Dear Mark,
> -----Original Message-----
> From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> Sent: Friday, April 15, 2005 9:41 PM
> To: John Fox
> Cc: 'R-Help'; 'Dren Scott'
> Subject: RE: [R] Pearson corelation and p-value for matrix
>
> John,
>
> Interesting test. Thanks for pointing that out.
>
> You are right, there is a knee-jerk reaction to avoid loops,
> especially nested loops.
>
> On the indexing of rows, I did that because Dren had
> indicated in his initial post:
>
> "I was trying to evaluate the pearson correlation and the p-values
> for an nxm matrix, where each row represents a vector.
> One way to do it would be to iterate through each row, and find its
> correlation value( and the p-value) with respect to the other rows."
>
> So I ran the correlations by row, rather than by column.
>
That's the second time yesterday that I responded to a posting without
reading it carefully enough -- a good lesson for me. I guess that Dren could
just apply my solution to the transpose of his matrix -- i.e.,
cor.pvalues(t(X)).
Sorry,
John
> Thanks again. Good lesson.
>
> Marc
>
> On Fri, 2005-04-15 at 21:36 -0400, John Fox wrote:
> > Dear Mark,
> >
> > I think that the reflex of trying to avoid loops in R is often
> > mistaken, and so I decided to try to time the two approaches (on a
> > 3GHz Windows XP system).
> >
> > I discovered, first, that there is a bug in your function -- you
> > appear to have indexed rows instead of columns; fixing that:
> >
> > cor.pvals <- function(mat)
> > {
> > cols <- expand.grid(1:ncol(mat), 1:ncol(mat))
> > matrix(apply(cols, 1,
> > function(x) cor.test(mat[, x[1]], mat[,
> x[2]])$p.value),
> > ncol = ncol(mat))
> > }
> >
> >
> > My function is cor.pvalues and yours cor.pvals. This is for a data
> > matrix with 1000 observations on 100 variables:
> >
> > > R <- diag(100)
> > > R[upper.tri(R)] <- R[lower.tri(R)] <- .5
> > > library(mvtnorm)
> > > X <- rmvnorm(1000, sigma=R)
> > > dim(X)
> > [1] 1000 100
> > >
> > > system.time(cor.pvalues(X))
> > [1] 5.53 0.00 5.53 NA NA
> > >
> > > system.time(cor.pvals(X))
> > [1] 12.66 0.00 12.66 NA NA
> > >
> >
> > I frankly didn't expect the advantage of my approach to be
> this large,
> > but there it is.
> >
> > Regards,
> > John
> >
> > --------------------------------
> > John Fox
> > Department of Sociology
> > McMaster University
> > Hamilton, Ontario
> > Canada L8S 4M4
> > 905-525-9140x23604
> > http://socserv.mcmaster.ca/jfox
> > --------------------------------
> >
> > > -----Original Message-----
> > > From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> > > Sent: Friday, April 15, 2005 7:08 PM
> > > To: John Fox
> > > Cc: 'Dren Scott'; R-Help
> > > Subject: RE: [R] Pearson corelation and p-value for matrix
> > >
> > > Here is what might be a slightly more efficient way to
> get to John's
> > > question:
> > >
> > > cor.pvals <- function(mat)
> > > {
> > > rows <- expand.grid(1:nrow(mat), 1:nrow(mat))
> > > matrix(apply(rows, 1,
> > > function(x) cor.test(mat[x[1], ], mat[x[2],
> > > ])$p.value),
> > > ncol = nrow(mat))
> > > }
> > >
> > > HTH,
> > >
> > > Marc Schwartz
> > >
> > > On Fri, 2005-04-15 at 18:26 -0400, John Fox wrote:
> > > > Dear Dren,
> > > >
> > > > How about the following?
> > > >
> > > > cor.pvalues <- function(X){
> > > > nc <- ncol(X)
> > > > res <- matrix(0, nc, nc)
> > > > for (i in 2:nc){
> > > > for (j in 1:(i - 1)){
> > > > res[i, j] <- res[j, i] <- cor.test(X[,i],
> X[,j])$p.value
> > > > }
> > > > }
> > > > res
> > > > }
> > > >
> > > > What one then does with all of those non-independent test
> > > is another
> > > > question, I guess.
> > > >
> > > > I hope this helps,
> > > > John
> > >
> > > > > -----Original Message-----
> > > > > From: [EMAIL PROTECTED]
> > > > > [mailto:[EMAIL PROTECTED] On Behalf Of Dren
> > > > > Scott
> > > > > Sent: Friday, April 15, 2005 4:33 PM
> > > > > To: r-help@stat.math.ethz.ch
> > > > > Subject: [R] Pearson corelation and p-value for matrix
> > > > >
> > > > > Hi,
> > > > >
> > > > > I was trying to evaluate the pearson correlation and the
> > > > > p-values for an nxm matrix, where each row represents
> a vector.
> > > One way to do
> > > > > it would be to iterate through each row, and find its
> > > > > correlation value( and the p-value) with respect to the other
> > > > > rows. Is there some function by which I can use the
> matrix as input?
> > > Ideally, the
> > > > > output would be an nxn matrix, containing the p-values
> > > between the
> > > > > respective vectors.
> > > > >
> > > > > I have tried cor.test for the iterations, but couldn't find a
> > > > > function that would take the matrix as input.
> > > > >
> > > > > Thanks for the help.
> > > > >
> > > > > Dren
> > >
> > >
> >
> > ______________________________________________
> > 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
>
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