similar to: VIM Syntax Highlighting

Hi list, I'm trying to get R to make a graph like the one shown in this pdf, where males are white bars and females are black bars. http://www.brrc.unr.edu/~knussear/mmgraph.pdf I tried barchart, but I couldnt get the bars to share a common x axis. Do you have any suggestions? Thanks for your help Ken

I have a dataset which I want to model using a Poisson distribution, with a given parameter. I would like to know what is the proper way to do a ''goodness of fit'' test using R. I know the steps I''d take if I were to do it ''manually'': grouping the numbers into classes, calculating the expected frequencies using ''ppois'', then

I noticed some very weird behaviour of the function: options(digits=n), where n is the number of digits you would expect to get in R calculations. Let's take a example: > options(digits=4) > getdata(caso.pool.k3.r3.e2) [1] 6.053 2.641 -3.639 14.259 6.082 Which works fine... now, trying again, with different data: > options(digits=4) > getdata(controle.pool.k3.r3.e2)

Dear All, I have a list, where several components are NULL, and I'd like to obtain that very same list without the NULL components (i.e., I do not want to unlist or otherwise loose the rest of the list structure). I can do that with a loop, but how could I do it without a loop? Thanks, Ram?n -- Ram?n D?az-Uriarte Bioinformatics Unit Centro Nacional de Investigaciones Oncol?gicas (CNIO)

Hi, my doubt is very simple. I'm sure I've seen someone using something like this before, but unfortunatelly my searches in the archives were useless. Well, I have some objects called after a name that has a number attached to it, varying. Let's say I have: > ls poly1 poly2 poly3 poly4 poly5 poly6 ... I would like to access these objects using a for(), in which I could do

Is it possible in R to subset a dataframe by more than one factor, all at once? For instance, I have the dataframe: >data p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 pred 1 0 1 0 0 0 0 0 0 0 0 0.5862069 4 0 0 0 0 0 0 0 0 0 1 0.5862069 5 0 0 0 0 0 0 1 0 0 0 0.5862069 6 0 0 0 0 0 0 0 1 0 0 0.5862069 7 0 0 1 0 0 0 0 0 0

Are there config files for vim that offer syntax highlighting of rjs templates? Charlie Bowman http://www.recentrambles.com -------------- next part -------------- An HTML attachment was scrubbed... URL: http://wrath.rubyonrails.org/pipermail/rails/attachments/20060329/486c9d84/attachment.html

I have made a vim syntax highlighting script for LLVM. It can be found at llvm/utils/vim/llvm.vim. You should copy this file to your ~/.vim/syntax/ directory and add the following lines to your .vimrc file to have it automatically load on a .ll file. augroup filetype au! BufRead,BufNewFile *.ll set filetype=llvm augroup END Misha claims to be making a script for all you emacs

Is it possible to use R as a data-mining tool? Here's the problem I've got. I have a couple of data sets consisting of results from a cDNA microarray experiment - the details about the biology don't really matter here, the same theory applies for any other data-mining task (that's why I thought it'd be more appropriate to post this on r-user). Each of these datasets consists

Hi, If I have a vector: junk <- c(2,0,0,3,0) and want to access, say, all the elements that are greater than zero. I just do: junk[which(junk>0)] Now, If I have a list: jlist <- list(NULL,c(1,0),NULL,c(1,2,3), NULL) and want to access all the elements that have length greater than zero, I know how to find the elements with: which(sapply(jlist,length)>0) But how do I get a

Dear R-devel & Translation Teams, In order to monitor the progress of the translation for the pt_BR team I wrote a script to summarize the status of the translations. It wasn't difficult to extend it to the other languages so I decided to set up a page with the summaries of the translation for all languages for which currently exist a translation.

Hi, Imagine a vector x with elements (1,2,1,1,3,5,3,3,1) and a vector y with elements (2,3). I need to find out what elements of x match any of the elements of y. Is there a simple command that will return a vector with elements (F,T,F,F,T,F,T,T,F). Ideally, I would like a solution that works with dataframe colums as well. I have tried x==y and it doesn't work. x==any(y) doesn't work

Dear r-useRs, After a couple of years in a 'R exile' of sorts, I've recently changed jobs and my current employer (an American multinational in the food manufacturing industry) is much more open than my past employer (which wouldn't even want to hear about anything that didn't begin with SAS...). So, after my insistence corporate IT is now considering adopting R as part of our

Dear r-useRs, After a couple of years in a 'R exile' of sorts, I've recently changed jobs and my current employer (an American multinational in the food manufacturing industry) is much more open than my past employer (which wouldn't even want to hear about anything that didn't begin with SAS...). So, after my insistence corporate IT is now considering adopting R as part of our

Hi, I am writing a for loop that creates one object, say 'outn' on every round of the loop. I would like the name of each object to include the index of the loop as in, for example: out1, out2, out3, ... And I would like the naming of the object to take place automatically as the loop moves through? Similarly, I would like to be able to call different objects (in1, in2, in3,

Hi, I have a vector 'data' of 58 probability values (bounded between 0 and 1) and want to draw a probability density function of these values. For this, I used the commands: data <- runif(58) a <- density(data, from=0, to=1) plot(a, type="l",lwd=3) But then, when I try to approximate the area under the plotted curve with the command: area <- sum(a$y)*(a$x[1]-a$y[2])

Hi I am trying to install vim-puppet for syntax highlight of puppet manifest file. I first installed vim-puppet and then vim. Then I ran "vim-addon-manager -w install puppet" I am still unable to see the colours on site.pp bala@WheezyMaster:~$ vim-addon-manager # Name User Status System Status editexisting removed removed justify

Hi, Anyone knows what would be a short way of extracting a month from a date in numeric or integer format? months("1979-12-20") returns "December" in character format. How could I get 12 in numeric or integer format? Thanks! G. [[alternative HTML version deleted]]

Hi, I'd like to get away from SAS, but I don't really know R well enough at this point to know if it would be good for this project. I tried to describe the essence of the project below without getting bogged down in details. It starts when I receive a data flat file. There's lots of columns, but the relevant ones are: custid (customer ID number) saledt (date of sale)

Full_Name: Fernando Henrique Ferraz Pereira da Rosa Version: 1.8.0 OS: Linux 2.4.21 Submission from: (NULL) (200.206.211.169) After installing R 1.8.0, the R DBI interface stopped working. I tracked it down as a problem in the 'methods' package, that comes in the default installation. Somehow the function '.valueClassTest' which is defined on package 'methods',