# RE: [R] Pearson corelation and p-value for matrix

From: John Fox <jfox_at_mcmaster.ca>
Date: Tue 19 Apr 2005 - 05:00:30 EST

Since cor(), on which Andy's solution is based, can compute pairwise-present correlations, you could adapt his function -- you'll have to adjust the df for each pair. Alternatively, you could probably save some time (programming time + computer time) by just using my solution:

> R <- diag(100)
> R[upper.tri(R)] <- R[lower.tri(R)] <- .5
> library(mvtnorm)
> X <- rmvnorm(6000, sigma=R)
> system.time(for (i in 1:50) cor.pvalues(X), gc=TRUE)
[1] 518.19 1.11 520.23 NA NA

I know that time is money, but nine minutes (on my machine) probably won't bankrupt anyone.

Regards,
John

John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
905-525-9140x23604
http://socserv.mcmaster.ca/jfox

> -----Original Message-----
> From: r-help-bounces@stat.math.ethz.ch
> [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Dren Scott
> Sent: Monday, April 18, 2005 12:41 PM
> To: 'R-Help'
> Subject: RE: [R] Pearson corelation and p-value for matrix
>
> Hi all,
>
> Thanks Andy, Mark and John for all the help. I really
> appreciate it. I'm new to both R and statistics, so please
> excuse any gaffes on my part.
>
> Essentially what I'm trying to do, is to evaluate for each
> row, how many other rows would have a p-value < 0.05. So,
> after I get my N x N p-value matrix, I'll just filter out
> values that are > 0.05.
>
> Each of my datasets (6000 rows x 100 columns) would consist
> of some NA's. The iterative procedure (cor.pvalues) proposed
> by John would yield the values, but it would take an
> inordinately long time (I have 50 of these datasets to
> process). The solution proposed by Andy, although fast, would
> not be able to incorporate the NA's.
>
> Is there any workaround for the NA's? Or possibly do you
> think I could try something else?
>
> Thanks very much.
>
> Dren
>
>
> "Liaw, Andy" <andy_liaw@merck.com> wrote:
> > From: John Fox
> >
> > Dear Andy,
> >
> > That's clearly much better -- and illustrates an effective strategy
> > for vectorizing (or "matricizing") a computation. I think I'll add
> > this to my list of programming examples. It might be a little
> > dangerous to pass ...
> > through to cor(), since someone could specify type="spearman", for
> > example.
>
> Ah, yes, the "..." isn't likely to help here! Also, it will
> only work correctly if there are no NA's, for example (or
> else the degree of freedom would be wrong).
>
> Best,
> Andy
>
> > Thanks,
> > John
> >
> > --------------------------------
> > John Fox
> > Department of Sociology
> > McMaster University
> > Hamilton, Ontario
> > 905-525-9140x23604
> > http://socserv.mcmaster.ca/jfox
> > --------------------------------
> >
> > > -----Original Message-----
> > > From: Liaw, Andy [mailto:andy_liaw@merck.com]
> > > Sent: Friday, April 15, 2005 9:51 PM
> > > To: 'John Fox'; MSchwartz@medanalytics.com
> > > Cc: 'R-Help'; 'Dren Scott'
> > > Subject: RE: [R] Pearson corelation and p-value for matrix
> > >
> > > We can be a bit sneaky and `borrow' code from cor.test.default:
> > >
> > > cor.pval <- function(x, alternative="two-sided", ...) { corMat <-
> > > cor(x, ...) n <- nrow(x) df <- n - 2 STATISTIC <-
> sqrt(df) * corMat
> > > / sqrt(1 - corMat^2) p <- pt(STATISTIC, df) p <- if
> (alternative ==
> > > "less") { p } else if (alternative == "greater") {
> > > 1 - p
> > > } else 2 * pmin(p, 1 - p)
> > > p
> > > }
> > >
> > > The test:
> > >
> > > > system.time(c1 <- cor.pvals(X), gcFirst=TRUE)
> > > [1] 13.19 0.01 13.58 NA NA
> > > > system.time(c2 <- cor.pvalues(X), gcFirst=TRUE)
> > > [1] 6.22 0.00 6.42 NA NA
> > > > system.time(c3 <- cor.pval(X), gcFirst=TRUE)
> > > [1] 0.07 0.00 0.07 NA NA
> > >
> > > Cheers,
> > > Andy
> > >
> > > > From: John Fox
> > > >
> > > > 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:MSchwartz@MedAnalytics.com]
> > > > > 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: r-help-bounces@stat.math.ethz.ch
> > > > > > > [mailto:r-help-bounces@stat.math.ethz.ch] 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
> > > > http://www.R-project.org/posting-guide.html
> > > >
> > > >
> > > >
> > >
> > >
> > >
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