Re: [R] Help with vector gymnastics
Philip - I don't know if this is the best way, but it gives you the output you want. Using your tf, vals - rle(ifelse(tf, 5*which(tf), 0)) vals$values[vals$values == 0] - vals$values[which(vals$values==0) - 1] inverse.rle(vals) [1] 5 5 5 5 25 30 30 Gladwin, Philip wrote: Hello, What is the best way of solving this problem? answer - ifelse(tf=TRUE, i * 5, previous answer) where as an initial condition tf[1] - TRUE For example if, tf - c(T,F,F,F,T,T,F) over i = 1 to 7 then the output of the function will be answer = 5 5 5 5 25 30 30 Thank you. Phil, __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Imputing missing values in time series
I think my example should work for you, but I couldn't think of a way to do this without an interative while loop. test - c(1,2,3,NA,4,NA,NA,5,NA,6,7,NA) while(any(is.na(test))) test[is.na(test)] - test[which(is.na(test))-1] test [1] 1 2 3 3 4 4 4 5 5 6 7 7 Horace Tso wrote: Folks, This must be a rather common problem with real life time series data but I don't see anything in the archive about how to deal with it. I have a time series of natural gas prices by flow date. Since gas is not traded on weekends and holidays, I have a lot of missing values, FDate Price 11/1/2006 6.28 11/2/2006 6.58 11/3/2006 6.586 11/4/2006 6.716 11/5/2006 NA 11/6/2006 NA 11/7/2006 6.262 11/8/2006 6.27 11/9/2006 6.696 11/10/20066.729 11/11/20066.487 11/12/2006NA 11/13/2006NA 11/14/20066.725 11/15/20066.844 11/16/20066.907 What I would like to do is to fill the NAs with the price from the previous date * gas used during holidays is purchased from the week before. Though real simple, I wonder if there is a function to perform this task. Some of the imputation functions I'm aware of (eg. impute, transcan in Hmisc) seem to deal with completely different problems. 2.5.0/Windows XP Thanks in advance. HT __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I avoid a loop?
One more variation on the solution, no idea how it compares in speed. Using your x ... ifelse(x, unlist(mapply(seq, to = rle(x)$lengths, from = 1)), 0) [1] 1 2 3 0 0 1 2 0 1 Feng, Ken wrote: Hi, I start with an array of booleans: x - c( TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, FALSE, TRUE ); I want to define an y - f(x) such that: y - c( 1, 2, 3, 0, 0, 1, 2, 0, 1 ); In other words, do a cumsum when I see a TRUE, but reset to 0 if I see a FALSE. I know I can do this with a very slow and ugly loop or maybe use apply, but I was hoping there are some R experts out there who can show me a cleaner/more elegant solution? Thanks in advance. - Ken __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Computing stats on common parts of multiple dataframes
Murali - I've come up with something that might with work, with gratutious use of the *apply functions. See ?apply, ?lappy, and ?mapply for how this would work. Basically, just set my.list equal to a list of data.frames you would like included. I made this to work with matrices first, so it does use as.matrix() in my function. Also, this could be turned into a general function so that you could specify a different function other than median. #Make my.list equal to a list of dataframes you want my.list - list(df1,df2) #What's the shortest? minrow - min(sapply(my.list,nrow)) #Chop all to the shortest tmp - lapply(my.list, function(x) x[(nrow(x)-(minrow-1)):nrow(x),]) #Do the computation, could change median to mean, or a user defined #function matrix(apply(mapply([,lapply(tmp,as.matrix), MoreArgs=list(1:(minrow*2))), 1, median), ncol=2) HTH, whether or not this is any better than your for loop solution is left up to you. Erik Murali Menon wrote: Folks, I have three dataframes storing some information about two currency pairs, as follows: R a EUR-USDNOK-SEK 1.231.33 1.221.43 1.261.42 1.241.50 1.211.36 1.261.60 1.291.44 1.251.36 1.271.39 1.231.48 1.221.26 1.241.29 1.271.57 1.211.55 1.231.35 1.251.41 1.251.30 1.231.11 1.281.37 1.271.23 R b EUR-USDNOK-SEK 1.231.22 1.211.36 1.281.61 1.231.34 1.211.22 R d EUR-USDNOK-SEK 1.271.39 1.231.48 1.221.26 1.241.29 1.271.57 1.211.55 1.231.35 1.251.41 1.251.33 1.231.11 1.281.37 1.271.23 The twist is that these entries correspond to dates where the *last* rows in each frame are today's entries, and so on backwards in time. I would like to create a matrix of medians (a median for each row and for each currency pair), but only for those rows where all dataframes have entries. My answer in this case should look like: EUR-USDNOK-SEK 1.251.41 1.251.33 1.231.11 1.281.37 1.271.23 where the last EUR-USD entry = median(1.27, 1.21, 1.27), etc. Notice that the output is of the same dimensions as the smallest dataframe (in this case 'b'). I can do it in a clumsy fashion by first obtaining the number of rows in the smallest matrix, chopping off the top rows of the other matrices to reduce them this size, then doing a for-loop across each currency pair, row-wise, to create a 3-vector which I then apply median() on. Surely there's a better way to do this? Please advise. Thanks, Murali Menon _ Valentine’s Day -- Shop for gifts that spell L-O-V-E at MSN Shopping __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remove component from list or data frame
Jason Horn wrote: Sorry to ask such a simple question, but I can't find the answer after extensive searching the docs and the web. How do you remove a component from a list? For example say you have: lst-c(5,6,7,8,9) How do you remove, for example, the third component in the list? Is the object lst really a list? Try is.list(lst) to check. To remove an element from a vector, use for example, lst[-3] lst[[3]]]-NULL generates an error: Error: more elements supplied than there are to replace If lst were actually a list, that command would work with the obvious syntax fix. So would lst[-3] though. Also, how do you remove a row from a data frame? For example, say you have: lst1-c(1,2,3,4,5) lst2-c(6,7,8,9,10) frame-data.frame(lst1,lst2) How do you remove, for example, the second row of frame? You use frame[-2, ] #remove second row, keep all columns. Thanks, - Jason __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R in Industry
Ben - Ben Fairbank wrote: To those following this thRead: There was a thread on this topic a year or so ago on this list, in which contributors mentioned reasons that corporate powers-that-be were reluctant to commit to R as a corporate statistical platform. (My favorite was There is no one to sue if something goes wrong.) One reason that I do not think was discussed then, nor have I seen discussed since, is the issue of the continuity of support. If one person has contributed disproportionately heavily to the development and maintenance of a package, and then retires or follows other interests, and the package needs maintenance (perhaps as a consequence of new operating systems or a new version of R), is there any assurance that it will be available? With a commercial package such as, say, SPSS, the corporate memory and continuance makes such continued maintenance likely, but is there such a commitment with R packages? If my company came to depend heavily on a fairly obscure R package (as we are contemplating doing), what guarantee is there that it will be available next month/year/decade? I know of none, nor would I expect one. But you would have the source code, so as long as someone knew R, you could maintain it, expand it, customize it, patch it yourselves, even if the original maintainer left the project. You can't say the same with a commercial package likely. As R says when it starts up, R is free software and comes with ABSOLUTELY NO WARRANTY. Ben Fairbank -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns Sent: Thursday, February 08, 2007 10:24 AM To: Albrecht,Dr. Stefan (AZ Private Equity Partner) Cc: r-help@stat.math.ethz.ch Subject: Re: [R] R in Industry From what I know Matlab is much more popular in fixed income than R, but R is vastly more popular in equities. R seems to be making quite a lot of headway in finance, even in fixed income to some degree. At least to some extent, this is probably logical behavior -- fixed income is more mathematical, and equities is more statistical. Matlab is easier to learn mainly because it has much simpler data structures. However, once you are doing something where a complex data structure is natural, then R is going to be easier to use and you are likely to have a more complete implementation of what you want. If speed becomes a limiting factor, then moving the heavy computing to C is a natural thing to do, and very easy with R. Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) Albrecht, Dr. Stefan (AZ Private Equity Partner) wrote: Dear all, I was reading with great interest your comments about the use of R in the industry. Personally, I use R as scripting language in the financial industry, not so much for its statistical capabilities (which are great), but more for programming. I once switched from S-Plus to R, because I liked R more, it had a better and easier to use documentation and it is faster (especially with loops). Now some colleagues of mine are (finally) eager to join me in my quantitative efforts, but they feel that they are more at ease with Matlab. I can understand this. Matlab has a real IDE with symbolic debugger, integrated editor and profiling, etc. The help files are great, very comprehensive and coherent. It also could be easier to learn. And, I was very astonished to realise, Matlab is very, very much faster with simple for loops, which would speed up simulations considerably. So I have trouble to argue for a use of R (which I like) instead of Matlab. The price of Matlab is high, but certainly not prohibitive. R is great and free, but maybe less comfortable to use than Matlab. Finally, after all, I have the impression that in many job offerings in the financial industry R is much less often mentioned than Matlab. I would very much appreciate any comments on my above remarks. I know there has been some discussions of R vs. Matlab on R-help, but these could be somewhat out-dated, since both languages are evolving quite quickly. With many thanks and best regards, Stefan Albrecht [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list
Re: [R] setting a number of values to NA over a data.frame.
John -
Your initial problem uses 0, but the example uses 1 for the value that
gets an NA. My solution uses 1 to fit with your example. There may be
a better way, but try something like
data1[3:5] - data.frame(lapply(data1[3:5], function(x) ifelse(x==1, NA,
x)))
The data1[3:5] is just a test subset of columns I chose from your data1
example. Notice it appears twice, once on each side of the assignment
operator.
In English, apply to each column of the data frame (which is a list) a
function that will return NA if the element is 1, and the value
otherwise, and then turn the modified lists into a data.frame, and save
it as data1.
See the help files for lapply and ifelse if you haven't seen those before.
Maybe someone has a better way?
Erik
John Kane wrote:
This is probably a simple problem but I don't see a
solution.
I have a data.frame with a number of columns where I
would like 0 - NA
thus I have df1[,144:157] - NA if df1[, 144: 157] ==0
and df1[, 190:198] - NA if df1[, 190:198] ==0
but I cannot figure out a way do this.
cata - c( 1,1,6,1,1,NA)
catb - c( 1,2,3,4,5,6)
doga - c(3,5,3,6,4, 0)
dogb - c(2,4,6,8,10, 12)
rata - c (NA, 9, 9, 8, 9, 8)
ratb - c( 1,2,3,4,5,6)
bata - c( 12, 42,NA, 45, 32, 54)
batb - c( 13, 15, 17,19,21,23)
id - c('a', 'b', 'b', 'c', 'a', 'b')
site - c(1,1,4,4,1,4)
mat1 - cbind(cata, catb, doga, dogb, rata, ratb,
bata, batb)
data1 - data.frame(site, id, mat1)
data1
# Obviously this works fine for one column
data1$site[data1$site ==1] - NA ; data1
but I cannot see how to do this with indices that
would allow me to do more than one column in the
data.frame.
At one point I even tried something like this
a - c(site)
data1$a[data1$a ==1] - NA
which seems to produce a corrupt data.frame.
I am sure it is simple but I don't see it.
Any help would be much appreciated.
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Re: [R] Use a text variable's value to specify another varaible?
Can I ask why you aren't just passing in the object to your function, but instead a text name for that object? Ben Fairbank wrote: Greetings guRus -- If a variable, e.g., 'varname', is a character string, e.g. varname - datavector, and I want to apply a function, such as table(), to datavector, what syntax or method will do so using only the variable varname? This seems similar to indirect addressing, but I have not seen a method for it in the R manuals. Is there a general name for such indirect reference that one might search for? (This came up while writing a function that takes the value of 'varname' from the keyboard and then applies functions to it.) With thanks for any solution, Ben Fairbank [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] relative frequency plot
See ?truehist in the MASS package. Philipp Pagel wrote: On Thu, Apr 27, 2006 at 10:48:39AM -0700, [EMAIL PROTECTED] wrote: Hi All, I want to use hist to get the relative frequency plot. But the range of ylab is greater than 1,which I think it should be less than 1 since it stands for the probability. I'm confused. Could you please help me with it? I was pretty confused by that, too at first. The solution is that freq=False cause hist to plot the DENSITY rather than frequency. And density is not necesssarily the same as relative frequency. Excerpt from ?hist: density: values f^(x[i]), as estimated density values. If 'all(diff(breaks) == 1)', they are the relative frequencies 'counts/n' and in general satisfy sum[i; f^(x[i]) (b[i+1]-b[i])] = 1, where b[i] = 'breaks[i]'. If you want relative distance try something like this: myhist = hist(x,breaks=52, plot=F) myhist$counts = myhist$counts / sum(myhist$counts) plot(myhist,main=NULL,border=TRUE,xlab=days,xlim=c(0,6),lty=2) Not exactly clean, though -- we are messing with the myhist object... cu Philipp __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] relative frequency plot
Martin - Of course you are right. The documentation for truehist (and hist) explains that fact nicely, which is why I thought to send him there. Sorry for any confusion. Thanks, Erik Martin Maechler wrote: Erik == Erik Iverson [EMAIL PROTECTED] on Thu, 27 Apr 2006 13:44:16 -0500 writes: Erik See ?truehist in the MASS package. Not in this case! truehist() also computes a density, and its values on the y axis are not probabilities, either! hist(*, freq = FALSE) is fully sufficient here -- the problem of the original poster was to understand that a density can have values larger than 1. It may be interesting and is somewhat disappointing for us teachers of statistics to see how many people have posted in the past on this exact topic, sometimes even more or less assuming that R was doing some things wrongly because it showed densities (or density estimates as here) with values larger than one... oh dear Mit der Dummheit kaempfen Goetter selbst vergebens. - Friedrich Schiller, Die Jungfrau von Orleans Martin Philipp Pagel wrote: On Thu, Apr 27, 2006 at 10:48:39AM -0700, [EMAIL PROTECTED] wrote: Hi All, I want to use hist to get the relative frequency plot. But the range of ylab is greater than 1,which I think it should be less than 1 since it stands for the probability. I'm confused. Could you please help me with it? I was pretty confused by that, too at first. The solution is that freq=False cause hist to plot the DENSITY rather than frequency. And density is not necesssarily the same as relative frequency. Excerpt from ?hist: density: values f^(x[i]), as estimated density values. If 'all(diff(breaks) == 1)', they are the relative frequencies 'counts/n' and in general satisfy sum[i; f^(x[i]) (b[i+1]-b[i])] = 1, where b[i] = 'breaks[i]'. If you want relative distance try something like this: myhist = hist(x,breaks=52, plot=F) myhist$counts = myhist$counts / sum(myhist$counts) plot(myhist,main=NULL,border=TRUE,xlab=days,xlim=c(0,6),lty=2) Not exactly clean, though -- we are messing with the myhist object... cu Philipp Erik __ Erik R-help@stat.math.ethz.ch mailing list Erik https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do Erik read the posting guide! Erik http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Were to find appropriate functions for a given task in R
Then, people tend to define its own functions (I'm doing this too), and a lack of standardization makes it difficult to keep everything into control. If you think of R as more of a language rather than a pre-packaged statistical program, I feel that helps. In the C++ world, there are people all over writing classes and functions, and many of these have close or duplicate functionality. As a programmer, you can decide which one to use for your needs, or program your own, or extend someone else's, the choice is yours. The same in the R world, you have choices, and there does not have to be necessarily only method to do each task. I don't feel there needs to be 'control', everyone can implement what they want. If you think of R only as a prepacked statistical program however, I can see getting frustrated that there are usually many ways to do the same thing. R is much more than this though. That's how I see it anyway. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Indexing a vector by a list of vectors
Hello R-help -
I have
vec - c(string1, string2, string3)
ind - list(c(1,2),c(1,2,3))
I want vec indexed by each vector in the list ind.
The first element of the list I want would be vec[c(1,2)],
the second element would be vec[c(1,2,3)], like the following.
[[1]]
[1] string1 string2
[[2]]
[1] string1 string2 string3
Using for loops, this is simple. For fun, I tried to implement it
without a for loop using some combination of *apply() functions and [.
I succeeded with
myfunc - function(x) {
do.call([,list(vec,x))
}
lapply(ind,myfunc)
I was not, however, able to get my desired result without defining my
own dummy function. Can anyone think of a way? As I said, I already
have a way that works, I'm just curious if there is a more 'elegant'
solution that does not rely on my having to define another function.
Seems like it should be possible.
Thanks, Erik Iverson
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Re: [R] Function dependency function
I had a similar need and found package mvbutils, function foodweb(). From the help file: 'foodweb' is applied to a group of functions (e.g. all those in a workspace); it produces a graphical display showing the hierarchy of which functions call which other ones. This is handy, for instance, when you have a great morass of functions in a workspace, and want to figure out which ones are meant to be called directly. 'callers.of(funs)' and 'callees.of(funs)' show which functions directly call, or are called directly by, 'funs'. Hope that helps, Erik Iverson Matthew Dowle wrote: Hi, Is there a function taking a function as an argument, which returns all the functions it calls, and all the the functions those functions call, and so on?I could use Rprof, but that would involve executing the function, which may miss some branches of code. I'd really like a function which looks at the source code to work out all the functions that could possibly be called. When I develop a function and release to production environment (or to some library) then I may need to release other functions I've developed which that function calls. As soon as the function call stack goes outside .GlobalEnv (for example into base) then the search can stop as I'm only interested in functions in .GlobalEnv (my own functions). Also useful would be the reverse function i.e. find all functions which could possibly call the function. This could be used to find functions which are never called and could be considered for deletion. Thanks, Matthew [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] tcltk loading in R-2.2.1 from src
Patrice - I had a very similar problem using TCL/TK 8.3. Below is the email I sent to my computing group at work about how I fixed it. Note that since my TCL/TK header (.h) files were in an odd location, the first step probably isn't relevant for you. But I bet the second step is. -- I believe I have found the solution to this problem. There were 2 steps I took to get a build of R in my home directory that properly uses Tcl/Tk 8.3 under Linux. The first was setting an environment variable to let R know where the tcl.h and tk.h files reside. This environment variable was TCLTK_CPPFLAGS and is set to -I/s/include . This can bet set in the config.site in R's build directory also. (NOTE to R-help: The above is site specific to my location, /s is for software on the network.) Second, there is some problem with the way R interacts with the tkConfig.sh file in /s/lib (or wherever your .sh file is located). It comes from the following line in tkConfig.sh. TK_XINCLUDES='# no special path needed' The # character, indicating a comment, is somehow misinterpreted by the configure script which breaks the Tcl/Tk functionality. I was able to get around this by simply removing the comment and leaving it as TK_XINCLUDES='' That got me a working version of the newest R using Tcl/Tk. I'm not sure if Tcl/Tk version 8.4 would still put that comment in there, I believe it's changed though, so you'd only have to follow the first step to get R compiled with Tcl/Tk support. I found reference to this problem on the R mailing list, it appears to only affect certain installations of Tcl/Tk. - HTH, Erik Iverson Patrice Seyed wrote: Hi, Having trouble loading tcltk in R 2.2.1 built from source. ./configure, make, make check, and make install run ok. library(tcltk) Error in firstlib(which.lib.loc, package) : Tcl/Tk support is not available on this system Error in library(tcltk) : .First.lib failed for 'tcltk' even though it is listed in library() output. I have the same problem even if i compile with options: ./configure --with-tcltk --with-tcl-config=/usr/lib/tclConfig.sh --with-tk-config=/usr/lib/tkConfig.sh Is there a dep for R 2.2.1 on a specific version of tcl? Any hints on this issue appreciated. running on linux (2.4.21-4) version: rpm -qa | grep tcl tcl-devel-8.3.5-92 tcl-8.3.5-92 Specifically, the package pbatR loads this library during installation. Thanks, Patrice __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
