.difftime(years(sdate))
sm - sm+d
while(sm 12){
sm - sm-12
sy - sy+1
}
sdate2 - MSTD(paste(sd,sm,sy,sep=/))
return(as.double(difftime(sdate2, sdate, units=days)))
}
MSTD - function(a){
if(class(a)[1] == dates) return (a)
return(dates(a, format=d/m/y))
}
[[alternative HTML version deleted]]
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Re: [R] Ubuntu vs. Windows
On 4/22/08, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Tue, 22 Apr 2008, Peter Dalgaard wrote: Doran, Harold wrote: Dear List: I am very much a unix neophyte, but recently had a Ubuntu box installed in my office. I commonly use Windows XP with 3 GB RAM on my machine and the Ubuntu machine is exactly the same as my windows box (e.g., processor and RAM) as far as I can tell. Now, I recently had to run a very large lmer analysis using my windows machine, but was unable to due to memory limitations, even after increasing all the memory limits in R (which I think is a 2gig max according to the FAQ for windows). So, to make this computationally feasible, I had to sample from my very big data set and then run the analysis. Even still, it would take something on the order of 45 mins to 1 hr to get parameter estimates. (BTW, SAS Proc nlmixed was even worse and kept giving execution errors until the data set was very small and then it ran for a long time) However, I just ran the same analysis on the Ubuntu machine with the full, complete data set, which is very big and lmer gave me back parameter estimates in less than 5 minutes. Because I have so little experience with Ubuntu, I am quite pleased and would like to understand this a bit better. Does this occur because R is a bit friendlier with unix somehow? Or, is this occuring because unix somehow has more efficient methods for memory allocation? Probably partly the latter and not the former (we try to make the most of what the OS offers in either case), but a more important difference is that we can run in 64 bit address space on non-Windows platforms (assuming that you run a 64 bit Ubuntu). Even with 64 bit Windows we do not have the 64 bit toolchain in place to build R except as a 32 bit program. Creating such a toolchain is beyond our reach, and although progress is being made, it is painfully slow (http://sourceforge.net/projects/mingw-w64/). Every now and then, the prospect of using commercial tools comes up, but they are not plug-compatible and using them would leave end users without the possibility of building packages with C code, unless they go out and buy the same toolchain. There is another possibility. lmer is heavy on matrix algebra, and so usually benefits considerably from an optimized BLAS. Under Windows you need to download one of those on CRAN (or build your own). I believe that under Ubuntu R will make use of one if it is already installed. Optimized BLAS is a possible explanation but it would depend on the Ubuntu package for the correct version of Atlas having been installed. I don't think those packages are installed by default. Even if they were installed, optimized BLAS are not always beneficial for lmer. Depending on the structure of the model, optimized BLAS, especially multithreaded BLAS, can actually slow lmer down. I think the difference is more likely due to swapping. A typical lmer call does considerable memory allocation at the beginning of the computation then keeps a stable memory footprint during the optimization of the deviance with respect to the model parameters. It does access essentially all the big chunks of memory in that footprint during the optimization. If the required memory is a bit larger than the available memory you get a lot of swapping, as I found out yesterday. I started an lmer run on a 64-bit Ubuntu machine forgetting that I had recently removed a defective memory module from that machine. It had only 2 GB of memory and about 8 GB of swap space. It spent a lot of time swapping. I definitely should have done that run on one of our servers that has much more real memory. Harold: Typing either cat /proc/meminfo or free in a shell window on your Ubuntu machine will tell you the amount of memory and swap space on the machine. If you start the lmer fit and switch to a terminal window where you run the program top you can watch the evolution of the memory usage by the R program. It will probably increase at the beginning of the run then stabilize. Harold __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 at step 1
Hi I Can't connect to the Oracle database, any tips? Has anybody actually got ROracle up and running on windows? unable to find an inherited method for function dbConnect, for signature OraDriver I can happily connect to the same database through RODBC. Oracle client, version 9.2 installed, amongst others. Sean library(DBI) library(ROracle) packageDescription(ROracle) Package: ROracle Version: 0.5-7 Date: 2006-02-13 Title: Oracle database interface for R Author: David A. James [EMAIL PROTECTED] Jake Luciani [EMAIL PROTECTED] Maintainer: David A. James [EMAIL PROTECTED] Description: Oracle database interface (DBI) driver for R. This is a DBI-compliant Oracle driver based on the ProC/C++ embedded SQL. It implements the DBI version 0.1-8 plus one extension. SaveImage: yes Depends: R (= 2.0.0), methods, DBI (= 0.1-8) License: LGPL version 2 or newer URL: http://stat.bell-labs.com/RS-DBI http://www.omegahat.org Packaged: Mon Feb 13 18:05:55 2006; dj Built: R 2.2.1; i386-pc-mingw32; 2006-02-13 18:05:59; windows -- File: C:/PROGRA~1/R/R-26~1.0/library/ROracle/DESCRIPTION packageDescription(DBI) Package: DBI Version: 0.2-4 Title: R Database Interface Author: R Special Interest Group on Databases (R-SIG-DB) Maintainer: David A. James [EMAIL PROTECTED] Depends: R (= 2.3.0), methods Imports: methods Description: A database interface (DBI) definition for communication between R and relational database management systems. All classes in this package are virtual and need to be extended by the various R/DBMS implementations. LazyLoad: yes License: LGPL (version 2 or later) Collate: DBI.R Util.R zzz.R Packaged: Tue Oct 16 21:43:31 2007; dj Built: R 2.6.0; ; 2007-10-17 12:21:14; windows -- File: C:/PROGRA~1/R/R-26~1.0/library/DBI/DESCRIPTION R.Version() $platform [1] i386-pc-mingw32 $arch [1] i386 $os [1] mingw32 $system [1] i386, mingw32 $status [1] $major [1] 2 $minor [1] 6.0 $year [1] 2007 $month [1] 10 $day [1] 03 $`svn rev` [1] 43063 $language [1] R $version.string [1] R version 2.6.0 (2007-10-03) sDB_un- scott sDB_pwd- tiger Oracle() An object of class OraDriver Slot Id: [1] 1952 dbDriver(Oracle) An object of class OraDriver Slot Id: [1] 1952 Ora- dbDriver(Oracle) Con - dbConnect(Ora, user = sDB_un, passwd = sDB_pwd) Error in function (classes, fdef, mtable) : unable to find an inherited method for function dbConnect, for signature OraDriver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 read in multiple files with unequal number of columns
Is this what you want? I am assuming that you will read the
dataframes into a list and then process them like below:
# put dataframe in a list -- would have read them in via a list
x - list(d, my.fake.data)
# determine maximum number of columns and then pad out the short one
# also use the column names of the largest one
col.max - max(sapply(x, ncol))
colNames - lapply(x, function(.data){
+ if (ncol(.data) == col.max) colnames(.data)
+ })[[1]]
new.data - lapply(x, function(.data){
+ if (ncol(.data) col.max){
+ .data[(ncol(.data) + 1):col.max] - NA
+ colnames(.data) - colNames
+ }
+ .data
+ })
all - do.call(rbind, new.data)
all
x y fac
1 1 1B
2 1 2B
3 1 3B
4 1 4B
5 1 5A
6 1 6A
7 1 7C
8 1 8C
9 1 9A
10 1 10C
11 1 2 NA
On Tue, Apr 22, 2008 at 9:05 AM, Tania Oh [EMAIL PROTECTED] wrote:
Dear all,
I want to read in 1000 files which contain varying number of columns.
For example:
file[1] contains 8 columns (mixture of characters and numbers)
file[2] contains 16 columns etc
I'm reading everything into one big data frame and when I try rbind, R
returns an error of
Error in rbind(deparse.level, ...) :
numbers of columns of arguments do not match
Below is my code:
all - NULL
all - as.data.frame(all)
##read in the contents of the files
for (f in 1:length(fnames)){
tmp - try(read.table(fnames[f], header=F, fill=T, sep=\t),
TRUE)
if (class(tmp) == try-error) {
next ## skip this file if it's empty/non-existent
}else{
## combine all the file contents into one big data frame
all - rbind(all, tmp)
}
}
Here is some example of what the data in the files:
L3 - LETTERS[1:3]
(d - data.frame(cbind(x=1, y=1:10), fac=sample(L3, 10, replace=TRUE)))
str(d)
'data.frame': 10 obs. of 3 variables:
$ x : num 1 1 1 1 1 1 1 1 1 1
$ y : num 1 2 3 4 5 6 7 8 9 10
$ fac: Factor w/ 3 levels A,B,C: 1 3 1 2 2 2 2 1 1 2
my.fake.data - data.frame(cbind(x=1, y=2))
str(my.fake.data)
'data.frame': 1 obs. of 2 variables:
$ x: num 1
$ y: num 2
all - rbind(d, my.fake.data)
Error in rbind(deparse.level, ...) :
numbers of columns of arguments do not match
I've searched the R-site but couldn't find any relevant solution.I
might have used the wrong keywords to search, so if this question has
been answered already, I'd be very grateful if someone could point me
to the post. Else any help/suggestions would be greatly appreciated.
Many thanks in advance,
tania
D.Phil student
Department of Physiology, Anatomy and Genetics
University of Oxford
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] Writing list object to a file
?save On Wed, Apr 23, 2008 at 2:51 AM, Arun Kumar Saha [EMAIL PROTECTED] wrote: Hi all, I am wondering how to write a 'list' object to a file. I already gone through some threads like http://mail.python.org/pipermail/python-list/2001-April/080639.html, however could not trace out any reliable solution. I tried following : write.table(calc, file=c:/data.csv) Error in data.frame(200501 = c(-0.000387071806652095, -0.000387221689252648, : arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27, 19, 23, 20, 11 Anybody can help? Regards, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer model building--include random effects?
On 4/22/08, Ista Zahn [EMAIL PROTECTED] wrote: Hello, This is a follow up question to my previous one http://tolstoy.newcastle.edu.au/R/e4/help/08/02/3600.html I am attempting to model relationship satisfaction (MAT) scores (measurements at 5 time points), using participant (spouseID) and couple id (ID) as grouping variables, and time (years) and conflict (MCI.c) as predictors. I have been instructed to include random effects for the slopes of both predictors as well as the intercepts, and then to drop non-significant random effects from the model. The instructor and the rest of the class is using HLM 6.0, which gives p- values for random effects, and the procedure is simply to run a model, note which random effects are not significant, and drop them from the model. I was hoping I could to something analogous by using the anova function to compare models with and without a particular random effect, but I get dramatically different results than those obtained with HLM 6.0. For example, I wanted to determine if I should include a random effect for the variable MCI.c (at the couple level), so I created two models, one with and one without, and compared them: m.3 - lmer(MAT ~ 1 + years + MCI.c + (1 + years | spouseID) + (1 + years + MCI.c | ID), data=Data, method = ML) m.1 - lmer(MAT ~ 1 + years + MCI.c + (1 + years + MCI.c | spouseID) + (1 + years + MCI.c | ID), data=Data, method = ML) anova(m.1, m.3) Data: Data Models: m.3: MAT ~ 1 + years + MCI.c + (1 + years | spouseID) + (1 + years + m.1: MCI.c | ID) m.3: MAT ~ 1 + years + MCI.c + (1 + years + MCI.c | spouseID) + (1 + m.1: years + MCI.c | ID) Df AIC BIC logLik Chisq Chi Df Pr(Chisq) m.3 12 5777.8 5832.7 -2876.9 m.1 15 5780.9 5849.5 -2875.4 2.9428 3 0.4005 The corresponding output from HLM 6.0 reads Random Effect Standard Variance dfChi- square P-value Deviation Component -- INTRCPT1, R0 6.80961 46.37075 60 112.809140.000 YEARS slope, R1 1.49329 2.22991 60 59.38729.500 MCI slope, R2 5.45608 29.76881 60 90.576150.007 -- To me, this seems to indicate that HLM 6.0 is suggesting that the random effect should be included in the model, while R is suggesting that it need not be. This is not (quite) a why do I get different results with X post, but rather an I'm worried that I might be doing something wrong post. Does what I've done look reasonable? Is there a better way to go about it The first thing I would try to determine is whether the model fit by HLM is equivalent to the model you have fit with lmer. The part of the HLM model output you have shown lists only variance components. It does not provide covariances or correlations. The lmer model is fitting a 3 by 3 symmetric positive definite variance-covariance matrix with a total of 6 parameters - 3 variances and 3 covariances. It may be that HLM is fitting a simpler model in which the covariances are all zero. The next question would be exactly how HLM is calculating that p-value. I wonder where the 60 degrees of freedom comes from. Do you happen to have 60 couples in the study? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ROracle error at step 1
Where did you get your ROracle package from? Built: R 2.2.1; i386-pc-mingw32; 2006-02-13 18:05:59; windows indicates it was too old. All methods-using packages need to have been installed recently, and definitely since 2.4.0. (It is unsafe to use one that has not been installed for the same 'y' in 2.y.z because of its propensity for extracting bits of the R under which the package was installed.) So 'step 1' is to install ROracle from the sources. On Wed, 23 Apr 2008, Creighton, Sean wrote: Hi I Can't connect to the Oracle database, any tips? Has anybody actually got ROracle up and running on windows? unable to find an inherited method for function dbConnect, for signature OraDriver I can happily connect to the same database through RODBC. Oracle client, version 9.2 installed, amongst others. Sean library(DBI) library(ROracle) packageDescription(ROracle) Package: ROracle Version: 0.5-7 Date: 2006-02-13 Title: Oracle database interface for R Author: David A. James [EMAIL PROTECTED] Jake Luciani [EMAIL PROTECTED] Maintainer: David A. James [EMAIL PROTECTED] Description: Oracle database interface (DBI) driver for R. This is a DBI-compliant Oracle driver based on the ProC/C++ embedded SQL. It implements the DBI version 0.1-8 plus one extension. SaveImage: yes Depends: R (= 2.0.0), methods, DBI (= 0.1-8) License: LGPL version 2 or newer URL: http://stat.bell-labs.com/RS-DBI http://www.omegahat.org Packaged: Mon Feb 13 18:05:55 2006; dj Built: R 2.2.1; i386-pc-mingw32; 2006-02-13 18:05:59; windows -- File: C:/PROGRA~1/R/R-26~1.0/library/ROracle/DESCRIPTION packageDescription(DBI) Package: DBI Version: 0.2-4 Title: R Database Interface Author: R Special Interest Group on Databases (R-SIG-DB) Maintainer: David A. James [EMAIL PROTECTED] Depends: R (= 2.3.0), methods Imports: methods Description: A database interface (DBI) definition for communication between R and relational database management systems. All classes in this package are virtual and need to be extended by the various R/DBMS implementations. LazyLoad: yes License: LGPL (version 2 or later) Collate: DBI.R Util.R zzz.R Packaged: Tue Oct 16 21:43:31 2007; dj Built: R 2.6.0; ; 2007-10-17 12:21:14; windows -- File: C:/PROGRA~1/R/R-26~1.0/library/DBI/DESCRIPTION R.Version() $platform [1] i386-pc-mingw32 $arch [1] i386 $os [1] mingw32 $system [1] i386, mingw32 $status [1] $major [1] 2 $minor [1] 6.0 $year [1] 2007 $month [1] 10 $day [1] 03 $`svn rev` [1] 43063 $language [1] R $version.string [1] R version 2.6.0 (2007-10-03) sDB_un- scott sDB_pwd- tiger Oracle() An object of class OraDriver Slot Id: [1] 1952 dbDriver(Oracle) An object of class OraDriver Slot Id: [1] 1952 Ora- dbDriver(Oracle) Con - dbConnect(Ora, user = sDB_un, passwd = sDB_pwd) Error in function (classes, fdef, mtable) : unable to find an inherited method for function dbConnect, for signature OraDriver __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Import_from_outputfile
Hello, I want to do some analyses with data originating from a hydrological model that are saved in multiple output files. I want to import the output-files in R without a treatment like preproceesing every output file. The output file is more or less a matrix, but with additional rows in the header, describing the type of data. My question is: 1. Is there a possibility to skip the rows or start with a certain row? I want to exclude row 2 and 3, so the first line become the header of the list below. This is the beginning of the output file to be imported. YYMMDDHH12345tot_average total runoff [mm per zone] (sum of QB, QI und QD)_submodel:unsatzon_model --------0.14666240.29580420.22250480.2279085 0.10712021. 2001 1 1 10.0552420.0373910.0515500.057681 0.0210700.046036 2001 1 1 20.0512880.0324040.0470270.056636 0.0210700.042736 2001 1 1 30.0524910.0329370.0480420.058415 0.0210700.043701 2001 1 1 40.0513890.0325080.0471360.056663 0.0210700.042812 2001 1 1 50.0507300.0322630.0465990.055585 0.0210700.042278 Would be a great help if there is a possibility. Thank you very much ! best regards Ole -- --- Ole RöÃler PhD - student Climatology and Landscape-Ecology Research Group Department of Geography University of Bonn Postal Address: Meckenheimer Allee 166 53115 Bonn Germany Tel.: +49-(0)228-737895 Fax.: +49-(0)228-737506 Email: [EMAIL PROTECTED] Homepage: [2]http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm - References 1. mailto:[EMAIL PROTECTED] 2. http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] lmer model building--include random effects?
On Apr 23, 2008, at 8:56 AM, Douglas Bates wrote: On 4/22/08, Ista Zahn [EMAIL PROTECTED] wrote: Hello, This is a follow up question to my previous one http://tolstoy.newcastle.edu.au/R/e4/help/08/02/3600.html I am attempting to model relationship satisfaction (MAT) scores (measurements at 5 time points), using participant (spouseID) and couple id (ID) as grouping variables, and time (years) and conflict (MCI.c) as predictors. I have been instructed to include random effects for the slopes of both predictors as well as the intercepts, and then to drop non-significant random effects from the model. The instructor and the rest of the class is using HLM 6.0, which gives p- values for random effects, and the procedure is simply to run a model, note which random effects are not significant, and drop them from the model. I was hoping I could to something analogous by using the anova function to compare models with and without a particular random effect, but I get dramatically different results than those obtained with HLM 6.0. For example, I wanted to determine if I should include a random effect for the variable MCI.c (at the couple level), so I created two models, one with and one without, and compared them: m.3 - lmer(MAT ~ 1 + years + MCI.c + (1 + years | spouseID) + (1 + years + MCI.c | ID), data=Data, method = ML) m.1 - lmer(MAT ~ 1 + years + MCI.c + (1 + years + MCI.c | spouseID) + (1 + years + MCI.c | ID), data=Data, method = ML) anova(m.1, m.3) Data: Data Models: m.3: MAT ~ 1 + years + MCI.c + (1 + years | spouseID) + (1 + years + m.1: MCI.c | ID) m.3: MAT ~ 1 + years + MCI.c + (1 + years + MCI.c | spouseID) + (1 + m.1: years + MCI.c | ID) Df AIC BIC logLik Chisq Chi Df Pr(Chisq) m.3 12 5777.8 5832.7 -2876.9 m.1 15 5780.9 5849.5 -2875.4 2.9428 3 0.4005 The corresponding output from HLM 6.0 reads Random Effect Standard Variance dfChi- square P-value Deviation Component -- INTRCPT1, R0 6.80961 46.37075 60 112.809140.000 YEARS slope, R1 1.49329 2.22991 60 59.38729 .500 MCI slope, R2 5.45608 29.76881 60 90.576150.007 -- To me, this seems to indicate that HLM 6.0 is suggesting that the random effect should be included in the model, while R is suggesting that it need not be. This is not (quite) a why do I get different results with X post, but rather an I'm worried that I might be doing something wrong post. Does what I've done look reasonable? Is there a better way to go about it The first thing I would try to determine is whether the model fit by HLM is equivalent to the model you have fit with lmer. The part of the HLM model output you have shown lists only variance components. It does not provide covariances or correlations. The lmer model is fitting a 3 by 3 symmetric positive definite variance-covariance matrix with a total of 6 parameters - 3 variances and 3 covariances. It may be that HLM is fitting a simpler model in which the covariances are all zero. Yes, I was also concerned that the model I fit in R may not be exactly the model fit by HLM. The estimates are similar but not exact. The model summaries from HLM and R are as follows: HLM OUTPUT: The outcome variable is MAT The model specified for the fixed effects was: Level-1Level-2 Level-3 Coefficients PredictorsPredictors - --- INTRCPT1, P0 INTRCPT2, B00INTRCPT3, G000 YEARS slope, P1 INTRCPT2, B10INTRCPT3, G100 * MCI slope, P2 INTRCPT2, B20INTRCPT3, G200 '*' - This variable has been centered around its group mean Summary of the model specified (in equation format) --- Level-1 Model Y = P0 + P1*(YEARS) + P2*(MCI) + E Level-2 Model P0 = B00 + R0 P1 = B10 + R1 P2 = B20 + R2 Level-3 Model B00 = G000 + U00 B10 = G100 + U10 B20 = G200 + U20 Run-time deletion has reduced the number of level-1 records to 716 For starting values, data from 716 level-1 and 120 level-2 records were used Iterations stopped due to small change in likelihood function *** ITERATION 1008 *** Sigma_squared =110.43050 Standard Error of Sigma_squared = 7.77797 Tau(pi) INTRCPT1,P0 46.37075 5.48151 -5.91342 YEARS,P1 5.48151 2.22991 5.80536 MCI,P2 -5.91342 5.80536 29.76881 Tau(pi) (as correlations) INTRCPT1,P0 1.000 0.539 -0.159 YEARS,P1
Re: [R] g(x,y) = f(x,y) - e(x)- e(y)?
Le mer. 23 avr. à 06:59, William Simpson a écrit :
Thanks Phipp very much for your help. I had meant, given that I'd
computed the matrix f[x,y] and the vector e[x], how to take the
difference. What is confusing is how to subtract a vector from a
matrix. I don't want the recycling rule.
Well, then, how do you define the difference between a matrix and a
vector if the vector is not recycled into a matrix?
Thanks to the column major order S uses and recycling, the difference
between an m x n matrix 'M' and a vector of length n 'v',
M - v,
yields the right thing, namely the difference between each column of
M and v.
HTH Vincent
Cheers
Bill
On Tue, Apr 22, 2008 at 9:53 AM, Philipp Pagel [EMAIL PROTECTED]
wrote:
g(x,y) = f(x,y) - e(x)- e(y)
These are continuous functions. I am not sure how to do this with
the
discrete equivalents in R.
Is this what you are looking for?
g - function(x, y) {
f(x,y) - e(x) - e(y)
}
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik
Technische Universität München
Wissenschaftszentrum Weihenstephan
85350 Freising, Germany
http://mips.gsf.de/staff/pagel
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[R] Luis Miguel Delgado Gomez/BBK está ausente d e la oficina.
Estaré ausente de la oficina desde el 22/04/2008 y no volveré hasta el 11/05/2008. Responderé a su mensaje cuando regrese. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing list object to a file
Arun Kumar Saha [EMAIL PROTECTED] wrote:
Hi all,
I am wondering how to write a 'list' object to a file. I already gone
through some threads like
http://mail.python.org/pipermail/python-list/2001-April/080639.html, however
could not trace out any reliable solution. I tried following :
write.table(calc, file=c:/data.csv)
Error in data.frame(200501 = c(-0.000387071806652095,
-0.000387221689252648, :
arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27, 19,
23, 20, 11
Anybody can help?
Remember that a list is not a rectangular table. It may have
components of various types and shapes. Therefore, it's unlikely
you'll be able to write a file that can be read easily by
another program.
That said, we do this occasionally for users who want to use
spreadsheets (yuck!) to analyze part of a list. We use the
following function, which simply prints the list to an ASCII
file with a long line length:
##
# File: Robj2txt.r
# Language: R
# Programmer:Michael H. Prager
# Date: July 7, 2004
# Synopsis:
# Function to write a complex R object to an ASCII file.
# Main use is to write objects created from .rdat files;
# however, could be used to save other objects as well.
# This can be used to create files for those who want to use
# a spreadsheet or other program on the data.
###
Robj2txt - function(x, file = paste(deparse(substitute(x)),
.txt, sep = )) {
#
# ARGUMENTS:
# x: R data object to save as ascii
# filename: Name of file to save. Default is name of x
with .txt extension
#
tmp.wid = getOption(width) # save current width
options(width = 1)# increase output width
sink(file)# redirect output to file
print(x) # print the object
sink()# cancel redirection
options(width = tmp.wid) # restore linewidth
return(invisible(NULL)) # return (nothing) from
function
}
##
--
Mike Prager, NOAA, Beaufort, NC
* Opinions expressed are personal and not represented otherwise.
* Any use of tradenames does not constitute a NOAA endorsement.
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Re: [R] Modeling presence only data in R
Dear Milton, you have 2 packages for modelling species distribution in R: grasper and Biomod (by Wilfried Thuiller), but they are all presence-absence models, so you must generate pseudoabsences for each species, following recommendations in papers (see Lobo 2007). On the other side, there are free gis tools to perform presence-only models (Biomapper, GARP, Openmodeller). Write me if you need if you want some advice. -- Alejandro González Fernández de Castro Departamento de Biodiversidad y Conservación Real Jardín Botánico (Consejo Superior de Investigaciones Científicas) Plaza de Murillo, 2 28014 Madrid - Spain 00 34 91 4203017 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Feed list of vectors to vioplot()?
Johannes Graumann wrote:
Hi,
I have a list of vectors and am trying to coerce them into something that
vioplot will take as groups of data to be plotted independently. Can
someone nudge me into the right direction?
Thanks, Joh
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I'll try less impolite and with more info ...
I'm writing a function that contains this:
# function(x, ... , morestuff){}
which is gathered into a list like so:
# mylist - list(x, ...)
down the line I'd like to output all elements from mylist in seperate
vioplots in a single coordiante system. Now I have a hell of a time with
that because vioplot does not accept a list as input but only a succession
of vectors ... how can I now force my list into that form so a can say
# vioplot(magicfunction(mylist))
and get a violin plot for each list-member?
Thanks for any hint,
Joh
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Re: [R] Modeling presence only data in R
Ecological Niche Factor Analysis (ENFA) is implemented in the adehabitat package available on CRAN. Cheers, Alejandro González wrote: Dear Milton, you have 2 packages for modelling species distribution in R: grasper and Biomod (by Wilfried Thuiller), but they are all presence-absence models, so you must generate pseudoabsences for each species, following recommendations in papers (see Lobo 2007). On the other side, there are free gis tools to perform presence-only models (Biomapper, GARP, Openmodeller). Write me if you need if you want some advice. -- Stéphane DRAY ([EMAIL PROTECTED] ) Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I 43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France Tel: 33 4 72 43 27 57 Fax: 33 4 72 43 13 88 http://biomserv.univ-lyon1.fr/~dray/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] g(x,y) = f(x,y) - e(x)- e(y)?
Thanks Phipp very much for your help. I had meant, given that I'd computed the matrix f[x,y] and the vector e[x], how to take the difference. What is confusing is how to subtract a vector from a matrix. I don't want the recycling rule. That function sounds like its describing how to subtract the vector from the matrix: g[1, 1] = f[1, 1] - e[1] - e[1] g[2, 3] = f[2, 3] - e[2] - e[3] (which implies that f is square) So maybe something like: g - f - outer(e, e, -) But a link to the paper and an example would be very helpful! Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Modeling presence only data in R
Dear Alejandro, Thank you very much for your advice. In fact I use GARP and Openmodeller, and they - up to I know - also generate pseudo-absence when modelling. I am searching some R solution to compare the results with OpenModeller outputs. Could you send-me a PDF (or the complete reference) of Lobo 2007? I never heared about grasper and Biomod. I will give a look on it. Kind regards, miltinho Brazil On 4/23/08, Alejandro González [EMAIL PROTECTED] wrote: Dear Milton, you have 2 packages for modelling species distribution in R: grasper and Biomod (by Wilfried Thuiller), but they are all presence-absence models, so you must generate pseudoabsences for each species, following recommendations in papers (see Lobo 2007). On the other side, there are free gis tools to perform presence-only models (Biomapper, GARP, Openmodeller). Write me if you need if you want some advice. -- Alejandro González Fernández de Castro Departamento de Biodiversidad y Conservación Real Jardín Botánico (Consejo Superior de Investigaciones Científicas) Plaza de Murillo, 2 28014 Madrid - Spain 00 34 91 4203017 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Modeling presence only data in R
Actually, the ENFA (function enfa, in the package adehabitat) is not really suitable for modelling environment suitability maps (though it is still useful to discover the patterns of the ecological niche with presence-only data). However, the package adehabitat also contains other functions allowing this kind of modelling: the Mahalanobis distances (function mahasuhab) and their factorial decomposition (function madifa) could be a useful approach. See: Calenge C, Darmon G, Basille M, Loison A, Jullien JM. (2008) The factorial decomposition of the Mahalanobis distances in habitat selection studies. Ecology, 89, 555-566. and references therein. You may also consider the GNESFA, a generalization of the MADIFA/ENFA approach (see the function gnesfa and references therein). Other commonly used approaches are also available in adehabitat. Best wishes, Clément Calenge. Stéphane Dray wrote: Ecological Niche Factor Analysis (ENFA) is implemented in the adehabitat package available on CRAN. Cheers, Alejandro González wrote: Dear Milton, you have 2 packages for modelling species distribution in R: grasper and Biomod (by Wilfried Thuiller), but they are all presence-absence models, so you must generate pseudoabsences for each species, following recommendations in papers (see Lobo 2007). On the other side, there are free gis tools to perform presence-only models (Biomapper, GARP, Openmodeller). Write me if you need if you want some advice. -- Clément CALENGE Observatoire des galliformes de montagne Office national de la chasse et de la faune sauvage Saint Benoist - 78610 Auffargis tel. (33) 01.30.46.54.14 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help needed: Plotting step by step.
Hello, I have generated 2 Poisson processes and want to plot them on a single graph in a step by step manner in order to be able to compare them. I tried plot and biplot but it does not help, I could connect two points by hand for point graph if they were 5 or 10 I have more than 200 such point to be connected and Poisson cluster makes it difficult for me to even read them properly. Can anyone tell me which is the function that can plot a stepwise graph for me? I did google over the plot for step fun but did not understand much of it. A simpler help would be more useful to me, as I am not a expert either in Statistics or R. Regards, Atul. -- Atul S. Kulkarni Graduate Student, Department of Computer Science, University Of Minnesota, Duluth, MN 55812. www.d.umn.edu/~kulka053 - Before you start some work, always ask yourself three questions - Why am I doing it, What the results might be and Will I be successful. Only when you think deeply and find satisfactory answers to these questions, go ahead. Chanakya quotes (Indian politician, strategist and writer, 350 BC-275 BC) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Comparing kendall's tau values?
Ashton, Gail ashtong at si.edu writes: I have 3 variables relating to the successful introductions of species to 95 different areas: introduction frequency; number of successes pre 1906; number of successes post 1906 The data are not normal, nor homo-skedatic, so I am using non-parametric statistics. I have calculated Kendall's tau between both introduction successes pre 1906 (tau=0.3903) and introduction successes post 1906 (tau=0.3317)- pre + post values are independent. Is it possible to test whether there is a significant difference between the tau values, ie one correlation is 'significantly' 'better'? Is there a way to calculate confidence intervals for tau? Thanks, Gail I think I would try bootstrapping/permutation tests in this case. Just a warning: nonparametric tests do _not_ necessarily do what you think in the case of heteroscedasticity -- they are often constructed to test for a difference in location parameter, assuming the same distribution around the location parameter (mean, median etc.) -- i.e. assuming homoscedasticity. Any chance of using GLMs (i.e., treating success as a binary variable)? Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Equivalent of intervals() in lmer
Dear Friends, (1) There may be a solution for those (e.g., experimental psychologists) who are *not at all* interested in generalizing the absolute level of the response variable (say, reaction time, rt) to other subjects, but *only* to generalize the effect of the manipulated variables within subjects (because that's what the theory speaks to) to other subjects. First, center the predictor(s) so as to remove any spurious correlation between mean and intercept, and then write the random effect w/o an intercept. Now the model will not address whether rt is different from 0, it will only estimate the slope and whether it's different from 0: require(lme4) require(gmodels) data(sleepstudy) ss - sleepstudy ss$days - with(ss, Days - mean(Days)) (fm1 - lmer(Reaction ~ days + (days|Subject), ss)) (fm3 - lmer(Reaction ~ days + (-1 + days|Subject), ss)) ci(fm1) ci(fm3) # CIs for days are about 14% smaller (2) When the predictor is not continuous, this approach doesn't work. A solution for a designed experiment is to plot CIs for differences, i.e., 5%LSDs (rather than plot the CIs on cell means). Those CIs are smaller and address the question of interest. Here is a one-way ANOVA: recall - c(10, 13, 13, 6, 8, 8, 11, 14, 14, 22, 23, 25, 16, 18, 20, 15, 17, 17, 1, 1, 4, 12, 15, 17, 9, 12, 12, 8, 9, 12) fr - data.frame(rcl = recall, time = factor(rep(c(1, 2, 5), 10)), subj = factor(rep(1:10, each = 3))) (fr.lmer - lmer(rcl ~ time -1 +(1 | subj), fr)) mm - unique(model.matrix(~ time -1, fr)) cm - mm[1, ] - mm[3, ] cm1 - mm[1, ] - mm[2, ] estimable(fr.lmer, cm = cm, conf.into = 0.95) estimable(fr.lmer, cm = cm1, conf.into = 0.95) plot mean \pm 2* 0.366 and call it a 5%LSD (uncorrected for multiple comparisons) In the case of a more-than-one-way ANOVA with interaction (say 2x2), choose which simple effects are of interest, get SEs for those; and then plot the four points with the CIs. These are not quite right for the difference between simple effects, but I don't know what to do about that. _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ On Apr 21, 2008, at 9:05 AM, Dieter Menne wrote: Douglas Bates bates at stat.wisc.edu writes: If you want to examine the three means then you should fit the model as lmer(rcl ~ time - 1 + (1 | subj), fr) True, but for the notorious error bars in plots that reviewers always request the 0.35 is probable more relevant than the 1.87. Which I think is justified in this case, but in most non-orthogonal designs with three or more factors, where we have a mixture of between/withing subject, there is no clear solution. What to do when required to produce error-bars that reasonably mirror p- values? It's easier with British Journals in the medical field that often have statistical professionals as reviewers, but many American Journals with their amateur physician/statisticians (why no t-test on raw data?) drive me nuts. Dieter #- library(lme4) recall - c(10, 13, 13, 6, 8, 8, 11, 14, 14, 22, 23, 25, 16, 18, 20, 15, 17, 17, 1, 1, 4, 12, 15, 17, 9, 12, 12, 8, 9, 12) fr - data.frame(rcl = recall, time = factor(rep(c(1, 2, 5), 10)), subj = factor(rep(1:10, each = 3))) fr.lmer - lmer(rcl ~ time -1 +(1 | subj), fr) summary(fr.lmer) fr.lmer - lmer(rcl ~ time +(1 | subj), fr) summary(fr.lmer) -- Fixed effects: Estimate Std. Error t value time1 11.000 1.879 5.853 time2 13.000 1.879 6.918 time5 14.200 1.879 7.556 Fixed effects: Estimate Std. Error t value (Intercept) 11. 1.8793 5.853 time2 2. 0.3507 5.703 time5 3.2000 0.3507 9.125 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Overall p-value from a factor in a coxph fit
Kåre Edvardsen [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Prof. Paul, Prof. Frank. Thank you very much for helping me out. The Design package did the trick. Here is how the anova table looks like without using the Design package: anova(Fit1) Analysis of Deviance Table Cox model: response is Surv(Time, cancer) Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev NULL 16783 5341.8 relativ 1 0.01678214995.0 hormone 3939.41677914055.6 As you see, no p-values reported Here is how it looks with after implementing Design: anova(Fit1) Wald Statistics Response: Surv(Time, cancer) Factor Chi-Square d.f.P relativ 6.08 1 0.0137 hormone 8.68 3 0.0339 (spacing added to hopefully look correct in monospaced font) I was a puzzled looking at those two outputs. In the first the change in deviance with addition of hormone was 939.4 with 3 df. In the second the Wald chi-square, which I presume should be similar to the difference in deviance, is only 8.68. Were these outputs from completely different models or data? Or are there pitfalls in applying Design functions to Surv objects created with other packages? -- David Winsemius Regards, Kare On Fri, 2008-04-18 at 11:03 -0500, Frank E Harrell Jr wrote: Paul Johnson wrote: On Fri, Apr 18, 2008 at 3:06 AM, Kåre Edvardsen [EMAIL PROTECTED] wrote : Hi all. If I run the simple regression when x is a categorical variable ( x - factor(x) ): MyFit -coxph( Surv(start, stop, event) ~ x ) How can I get the overall p-value on x other than for each dummy variable? anova(MyFit) does NOT provide that information as previously suggested on the li st. It should work... Here's a self contained example showing that anova does give the desired significance test for an lm model. y - rnorm(100) x - gl(5,20) mod - lm(y~x) anova(mod) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(F) x 4 6.575 1.644 1.5125 0.2047 Residuals 95 103.237 1.087 If you provide a similar self contained example leading up to a coxph , I would be glad to investigate your question. You don't give enough information for me to tell which version of coxph you are running, an d from what package. Suppose I guess that you are using the coxph from the package survival. If so, it appears to me there is a bug in that package at the moment. The methods anova.coxph and drop1.coxph did exist at one time, until very recently. There is a thread in r-help (which I foun d by typing RSiteSearch(anova.coxph) ) discussing recent troubles with anova.coxph. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/118481.html As you see from the discussion in that thread, there used to be an anova method for coxph, and in the version of survival I have now, there is no such method. The version I have is 2.34-1, Date: 2008-03-31. Here's what I see after I run example(coxph) in order to create som e coxph objects, on which I can test the diagnostics: drop1(test2) Error in terms.default(terms1) : no terms component anova(test2) Error in UseMethod(anova) : no applicable method for anova In that survival package, I do find anova.survreg, but not anova.coxph. If you are using the survival package, I'd suggest you contact Thomas Lumley directly, since he maintains it. I think if you had reported the exact error you saw, it would have been easier for me to diagnose the trouble. HTH pj In the meantime you can do library(Design) f - cph( . . . ) anova(f) # multiple d.f. Wald statistics including tests of nonlinearity cph uses coxph but anova.Design is separate from the survival package. Frank [[alternative HTML version deleted]] --===0807199787== __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Attachment decoded: untitled-3.txt --===0807199787==-- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Density estimation
Look at the logspline package. It has a different way of estimating densities that allows for limits to be specified (i.e. probability is 0 beyond the point(s) you specify). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gunther Jansen Sent: Wednesday, April 23, 2008 4:59 AM To: r-help@r-project.org Subject: [R] Density estimation Hi, I am analysing a dataset containing genetic distances within and between species. I want to show a overlap of the distributions of the intra- and interspecific values; on a second graph I use a cut-off value to determine these boundaries. As the dataset contains 30 000 values, I would like to do this with a simple line rather than superimposed histograms. Hence, density plots. With the standard settings of plot(density(x)), I receive the desirable result, except that the function extends slightly in negative x values (which is impossible, distance values are always positive). Furthermore, in the second figure I supplied the cut-off value a priori, so overlap between the two classes is zero by definition. Is there a way to visualise this information in another way, or to readjust the density parameters so intraspecific values below zero have zero probability in the first case, and there is no overlap in the second case? Thanks, gunther __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] nested time series data with measurement error
Dear R helpers, I am trying to fit a model with the main objective of assessing differences, rather than predicting. The treatment was applied to half of the subjects after 9 months of measurement. Then 9 more months of data were collected. The variable month has measurement error due to the complexities of data collection. The dependent variable is numeric and continuous. This would be a simple difference between means were it not for the repeated measurements and measurement error. I am wondering what the R gurus suggest as to which R function might best model these data. I have been looking at tsls (two-stage least squares) in the sem package and pls (partial least squares). I am not sure about their compatibility with repeated measurements and time series. The response curve is also likely to be non-linear. The following script approximates the data, including sample size. In the real data, there are 18 months spread out over a 36 month period. #generate data tree- gl(6,18,label = paste(tree,1:6)) month - gl(18,1,length = 108,label = paste(month,1:18), ordered = TRUE) trtmt - gl(2, 54, length = 108, label = paste(trt,1:2)) pre.post - gl(2,9,length = 108, label = c(pre,post)) response - runif(108, min = -28, max = -25) help - data.frame(tree,month,trtmt,pre.post, response) Thank you in advance for your assistance. Toby Gass Graduate Degree Program in Ecology Department of Forest, Rangeland, and Watershed Stewardship Warner College of Natural Resources Colorado State University Fort Collins, CO 80523 email: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pdf() and histogram() in function call
Here is a function I wrote. It runs no problem, but generate empty pdf
files.
I can't find what is the problem.
create.pdf- function(x, dir)
{
dir.create(dir, showWarnings = FALSE)
plist- c(a, b , c, d)
for(j in plist)
{
filedir- paste(dir, /, j, .pdf, sep=)
form1- as.formula(paste(~ , j, | var1, sep= ))
form2- as.formula(paste(~ , j, | var2, sep= ))
form3- as.formula(paste(~ , j, | var3, sep= ))
pdf(filedir)
histogram(form1,data=x,type=count, xlab=j,main=Histogram conditioned
on the levels of var1)
histogram(form2,data=x, type=count,xlab=j,main=Histogram conditioned
on the levels of var2)
histogram(form3, data=x, type=count, xlab=j, main=Histogram
conditioned on the levels of var3 )
dev.off()
}
}
I have tried to find the problem, but no luck! Seriously need HELP! Thanks.
_
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Re: [R] g(x,y) = f(x,y) - e(x)- e(y)?
It is not clear exactly what you want to do by subtracting a vector from a
matrix, but look at the sweep function, it may do what you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of William Simpson
Sent: Wednesday, April 23, 2008 4:59 AM
To: Philipp Pagel
Cc: r-help@r-project.org
Subject: Re: [R] g(x,y) = f(x,y) - e(x)- e(y)?
Thanks Phipp very much for your help. I had meant, given that
I'd computed the matrix f[x,y] and the vector e[x], how to
take the difference. What is confusing is how to subtract a
vector from a matrix. I don't want the recycling rule.
Cheers
Bill
On Tue, Apr 22, 2008 at 9:53 AM, Philipp Pagel
[EMAIL PROTECTED] wrote:
g(x,y) = f(x,y) - e(x)- e(y)
These are continuous functions. I am not sure how to do
this with
the discrete equivalents in R.
Is this what you are looking for?
g - function(x, y) {
f(x,y) - e(x) - e(y)
}
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für Genomorientierte Bioinformatik Technische
Universität
München Wissenschaftszentrum Weihenstephan 85350
Freising, Germany
http://mips.gsf.de/staff/pagel
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Re: [R] help needed: Plotting step by step.
Can you show us the code you used for the 5 to 10 points? (either generate some random data, or use a sample dataset). -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Atul Kulkarni Sent: Wednesday, April 23, 2008 9:07 AM To: r-help@r-project.org Subject: [R] help needed: Plotting step by step. Hello, I have generated 2 Poisson processes and want to plot them on a single graph in a step by step manner in order to be able to compare them. I tried plot and biplot but it does not help, I could connect two points by hand for point graph if they were 5 or 10 I have more than 200 such point to be connected and Poisson cluster makes it difficult for me to even read them properly. Can anyone tell me which is the function that can plot a stepwise graph for me? I did google over the plot for step fun but did not understand much of it. A simpler help would be more useful to me, as I am not a expert either in Statistics or R. Regards, Atul. -- Atul S. Kulkarni Graduate Student, Department of Computer Science, University Of Minnesota, Duluth, MN 55812. www.d.umn.edu/~kulka053 - Before you start some work, always ask yourself three questions - Why am I doing it, What the results might be and Will I be successful. Only when you think deeply and find satisfactory answers to these questions, go ahead. Chanakya quotes (Indian politician, strategist and writer, 350 BC-275 BC) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Course*** R/S-Plus Advanced Programming ****by XLSolutions Corp / May 2008 in San Francisco
Announcing R/Splus Advanced Programming course: May 29-30, 2008 in San Francisco. *** half of the seats are already taken *** Please email for earlybird rates: Payments due after the class. R/Splus Advanced Programming Course Outline: Day 1 - Overview of R/S fundamentals: Syntax and Semantics - Class and Inheritance - Concepts, Construction and good use of Language Objects - Coercion and Efficiency - Object-oriented Programming in R and S-Plus - Taking advantage of fast objects and fast functions - Advanced Manipulation tools: Parse, Deparse, Substitute, etc. - How to fully take advantage of Vectorization - Generic and Method Functions - Search path, Databases and Frames Visibility (S-plus) Day 2 - Working with Large Objects - Handling Properly Recursion and Iterative Calculations - Managing loops; For (S-Plus) and for() loops - Consequences of Lazy Evaluation - Efficient Code Practices for Large Computations - Memory Management and Resource Monitoring - Writing R and S-plus Functions to call Compiled Code - Writing and Debugging Compiled Code for S-plus and R system - Connecting R to External Data Sources - Macros in R - Understanding the Structure of Model fitting Functions in R and S-plus - Designing and Packaging Efficiently a new model Function Payment due AFTER the class Email us for group discounts. Email Sue Turner: [EMAIL PROTECTED] Phone: 206-686-1578 Visit us: www.xlsolutions-corp.com/courselist.htm Please let us know if you and your colleagues are interested in this class to take advantage of group discount. Register now to secure your seat! Cheers, Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] pdf() and histogram() in function call
Read FAQ 7.22
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of qian z
Sent: Wednesday, April 23, 2008 10:49 AM
To: r-help@r-project.org
Subject: [R] pdf() and histogram() in function call
Here is a function I wrote. It runs no problem, but
generate empty pdf
files.
I can't find what is the problem.
create.pdf- function(x, dir)
{
dir.create(dir, showWarnings = FALSE)
plist- c(a, b , c, d)
for(j in plist)
{
filedir- paste(dir, /, j, .pdf, sep=)
form1- as.formula(paste(~ , j, | var1, sep= ))
form2- as.formula(paste(~ , j, | var2, sep= ))
form3- as.formula(paste(~ , j, | var3, sep= ))
pdf(filedir)
histogram(form1,data=x,type=count,
xlab=j,main=Histogram conditioned
on the levels of var1)
histogram(form2,data=x,
type=count,xlab=j,main=Histogram conditioned
on the levels of var2)
histogram(form3, data=x, type=count, xlab=j, main=Histogram
conditioned on the levels of var3 )
dev.off()
}
}
I have tried to find the problem, but no luck! Seriously
need HELP! Thanks.
_
Be a better friend, newshound, and know-it-all with Yahoo!
Mobile. [1]Try it
now.
References
1.
http://us.rd.yahoo.com/evt=51733/*http://mobile.yahoo.com/;_yl
t=Ahu06i62sR8HDtDypao8Wcj9tAcJ
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Feed list of vectors to vioplot()?
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
It is hard to provide a solution if we do not understand the problem
to be solved. Sample data would be helpful along with an
understanding of what you would expect for output.
On Wed, Apr 23, 2008 at 10:11 AM, Johannes Graumann
[EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
Hi,
I have a list of vectors and am trying to coerce them into something that
vioplot will take as groups of data to be plotted independently. Can
someone nudge me into the right direction?
Thanks, Joh
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minimal, self-contained, reproducible code.
I'll try less impolite and with more info ...
I'm writing a function that contains this:
# function(x, ... , morestuff){}
which is gathered into a list like so:
# mylist - list(x, ...)
down the line I'd like to output all elements from mylist in seperate
vioplots in a single coordiante system. Now I have a hell of a time with
that because vioplot does not accept a list as input but only a succession
of vectors ... how can I now force my list into that form so a can say
# vioplot(magicfunction(mylist))
and get a violin plot for each list-member?
Thanks for any hint,
Joh
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Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] help needed: Plotting step by step.
Maybe like this? proc.1 - rexp(200,1) proc.2 - rexp(200,2) plot( ecdf( proc.1 ), xlim=range( proc.1, proc.2 ) ) plot( ecdf( proc.2 ), add=T, col.points='red' ) See ?ecdf ?plot.stepfun HTH, Chuck On Wed, 23 Apr 2008, Greg Snow wrote: Can you show us the code you used for the 5 to 10 points? (either generate some random data, or use a sample dataset). -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Atul Kulkarni Sent: Wednesday, April 23, 2008 9:07 AM To: r-help@r-project.org Subject: [R] help needed: Plotting step by step. Hello, I have generated 2 Poisson processes and want to plot them on a single graph in a step by step manner in order to be able to compare them. I tried plot and biplot but it does not help, I could connect two points by hand for point graph if they were 5 or 10 I have more than 200 such point to be connected and Poisson cluster makes it difficult for me to even read them properly. Can anyone tell me which is the function that can plot a stepwise graph for me? I did google over the plot for step fun but did not understand much of it. A simpler help would be more useful to me, as I am not a expert either in Statistics or R. Regards, Atul. -- Atul S. Kulkarni Graduate Student, Department of Computer Science, University Of Minnesota, Duluth, MN 55812. www.d.umn.edu/~kulka053 - Before you start some work, always ask yourself three questions - Why am I doing it, What the results might be and Will I be successful. Only when you think deeply and find satisfactory answers to these questions, go ahead. Chanakya quotes (Indian politician, strategist and writer, 350 BC-275 BC) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ccf and covariance
Hi. It's my understanding that a cross-correlation function of vectors x and y at lag zero is equivalent to their correlation (or covariance, depending on how the ccf is defined). If this is true, could somebody please explain why I get an inconsistent result between cov() and ccf(type = covariance), but a consistent result between cor() and ccf(type = correlation)? Or have I misunderstood what is a cross-correlation? (unfortunately, I can't seem to get a look at the ccf code, since I think it's buried in some C function outside of the main environment) Thanks very much. --Bob Farmer PhD candidate, Dalhousie University Halifax, NS, Canada Example: d1-data.frame(matrix(ldeaths, nrow = 6, byrow = T)) seventy_4-as.numeric(d1[1,]) seventy_5-as.numeric(d1[2,]) ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = covariance ) cov(seventy_4, seventy_5) #inconsistent ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = correlation ) cor(seventy_4, seventy_5) #consistent __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] significant variables in GPLS ?
Hello, I am using the gpls package for modelling vegetation classes. My problem is that I now want to know which input variables are significant for the modelling of the classes to recalculate the equation again with just the selected variables. I think I can analyse the significance of the variables via their weights. I used the gpls1a term for two group classification. Here my code: -- library(gpls) # spex_Y-read.csv(F:/GPLS/spex_Y.csv, header=TRUE, sep=;)#, row.names=ID) # spex_X-read.csv(F:/GPLS/spex_X.csv, header=TRUE, sep=;)#, row.names=ID) # test - glpls1a(spex_X, spex_Y$A_mell,K.prov=7, br=FALSE) names(test) [1] coefficients convergenceniter family [5] link levs bias.reduction coefficients = regression coefficients convergence = whether convergence is achieved niter total = number of iterations bias.reduction = whether Firth's procedure is used link = link function, logit is the only one practically implemented now - But the values (coefficients, convergence, niter, family, link, levs, bias.reduction) I have got from the gpls1a do not contain any information about the significance of my input variables. Does anybody have an idea how I can get information about the significance of my input values? This would really help me a lot. -- Jetzt dabei sein: http://www.shortview.de/[EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ccf and covariance
On Wed, 23 Apr 2008, Bob Farmer wrote: Hi. It's my understanding that a cross-correlation function of vectors x and y at lag zero is equivalent to their correlation (or covariance, depending on how the ccf is defined). The ratio of your values is MASS::fractions(282568.5/259021) [1] 12/11 ? Do you recognize it? There is an explanation in MASS4, p. 390, for example. If this is true, could somebody please explain why I get an inconsistent result between cov() and ccf(type = covariance), but a consistent result between cor() and ccf(type = correlation)? Or have I misunderstood what is a cross-correlation? (unfortunately, I can't seem to get a look at the ccf code, since I think it's buried in some C function outside of the main environment) It is in the R sources, not 'buried' at all - that is what 'Open Source' means. You can browse them at https://svn.r-project.org/R/trunk, or download them for study. Thanks very much. --Bob Farmer PhD candidate, Dalhousie University Halifax, NS, Canada Example: d1-data.frame(matrix(ldeaths, nrow = 6, byrow = T)) seventy_4-as.numeric(d1[1,]) seventy_5-as.numeric(d1[2,]) ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = covariance ) cov(seventy_4, seventy_5) #inconsistent ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = correlation ) cor(seventy_4, seventy_5) #consistent __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Feed list of vectors to vioplot()?
posted mailed
Thanks for the very appropriate scolding.
Here's my example (based on ?vioplot):
mu-2
si-0.6
bimodal-c(rnorm(1000,-mu,si),rnorm(1000,mu,si))
uniform-runif(2000,-4,4)
normal-rnorm(2000,0,3)
# Working just fine
myfunction1 - function(x, ...){vioplot(x,...)}
myfunction1(bimodal,uniform,normal)
# What I (believe to) need
myfunction2 - function(x, ...){
mylist - list(x, ...)
# plenty of lapply stuff
vioplot(mylist)
}
myfunction2(bimodal,uniform,normal)
-- Error in min(data) : invalid 'type' (list) of argument
I'm at a loss on how to disentangle mylist - unlist just gives me one big
vector back rather than the 3 I want to have back ...
Thanks for your patience, Joh
jim holtman wrote:
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
It is hard to provide a solution if we do not understand the problem
to be solved. Sample data would be helpful along with an
understanding of what you would expect for output.
On Wed, Apr 23, 2008 at 10:11 AM, Johannes Graumann
[EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
Hi,
I have a list of vectors and am trying to coerce them into something
that vioplot will take as groups of data to be plotted independently.
Can someone nudge me into the right direction?
Thanks, Joh
__
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http://www.R-project.org/posting-guide.html and provide commented,
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I'll try less impolite and with more info ...
I'm writing a function that contains this:
# function(x, ... , morestuff){}
which is gathered into a list like so:
# mylist - list(x, ...)
down the line I'd like to output all elements from mylist in seperate
vioplots in a single coordiante system. Now I have a hell of a time with
that because vioplot does not accept a list as input but only a
succession of vectors ... how can I now force my list into that form so a
can say
# vioplot(magicfunction(mylist))
and get a violin plot for each list-member?
Thanks for any hint,
Joh
__
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http://www.R-project.org/posting-guide.html and provide commented,
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[R] Can I get rid of this for loop using apply?
Hey all,
The code below creates a partial dependence plot for the variable x1 in the
linear model y ~ x1 + x1^2 + x2.
I have noticed that the for loop in the code takes a long time to run if the
size of the data is increased. Is there a way to change the for loop into
an apply statement? The tricky part is that I need to change the values of
x1 in each step of the loop to give me the appropriate dataset to make
predictions on cbind(m[,-match(x1,names(m))],x1=a[1,i+1]) . If I try
and add the 1112 columns to the dataset a priori and use apply, the code
won't work because the predict function needs the column labeled x1. I
realize I could just grab the form of the linear function and use that
instead of predict(), but I don't want to do that because I want to make
this code applicable to generic model fits.
#create fake data and fit a simple linear regression model
x1 - rep(c(1,3,4),100)
x2 - rep(c(1:6),50)
y - 40+2*x1^.5 - 6*x2 + rnorm(100,0,2)
m - as.data.frame(cbind(y,x1,x2))
lm1 - lm(y~x1+I(x1^2)+x2,data=m)
#super small version of R code for partial dependence plot
a - rbind(c(0:)*(max(m$x1)-min(m$x1))/ +
min(m$x1),c(0:)*0-9)
for(i in c(0:))
{
a[2,i+1] -
mean(predict(lm1,cbind(m[,-match(x1,names(m))],x1=a[1,i+1])))
}
plot(a[1,],a[2,],xlab=x1,ylab=Response,type=l,main=Partial Dependence
Plot)
Many thanks,
Mike
[[alternative HTML version deleted]]
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[R] Minimise a parameter of a given function f, with f 0
Hi, I need to find the minimum value of the parameter, s, such that the function f(s,t) 0 (where -Inf t Inf) I've looked into optim, constrOptim and others but they don't seem to do this. Des anyone have some suggestions? Thanks in advance. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ccf and covariance
Thanks to Prof. Ripley and Phil Spector for pointing out that the autocorrelation functions must use a nontraditional definition of the covariance, involving a denominator of n (instead of n-1) in order to satisfy an assumption of second-order stationarity in the (unbiased) covariance estimators of a time series. In terms of getting at the source code for the (apparently compiled) R_acf, however, I've had no luck. While https://svn.r-project.org/R/trunk seems to be able to show me the source code for otherwise obscured (in the R console) functions like print() (e.g. https://svn.r-project.org/R/trunk/src/library/base/R/print.R ), I can't seem to find the C code (R_acf?) called in this section: array(.C(R_acf, as.double(x), as.integer(sampleT), as.integer(nser), as.integer(lag.max), as.integer(type == correlation), acf = double((lag.max + 1) * nser * nser), NAOK = TRUE) of acf(). For instance, in https://svn.r-project.org/R/trunk/src/library/stats/src/ there is (seemingly) no R_acf.C or stats.C file that I would expect to see. I apologize in advance if this question is elementary or naive -- this is my first time dealing with the source code. Thanks again. --Bob Farmer On Wed, Apr 23, 2008 at 3:31 PM, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Wed, 23 Apr 2008, Bob Farmer wrote: Hi. It's my understanding that a cross-correlation function of vectors x and y at lag zero is equivalent to their correlation (or covariance, depending on how the ccf is defined). The ratio of your values is MASS::fractions(282568.5/259021) [1] 12/11 ? Do you recognize it? There is an explanation in MASS4, p. 390, for example. If this is true, could somebody please explain why I get an inconsistent result between cov() and ccf(type = covariance), but a consistent result between cor() and ccf(type = correlation)? Or have I misunderstood what is a cross-correlation? (unfortunately, I can't seem to get a look at the ccf code, since I think it's buried in some C function outside of the main environment) It is in the R sources, not 'buried' at all - that is what 'Open Source' means. You can browse them at https://svn.r-project.org/R/trunk, or download them for study. Thanks very much. --Bob Farmer PhD candidate, Dalhousie University Halifax, NS, Canada Example: d1-data.frame(matrix(ldeaths, nrow = 6, byrow = T)) seventy_4-as.numeric(d1[1,]) seventy_5-as.numeric(d1[2,]) ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = covariance ) cov(seventy_4, seventy_5) #inconsistent ccf(x=seventy_4, y=seventy_5, plot = F, lag.max = 0, type = correlation ) cor(seventy_4, seventy_5) #consistent __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Feed list of vectors to vioplot()?
Off-list it was pointed out to me that the trivial solution to this would
look like this:
myfunction2 - function(x, ...){
mylist - list(x, ...)
# plenty of lapply stuff
do.call(vioplot,mylist)
}
Thanks for everybodies patience,
Joh
Johannes Graumann wrote:
posted mailed
Thanks for the very appropriate scolding.
Here's my example (based on ?vioplot):
mu-2
si-0.6
bimodal-c(rnorm(1000,-mu,si),rnorm(1000,mu,si))
uniform-runif(2000,-4,4)
normal-rnorm(2000,0,3)
# Working just fine
myfunction1 - function(x, ...){vioplot(x,...)}
myfunction1(bimodal,uniform,normal)
# What I (believe to) need
myfunction2 - function(x, ...){
mylist - list(x, ...)
# plenty of lapply stuff
vioplot(mylist)
}
myfunction2(bimodal,uniform,normal)
-- Error in min(data) : invalid 'type' (list) of argument
I'm at a loss on how to disentangle mylist - unlist just gives me one
big vector back rather than the 3 I want to have back ...
Thanks for your patience, Joh
jim holtman wrote:
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
It is hard to provide a solution if we do not understand the problem
to be solved. Sample data would be helpful along with an
understanding of what you would expect for output.
On Wed, Apr 23, 2008 at 10:11 AM, Johannes Graumann
[EMAIL PROTECTED] wrote:
Johannes Graumann wrote:
Hi,
I have a list of vectors and am trying to coerce them into something
that vioplot will take as groups of data to be plotted independently.
Can someone nudge me into the right direction?
Thanks, Joh
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
I'll try less impolite and with more info ...
I'm writing a function that contains this:
# function(x, ... , morestuff){}
which is gathered into a list like so:
# mylist - list(x, ...)
down the line I'd like to output all elements from mylist in seperate
vioplots in a single coordiante system. Now I have a hell of a time with
that because vioplot does not accept a list as input but only a
succession of vectors ... how can I now force my list into that form so
a can say
# vioplot(magicfunction(mylist))
and get a violin plot for each list-member?
Thanks for any hint,
Joh
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] combining two (or more) tables by creating another dimension
Dear R community, I wish to combine two tables in one by adding an additional dimension: e.g.: t1-as.table(matrix(rnorm(40),nrow=4,ncol=10));rownames(t1)-c(rowone,rowtwo,rowthree,rowfour) t1 A B C D E F G H I J rowone -1.04203810 -0.05148987 -1.74162922 0.02683198 -0.28622512 -0.87690444 -0.12335543 -0.60394319 -0.15923255 0.53235864 rowtwo -1.01168507 1.07785313 -0.17483300 -0.60747916 1.12944895 0.89512881 0.13972544 0.05006999 1.08190614 1.65985042 rowthree -1.52667891 -0.50032166 2.04588634 -1.55943514 1.58400939 -0.03039314 -0.21145199 0.11526848 2.24376313 -0.74787863 rowfour 0.39287271 -0.55125576 -0.74755340 -1.73366333 0.18886435 0.60942387 1.03573198 -1.93694305 0.32919872 -0.09073655 t2-as.table(matrix(rnorm(40),nrow=4,ncol=10));rownames(t1)-c(rowone,rowtwo,rowthree,rowfour) t2 A B C D E F G H I J A -0.23533769 -1.13888448 0.33531402 0.53159432 0.37499285 0.14237453 0.38195561 -1.06769856 -0.43285636 0.50321651 B -1.96015770 -0.70381431 0.24031069 -0.65790501 -0.87634039 0.26629589 0.66948553 -0.78660318 0.86132576 -1.03440605 C 1.48375673 -0.07710868 -0.18416378 -2.39868210 -0.09285612 0.88384443 1.34379990 1.15250558 -1.26557860 -0.63810608 D -1.45636216 0.03015655 1.68455454 1.19187925 0.10197788 0.54773000 -0.19173498 0.74170610 1.10240139 1.32562483 How can t1 and t2 be combined in an object with dim(object)=c(2,4,10)? Thank you and wishing you a great day! Georg. * Georg B. Ehret Johns Hopkins Baltimore, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help on coxph.wtest
Hi, i need to use pspline. In this pspline function coxph.wtest was used. When I try to make some change to this function by pulling out the pspline function, it turns out R gave me an error msg, saying coxph.wtest cannot be found. Even if i dont change anything in pspline and just rename it and run the function, it did not work out. Can any one help me with this? is there anyway to get the coxph.wtest function? thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multidimensional contingency tables
It seems to me that a combination of ftable and xtabs works fine:
prob1- data.frame(victim=c(rep('white',4),rep('black',4)),
+ perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)),
+ death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97))
prob1
victim perp death count
1 white white yes19
2 white whiteno 132
3 white black yes11
4 white blackno52
5 black white yes 0
6 black whiteno 9
7 black black yes 6
8 black blackno97
ftable(xtabs(count ~ victim + perp + death, data = prob1))
death no yes
victim perp
black black97 6
white 9 0
white black52 11
white 132 19
Best,
Giovanni
Date: Tue, 22 Apr 2008 08:24:36 -0500
From: hadley wickham [EMAIL PROTECTED]
Sender: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Precedence: list
On Mon, Apr 21, 2008 at 9:46 PM, Robert A. LaBudde [EMAIL PROTECTED] wrote:
How does one ideally handle and display multidimenstional contingency
tables in R v. 2.6.2?
E.g.:
prob1- data.frame(victim=c(rep('white',4),rep('black',4)),
+ perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)),
+ death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97))
prob1
victim perp death count
1 white white yes19
2 white whiteno 132
3 white black yes11
4 white blackno52
5 black white yes 0
6 black whiteno 9
7 black black yes 6
8 black blackno97
The xtabs() function doesn't seem appropriate, as it has no means of
using 'count'.
This must be a common problem.
You can also use the reshape package (http://had.co.nz/reshape)
cast(prob1, victim ~ perp, sum, value=count)
cast(prob1, victim ~ perp ~ death, sum, value=count)
cast(prob1, death + victim ~ perp, sum, value=count)
etc.
Hadley
--
http://had.co.nz/
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and provide commented, minimal, self-contained, reproducible code.
--
Giovanni Petris [EMAIL PROTECTED]
Associate Professor
Department of Mathematical Sciences
University of Arkansas - Fayetteville, AR 72701
Ph: (479) 575-6324, 575-8630 (fax)
http://definetti.uark.edu/~gpetris/
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[R] BB - a new package for solving nonlinear system of equations and for optimization with simple constraints
Hi, We (Paul Gilbert and I) have just released a new R package on CRAN called BB (stands for Barzilai-Borwein) that provides functionality for solving large-scale (and small-scale) nonlinear system of equations. Until now, R didn't have any functionality for solving nonlinear systems. We hope that this package fills that need. We also have an implementation of the spectral projected gradient method for the optimization of (smooth) nonlinear objective functions, subject to simple constraints that can be defined as a projection mapping. This function, spg(), is suited for large-scale optimization. It generally performs better than the conjugate-gradient methods in optim, and complements the low-memory BFGS method for constraints that can be defined by the user as projection mapping. Best regards, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] filled contour plots
hello everyone,
I was wondering if anybody can help me solve 2
problems related to the function filled.contour.
I am entering the following R command:
filled.contour(xx,yy,P1, nlevels=20,color=cm.colors,
plot.axes={
contour(xx,yy,P1,add=T,col=grey,
nlevels=20, drawlabels=F)
axis(1,1:length(xx),labels=xlabels)
axis(2,1:length(yy),labels=ylabels)
})
which results in the figure attached. Overall the
figure looks great. However, my 2 question are:
1) how can I make sure that the contour lines match
the shading? you can see from the figure (center
left) there's a contour missing.
2) how can I get smoother contours/shading, without so
many sharp angles?
I appreciate any help from you.
thank you so very much for the attention.
F. DeSales
Be a better friend, newshound, and
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[R] filled contour plots
sorry. the figure did not go through. sending again as PDF. hope it will work this time. Thanks Be a better friend, newshound, and __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Compiling msm on Fedora Core Linux
Greetings all, I'm trying to install the msm package and it is failing on compilation. The problem seems to be the analyticp component? Any advice on how to get it to work? error message is below. I'm running R version 2.6.2 (2008-02-08) x86_64-redhat-linux-gnu on a dell precision 690 with Fedora Core 8. Thanks, Adam install.packages(msm,lib=/usr/lib64/R/library) trying URL 'http://cran.mirrors.hoobly.com/src/contrib/msm_0.8.tar.gz' Content type 'application/x-tar' length 577213 bytes (563 Kb) opened URL == downloaded 563 Kb * Installing *source* package 'msm' ... ** libs gcc -m64 -std=gnu99 -I/usr/include/R -I/usr/include/R -I/usr/local/include-fpic -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m64 -mtune=generic -c analyticp.c -o analyticp.o In file included from analyticp.c:9: msm.h:3:15: error: R.h: No such file or directory msm.h:4:26: error: R_ext/Applic.h: No such file or directory analyticp.c: In function 'AnalyticP': analyticp.c:51: warning: implicit declaration of function 'Calloc' analyticp.c:51: error: expected expression before 'double' analyticp.c:51: warning: cast to pointer from integer of different size analyticp.c:52: error: expected expression before 'double' analyticp.c:52: warning: cast to pointer from integer of different size analyticp.c:63: warning: implicit declaration of function 'error' analyticp.c:91: warning: implicit declaration of function 'Free' make: *** [analyticp.o] Error 1 ERROR: compilation failed for package 'msm' ** Removing '/usr/lib64/R/library/msm' The downloaded packages are in /tmp/RtmpaoUZiC/downloaded_packages Updating HTML index of packages in '.Library' Warning messages: 1: In install.packages(msm, lib = /usr/lib64/R/library) : installation of package 'msm' had non-zero exit status 2: In tools:::unix.packages.html(.Library) : cannot create HTML package index -- Adam Wilson http://hydrodictyon.eeb.uconn.edu/people/wilson/ Department of Ecology and Evolutionary Biology BioPharm 223 University of Connecticut Tel: 860.486.4157 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 64 bit Linux/Ubuntu memory limit problem?
Dear list, I've recently installed R on a 64 bit machine with 8 GB of RAM. I set this computer up as a dual-boot system, with windows XP 64 and Ubuntu 7.10. I downloaded the Linux 64 bit version of R and installed it. I'm trying to run rather large Random forest models and was running into severe memory limitations with my old windows 32 bit machines. When I run Random Forest models on my new machine, I get the error message: Error: cannot allocate vector of size 2.1 GB. I'm a new Linux user, but I thought that R under Linux would by default take advantage of all available memory. I've searched previous posts and found only 1 thread (posted by David Katz) that didn't really help. Do I need to specify that the Ubuntu/Linux OS use all 8 Gigs of RAM? Could I have done something wrong when I downloaded and installed R? I'd appreciate any advice. Thanks in advance, Zack [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can I get rid of this for loop using apply?
The answer to my post is yes (which I just figured out).
Solution:
#super small version of R code for pd plot using apply
a - rbind(c(0:)*(max(m$x1)-min(m$x1))/ +
min(m$x1),c(0:)*0-9)
b - matrix(rep(c(0:)*(max(m$x1)-min(m$x1))/ + min(m$x1), nrow(m)),
nrow(m), 1112, byrow=T)
a[2,] - apply(b,2,FUN=function(x)
{mean(predict(lm1,cbind(m[,-match(x1,names(m))],x1=x))) })
plot(a[1,],a[2,],xlab=x1,ylab=Response,type=l,main=Partial Dependence
Plot)
Mike Dugas
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[R] filled contour plots
The see the figure I refer to on my email, please
use this link:
http://us.f13.yahoofs.com/bc/480fa972_411e/bc/My+Documents/filled_contour_plot.pdf?bfy26DIBonZ1vybc
thanks
hello everyone,
I was wondering if anybody can help me solve 2
problems related to the function filled.contour.
I am entering the following R command:
filled.contour(xx,yy,P1, nlevels=20,color=cm.colors,
plot.axes={
contour(xx,yy,P1,add=T,col=grey,
nlevels=20, drawlabels=F)
axis(1,1:length(xx),labels=xlabels)
axis(2,1:length(yy),labels=ylabels)
})
which results in the figure attached. Overall the
figure looks great. However, my 2 question are:
1) how can I make sure that the contour lines match
the shading? you can see from the figure (center
left) there's a contour missing.
2) how can I get smoother contours/shading, without
so
many sharp angles?
I appreciate any help from you.
thank you so very much for the attention.
F. DeSales
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Re: [R] help on coxph.wtest
Hi i need to use pspline. In this pspline function coxph.wtest was used. When I try to make some change to this function by pulling out the pspline function, it turns out R gave me an error msg, saying coxph.wtest cannot be found. Even if i dont change anything in pspline and just rename it and run the function, it did not work out. Can any one help me with this? is there anyway to get the coxph.wtest function? thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can I get rid of this for loop using apply?
On Wed, Apr 23, 2008 at 4:23 PM, Mike Dugas [EMAIL PROTECTED] wrote:
The answer to my post is yes (which I just figured out).
Switching from for to apply isn't going to speed up your code. If you
carefully read the source code of apply, you'll see the guts of the
work is done by:
for (i in 1:d2) {
tmp - FUN(array(newX[, i], d.call, dn.call), ...)
if (!is.null(tmp))
ans[[i]] - tmp
}
i.e apply uses for internally. The reason to use apply instead of a
for loop is so that you can better express the intent of your
algorithm.
Hadley
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http://had.co.nz/
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[R] printing to the console in color
Hi, Is there a function that would allow me to print things in different colors to the CONSOLE, say one line in blue and another line in green? Right now, the print() function only prints in navy blue... Thanks! -- View this message in context: http://www.nabble.com/printing-to-the-console-in-color-tp16834851p16834851.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help on coxph.wtest
Hi, i need to use pspline. In this pspline function coxph.wtest was used. When I try to make some change to this function by pulling out the pspline function, it turns out R gave me an error msg, saying coxph.wtest cannot be found. Even if i dont change anything in pspline and just rename it and run the function, it did not work out. Can any one help me with this? is there anyway to get the coxph.wtest function? thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
Hello,
I have been trying to apply some of the different
ways suggested in the past to replace empty values in
a vector for a letter, without success.
Actually, I have a data frame that might contain
empty values, I need to replace all those empty values
for the letter 'N'.
This was my latest try:
dat-read.csv('myfile.csv', na.strings=)
Be a better friend, newshound, and
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[R] filled contour plots (link)
I have moved the figure to the link below. thank you Be a better friend, newshound, and __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] select rows from data based on a vector of char strings
Try this, x=label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 yourdata=read.table(textConnection(x),header=TRUE) attached(yourdata) flist-c(fun,food) yourdata[label %in% flist,] I hope this helps, Jorge On Wed, Apr 23, 2008 at 3:13 AM, Dirkheld [EMAIL PROTECTED] wrote: Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 etc And I have a vector flist-c(fun,food) Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food 3 3 When I do 'data$label==flist[1]' I get 'F T F F', so it works for one item in the char vector flist. But, when I do 'data$label==flist' I get 'F F F F' while I expected 'F T F T'. It seems that I can't perform this action with a vector of charstrings? Is there an other way to do so? -- View this message in context: http://www.nabble.com/select-rows-from-data-based-on-a-vector-of-char-strings-tp16832735p16832735.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mysterious (to me) automatic loading of packages when R starts up
Hello, I have recently noticed that in one of my work areas, a number of packages are automatically loaded without my explicitly requesting them to be loaded (see below). This only happens in one particular workspace (located in a folder). I suspect (without any real evidence) that this may be related to my creating plots using the ggplot2 package earlier in this workspace. I am running R 2.7.0 under Vista but I noticed the same behavior under R 2.6.2 last week. Can someone please tell me why this is happening and how I can prevent this automatic package loading in this workspace? Thanks, Duncan R version 2.7.0 (2008-04-22) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. Loading required package: proto Loading required package: grid Loading required package: reshape [Previously saved workspace restored] __ Dr. Duncan Mackay School of Biological Sciences Flinders University GPO Box 2100 Adelaide S.A. 5001 AUSTRALIA Phone61-8-82012627 FAX 61-8-82013015 http://www.scieng.flinders.edu.au/biology/people/academic/mackay_d/index.htm l __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] select rows from data based on a vector of char strings
u... One of the S language's great strengths is the variety and flexibility of it's data structures. It pays to familiarize yourself with them. So if x is already a dataframe (dataset is a rather unhelpful description), then row.names(x) - x$label x[c(fun,food),] will extract the rows you want (to be suitably assigned by you, of course). Incidentally, the row names typically could have been set when you first read the data in (e.g. via ?read.table() and friends). If x is not a dataframe, then your should make it one, and ?data.frame will tell you how, depending on what it actually is. I repeat: learn about R's data structures and how to use them if you wish to take full advantage of the language and its power (and avoid the unnecessary, though correct (mod the typo), manipulations described in the reply below). -- Bert Gunter -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jorge Ivan Velez Sent: Wednesday, April 23, 2008 4:48 PM To: Dirkheld Cc: r-help@r-project.org Subject: Re: [R] select rows from data based on a vector of char strings Try this, x=label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 yourdata=read.table(textConnection(x),header=TRUE) attached(yourdata) flist-c(fun,food) yourdata[label %in% flist,] I hope this helps, Jorge On Wed, Apr 23, 2008 at 3:13 AM, Dirkheld [EMAIL PROTECTED] wrote: Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 etc And I have a vector flist-c(fun,food) Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food 3 3 When I do 'data$label==flist[1]' I get 'F T F F', so it works for one item in the char vector flist. But, when I do 'data$label==flist' I get 'F F F F' while I expected 'F T F T'. It seems that I can't perform this action with a vector of charstrings? Is there an other way to do so? -- View this message in context: http://www.nabble.com/select-rows-from-data-based-on-a-vector-of-char-string s-tp16832735p16832735.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] select rows from data based on a vector of char strings
I'm sorry about that. Second line should be attach(yourdata) Cheers, Jorge On Wed, Apr 23, 2008 at 7:48 PM, Jorge Ivan Velez [EMAIL PROTECTED] wrote: Try this, x=label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 yourdata=read.table(textConnection(x),header=TRUE) attached(yourdata) flist-c(fun,food) yourdata[label %in% flist,] I hope this helps, Jorge On Wed, Apr 23, 2008 at 3:13 AM, Dirkheld [EMAIL PROTECTED] wrote: Hi, I have loaded a dataset in R : data = label freq1 freq2 news 54 35 fun 37 21 milk19 7 food 3 3 etc And I have a vector flist-c(fun,food) Now I want to use the vector 'flist' for selecting these values from 'data' so that I get the following dataset : label freq1 freq2 fun 37 21 food 3 3 When I do 'data$label==flist[1]' I get 'F T F F', so it works for one item in the char vector flist. But, when I do 'data$label==flist' I get 'F F F F' while I expected 'F T F T'. It seems that I can't perform this action with a vector of charstrings? Is there an other way to do so? -- View this message in context: http://www.nabble.com/select-rows-from-data-based-on-a-vector-of-char-strings-tp16832735p16832735.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can I get rid of this for loop using apply?
On Wed, Apr 23, 2008 at 7:31 PM, Mike Dugas [EMAIL PROTECTED] wrote: Thanks for the help. That explains why my time testing showed no difference. Is there any way to speed up the program? It is unbearably slow if I increase the number of loops. Could you explain exactly what you're trying to do with your code? It's a little hard to understand. Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Can I get rid of this for loop using apply?
Thanks for the help. That explains why my time testing showed no
difference. Is there any way to speed up the program? It is unbearably
slow if I increase the number of loops.
Mike
On Wed, Apr 23, 2008 at 6:23 PM, hadley wickham [EMAIL PROTECTED] wrote:
On Wed, Apr 23, 2008 at 4:23 PM, Mike Dugas [EMAIL PROTECTED] wrote:
The answer to my post is yes (which I just figured out).
Switching from for to apply isn't going to speed up your code. If you
carefully read the source code of apply, you'll see the guts of the
work is done by:
for (i in 1:d2) {
tmp - FUN(array(newX[, i], d.call, dn.call), ...)
if (!is.null(tmp))
ans[[i]] - tmp
}
i.e apply uses for internally. The reason to use apply instead of a
for loop is so that you can better express the intent of your
algorithm.
Hadley
--
http://had.co.nz/
[[alternative HTML version deleted]]
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Re: [R] Can I get rid of this for loop using apply?
Sure, I am creating a partial dependence plot (reference Friedman's stochastic gradient paper from, I want to say, 2001). The idea is to find the relationship between one of the predictors, say x1, and y by creating the following plot: take a random sample of actual data points, hold other predictors fixed (x2-xp), vary x1 across its range, create a string of predictions for each value of x1, repeat for all observations in sample, and finally average all the predictions for each value of x1. If you think about it, this plot solves Simpson's paradox under fairly mild conditions. The code I wrote does this using predict() which is useful for modeling approaches like GAMs. Mike On Wed, Apr 23, 2008 at 8:47 PM, hadley wickham [EMAIL PROTECTED] wrote: On Wed, Apr 23, 2008 at 7:31 PM, Mike Dugas [EMAIL PROTECTED] wrote: Thanks for the help. That explains why my time testing showed no difference. Is there any way to speed up the program? It is unbearably slow if I increase the number of loops. Could you explain exactly what you're trying to do with your code? It's a little hard to understand. Hadley -- http://had.co.nz/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replecing empty values with a letter (continuation)
Judith Flores [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Sorry, I hit the send botton by accident. Here is my
code so far:
dat-read.csv('myfile.csv', na.strings=)
dat[is.na(dat]-'N'
But it doesn't replace the NA for the letter 'N'.
What happens when you just type:
dat[is.na(dat)] #?
My guess is that you don't have any, since na.strings= told
read.table that there were no values that should be assigned NA. It
is possible to have a value of NA that is not NA.
x - c(NA, 1 , 2)
#NA is not NA
is.na(x)
[1] FALSE FALSE FALSE
txt - nnn 1 2 3 4 5
xt- read.table(textConnection(txt), header=FALSE)
xt
V1 V2 V3 V4 V5 V6
1 nnn 1 2 3 4 5
xt- read.table(textConnection(txt), header=FALSE,na.strings=nnn)
xt
V1 V2 V3 V4 V5 V6
1 NA 1 2 3 4 5
#That is a TRUE NA
is.na(xt)
V1V2V3V4V5V6
[1,] TRUE FALSE FALSE FALSE FALSE FALSE
xt[is.na(xt)] - N
xt
V1 V2 V3 V4 V5 V6
1 N 1 2 3 4 5
--
David Winsemius
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Re: [R] Replecing empty values with a letter (continuation)
Judith
I think you'll find the issue is that read.csv creates a data.frame and
your subsetting syntax
dat[is.na(dat)] is assuming it to be a matrix. If this is the case, you
have two options:
Coerce dat into a matrix with as.matrix(dat);
Use apply() to replace the missing data for each column in turn ...
apply(dat, 2, function(x) x[is.na(x)] - 'N')
HTH ...
Peter Alspach
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Judith Flores
Sent: Thursday, 24 April 2008 11:20 a.m.
To: RHelp
Subject: [R] Replecing empty values with a letter (continuation)
Sorry, I hit the send botton by accident. Here is my code so far:
dat-read.csv('myfile.csv', na.strings=) dat[is.na(dat]-'N'
But it doesn't replace the NA for the letter 'N'.
I am using R v 2/7/0, running it under Windows XP.
Thank you,
Judith
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Import_from_outputfile
One way is to first read in the data with readLines. This will create a character vector of the input lines. You can then remove the lines you want and then use textConnect to read in the modified data: x - readLines(file) x - x[-c(2,3)] # delete lines two and three input - read.table(textConnection(x), header=TRUE) On Wed, Apr 23, 2008 at 9:29 AM, Ole Roessler [EMAIL PROTECTED] wrote: Hello, I want to do some analyses with data originating from a hydrological model that are saved in multiple output files. I want to import the output-files in R without a treatment like preproceesing every output file. The output file is more or less a matrix, but with additional rows in the header, describing the type of data. My question is: 1. Is there a possibility to skip the rows or start with a certain row? I want to exclude row 2 and 3, so the first line become the header of the list below. This is the beginning of the output file to be imported. YYMMDDHH12345tot_average total runoff [mm per zone] (sum of QB, QI und QD)_submodel:unsatzon_model --------0.14666240.29580420.22250480.2279085 0.10712021. 2001 1 1 10.0552420.0373910.0515500.057681 0.0210700.046036 2001 1 1 20.0512880.0324040.0470270.056636 0.0210700.042736 2001 1 1 30.0524910.0329370.0480420.058415 0.0210700.043701 2001 1 1 40.0513890.0325080.0471360.056663 0.0210700.042812 2001 1 1 50.0507300.0322630.0465990.055585 0.0210700.042278 Would be a great help if there is a possibility. Thank you very much ! best regards Ole -- --- Ole Röà ler PhD - student Climatology and Landscape-Ecology Research Group Department of Geography University of Bonn Postal Address: Meckenheimer Allee 166 53115 Bonn Germany Tel.: +49-(0)228-737895 Fax.: +49-(0)228-737506 Email: [EMAIL PROTECTED] Homepage: [2]http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm - References 1. mailto:[EMAIL PROTECTED] 2. http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Import_from_outputfile
You can issue a read for just the headers and then skip first 3 lines so as to issue a read for just the body. fn - myfile.dat DF - read.table(fn, skip = 3, na.strings = --, col.names = read.table(fn, nrows = 1, as.is = TRUE)) On Wed, Apr 23, 2008 at 9:29 AM, Ole Roessler [EMAIL PROTECTED] wrote: Hello, I want to do some analyses with data originating from a hydrological model that are saved in multiple output files. I want to import the output-files in R without a treatment like preproceesing every output file. The output file is more or less a matrix, but with additional rows in the header, describing the type of data. My question is: 1. Is there a possibility to skip the rows or start with a certain row? I want to exclude row 2 and 3, so the first line become the header of the list below. This is the beginning of the output file to be imported. YYMMDDHH12345tot_average total runoff [mm per zone] (sum of QB, QI und QD)_submodel:unsatzon_model --------0.14666240.29580420.22250480.2279085 0.10712021. 2001 1 1 10.0552420.0373910.0515500.057681 0.0210700.046036 2001 1 1 20.0512880.0324040.0470270.056636 0.0210700.042736 2001 1 1 30.0524910.0329370.0480420.058415 0.0210700.043701 2001 1 1 40.0513890.0325080.0471360.056663 0.0210700.042812 2001 1 1 50.0507300.0322630.0465990.055585 0.0210700.042278 Would be a great help if there is a possibility. Thank you very much ! best regards Ole -- --- Ole Röà ler PhD - student Climatology and Landscape-Ecology Research Group Department of Geography University of Bonn Postal Address: Meckenheimer Allee 166 53115 Bonn Germany Tel.: +49-(0)228-737895 Fax.: +49-(0)228-737506 Email: [EMAIL PROTECTED] Homepage: [2]http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm - References 1. mailto:[EMAIL PROTECTED] 2. http://www.giub.uni-bonn.de/loeffler/people/people_sites/roessler.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting two functions
by the way, i want both in the same figure, not side by side or others.. tks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting two functions
Hi Manoel,
Try this:
x-seq(-10, 10, l=100)
plot(x^2,type=l,col=2,ylim=range(x^2,x^5),ylab='f(x)')
points(x^5,type=l,col=4)
legend('topleft',c(expression(x^2),expression(x^5)),col=c(2,4),lty=1)
See ?plot
HTH,
Jorge
On Wed, Apr 23, 2008 at 11:23 PM, Manoel Santos [EMAIL PROTECTED]
wrote:
i wanna compare functions
to be simple , let's say i want x^2 and x^5 in same plot ( it's not the
case but if i get it i'll understand for others )
how i do it?
x-seq(-10, 10, l=100)
plot(x^2)
and?
tks
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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Re: [R] plotting two functions
On Thu, 24 Apr 2008, Manoel Santos wrote: i wanna compare functions to be simple , let's say i want x^2 and x^5 in same plot ( it's not the case but if i get it i'll understand for others ) how i do it? See ?curve example( curve ) HTH, Chuck x-seq(-10, 10, l=100) plot(x^2) and? tks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:[EMAIL PROTECTED] UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 with which
Hi, I'm having trouble with the which or the seq function, I'm not sure. Here's an example : lat=seq(1,2,by=0.1) lat [1] 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 which(lat==1) [1] 1 which(lat==1.1) [1] 2 which(lat==1.2) [1] 3 which(lat==1.3) [1] 4 which(lat==1.4) [1] 5 which(lat==1.5) [1] 6 which(lat==1.6) [1] 7 which(lat==1.7) *integer(0)* which(lat==1.8) [1] 9 which(lat==1.9) [1] 10 which(lat==2) [1] 11 This doesn't seem to happen with integers. Am I missing something ?? Is there a better function for non-integers ? Thanks a lot, Melanie __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] problem with which
Melanie Abecassis [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Hi, I'm having trouble with the which or the seq function, I'm not sure. Here's an example : lat=seq(1,2,by=0.1) lat [1] 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 which(lat==1) [1] 1 which(lat==1.1) snip [1] 7 which(lat==1.7) *integer(0)* This doesn't seem to happen with integers. Am I missing something ?? Is there a better function for non-integers ? Read the FAQ (Section 7.31 Why doesn't R think these numbers are equal? http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f and look at this: lat - 1.7 [1] -7.00e-01 -6.00e-01 -5.00e-01 -4.00e-01 [5] -3.00e-01 -2.00e-01 -1.00e-01 2.220446e-16 [9] 1.00e-01 2.00e-01 3.00e-01 -- David Winsemius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] problem with which
G'day Melanie, On Wed, 23 Apr 2008 17:46:56 -1000 Melanie Abecassis [EMAIL PROTECTED] wrote: This doesn't seem to happen with integers. Am I missing something ?? Yes, FAQ 7.31: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f Is there a better function for non-integers ? which(sapply(tt, function(x) isTRUE(all.equal(x,1.7 [1] 8 seems to work. HTH. Cheers, Berwin === Full address = Berwin A TurlachTel.: +65 6515 4416 (secr) Dept of Statistics and Applied Probability+65 6515 6650 (self) Faculty of Science FAX : +65 6872 3919 National University of Singapore 6 Science Drive 2, Blk S16, Level 7 e-mail: [EMAIL PROTECTED] Singapore 117546http://www.stat.nus.edu.sg/~statba __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plotting two functions
i wanna compare functions to be simple , let's say i want x^2 and x^5 in same plot ( it's not the case but if i get it i'll understand for others ) how i do it? x-seq(-10, 10, l=100) plot(x^2) and? tks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compiling msm on Fedora Core Linux
The problem seems to be in the header paths: gcc -m64 -std=gnu99 -I/usr/include/R -I/usr/include/R -I/usr/local/include Is R.h not in /usr/include/R? My guess is that you installed an R rpm and not the R-devel rpm. But if you installed from rpms, this is a Fedora support issue -- this community is not responsible for Fedora's re-packaging of R. On Wed, 23 Apr 2008, Adam Wilson wrote: Greetings all, I'm trying to install the msm package and it is failing on compilation. The problem seems to be the analyticp component? Any advice on how to get it to work? error message is below. I'm running R version 2.6.2 (2008-02-08) x86_64-redhat-linux-gnu on a dell precision 690 with Fedora Core 8. Thanks, Adam install.packages(msm,lib=/usr/lib64/R/library) trying URL 'http://cran.mirrors.hoobly.com/src/contrib/msm_0.8.tar.gz' Content type 'application/x-tar' length 577213 bytes (563 Kb) opened URL == downloaded 563 Kb * Installing *source* package 'msm' ... ** libs gcc -m64 -std=gnu99 -I/usr/include/R -I/usr/include/R -I/usr/local/include-fpic -O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector --param=ssp-buffer-size=4 -m64 -mtune=generic -c analyticp.c -o analyticp.o In file included from analyticp.c:9: msm.h:3:15: error: R.h: No such file or directory msm.h:4:26: error: R_ext/Applic.h: No such file or directory analyticp.c: In function 'AnalyticP': analyticp.c:51: warning: implicit declaration of function 'Calloc' analyticp.c:51: error: expected expression before 'double' analyticp.c:51: warning: cast to pointer from integer of different size analyticp.c:52: error: expected expression before 'double' analyticp.c:52: warning: cast to pointer from integer of different size analyticp.c:63: warning: implicit declaration of function 'error' analyticp.c:91: warning: implicit declaration of function 'Free' make: *** [analyticp.o] Error 1 ERROR: compilation failed for package 'msm' ** Removing '/usr/lib64/R/library/msm' The downloaded packages are in /tmp/RtmpaoUZiC/downloaded_packages Updating HTML index of packages in '.Library' Warning messages: 1: In install.packages(msm, lib = /usr/lib64/R/library) : installation of package 'msm' had non-zero exit status 2: In tools:::unix.packages.html(.Library) : cannot create HTML package index -- Adam Wilson http://hydrodictyon.eeb.uconn.edu/people/wilson/ Department of Ecology and Evolutionary Biology BioPharm 223 University of Connecticut Tel: 860.486.4157 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] printing to the console in color
If this is RGui on Windows, no. It supports only two colours of text (which you can set in the preferences dialiog box). More generally, you send the appopriate escapes to your console to change colour, and the console/terminal documentation will tell you what those are. On Wed, 23 Apr 2008, Applejus wrote: Hi, Is there a function that would allow me to print things in different colors to the CONSOLE, say one line in blue and another line in green? Right now, the print() function only prints in navy blue... Thanks! -- View this message in context: http://www.nabble.com/printing-to-the-console-in-color-tp16834851p16834851.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do. We ask for minimal information that you have not supplied. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Coefficient of determination in a regression model with AR(1) residuals
Dear R-users, I used lm() to fit a standard linear regression model to a given data set, which led to a coefficient of determination (R^2) of about 0.96. After checking the residuals I realized that they follow an autoregressive process (AR) of order 1 (and therefore contradicting the i.i.d. assumption of the regression model). I then used gls() [library nlme] to fit a linear regression model with AR(1)-residuals. The residuals look perfect (residual plot, ACF, PACF, QQPlot, Ljung- Box test). As mentioned on http://en.wikipedia.org/wiki/ Coefficient_of_determination (citation [2008-04-24]: For cases other than fitting by ordinary least squares, the R^2 statistic can be calculated as above and later: Values for R^2 can be calculated for any type of predictive model), I tried to calculate the standard R^2 for the model with AR(1) residuals. However, I ended up with R^2 larger than 1! As mentioned on the German wikipedia page (http://de.wikipedia.org/ wiki/Bestimmtheitsmaß), in models fitted using Maximum Likelihood Estimation (MLE), the coefficient of determination does _not_ exist (citation [2008-04-24]: Bei bestimmten statistischen Modellen, z.B. bei Maximum-Likelihood-Schätzungen, existiert das Bestimmtheitsmaß R^2 nicht). Any comments on that? The German Wikipedia page mentions McFadden's pseudo-coefficient of determination, the English Wikipedia page the one of Nagelkerke. I know there are others, too. Is there a general agreement on which coefficient of determination (or goodness-of-fit measure in general) to use for a regression model with autocorrelated errors? Is there a possibility to compare (non-graphically) the standard regression model with the model with AR(1) residuals to justify the better fit of the latter? Any comments are appreciated. Best regards. Marius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
