RE: [R] Matrix problems

It 'unwraps' the 200,000 x 2 matrix into a vector with 400,000 elements and giving you the location/subscripts of this vector that matches the criterion. Instead you should use apply(mat1, 2, function(y) which( y == 2 )) which gives you the rows in each column that corresponds to 185.

RE: [R] r-help using random generating

Yes it looks like a really good student exercise into the sampling properties of a distribution. What you need to do is initialize vectors beforehand that will contain the mean and variance values. Then simply use a for() loop that generates 5 random numbers. Use the mean(), var() and store the

[R] X11 connection error in web cgi mode only

Dear all, I am trying to create a web interface using Perl-CGI to call R plots and to display them. The following codes works perfectly fine when I copy and paste into the console directly or if I save it into script.file and then R --no-save script.file producing the graphs. jpeg(graph.jpeg,

[R] Execution halts after some time.

Dear all, I fit independent GLMs for a 2x2 factorial problem on the data matrix of size 9500 x 12 (genes x arrays) and get 9500 observed t-values using the apply() function. Now, I wish to get the permutated p-values. Therefore I random sample the class labels and perform the glm fitting to get

RE: [R] Simple question about export

Well, you did not specify the object to save in write.table(). Suppose you want to save the following: x - matrix( c(1:25), nrow=5) write.table(x, file=c:/my.output.file.txt) If all goes well, you should see the prompt without any message. Also use / rather than \ as it is reserved for escape

[R] Using evaluate-deparse-substitute

Being the lazy soul I am, I wish to write a function to replace saying ls(pattern=...) everytime. Here is what I have: lsp - function(x){ y - eval(deparse(substitute(x))) print(y) # CHECK print( ls(pattern = eval(y)) )# TRY 1

RE: [R] Interpolation

Here is another way of doing it: x - c(1990, 1994, 1995, 1997) all.x - seq(min(x), max(x)) complementary.x - setdiff(all.x, x) -Original Message- From: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED]] Sent: Wednesday, February 12, 2003 9:39 PM To: Remigijus Lapinskas Cc: [EMAIL PROTECTED]

RE: [R] to modify a matrix

May I suggest you learn the handy function called which(). yes.rows - which(x[,1]==3)# Row that meets this condition x[ yes, 2] - x[ yes.rows, 2] * 5 # Perform the desired change for the row that meets this condition. At 12:04 AM 2/9/2003 +0900, Mitsuo Igarashi wrote: Hi All. I am

RE: [R] Clustering partition and memory

What distance metric are you using? See if clara() in cluster library etc is more appropriate. -Original Message- From: Vincent Stoliaroff [mailto:[EMAIL PROTECTED]] Sent: Sunday, February 09, 2003 7:33 PM To: [EMAIL PROTECTED] Subject: [R] Clustering partition and memory Dear

[R] Finding inflexion points from data

Dear all, I have a measurements from an experiment. Graphically I can see there is one inflexion point but would like to automate this process accurately. I am wondering if someone had written such a code (which I think is similar to turnpoints() in library pastecs). Thank you. Regards, Adai.

RE: [R] key in margin area

Well, if you don't want to play with the margins, then I guess you have to use the layout() function. You will want something similar to the following if you want legends at the right: def.par - par(no.readonly = TRUE)# save default nf - layout( matrix( c(1,2), nrow=1), c(4,1) ) # If you want

[R] Rearranging subtrees and Eisen Cluster

I am attempting to replicate what Cluster 3.0 and Treeview (both by Mike Eisen) to cluster both microarray genes and arrays does using R with hclust. I basically utilized the plot.mat function in sma library with some layout() and hclust(). 1. Can I know if some has already written such a

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