[R] Pearson corelation and p-value for matrix
John Fox
jfox at mcmaster.ca
Tue Apr 19 01:10:25 CEST 2005
Dear Andy,
Very nice! (My point was that if this is a one-time thing, for Dren to
puzzle over it is probably more time-consuming than simply doing it
inefficiently.)
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: Liaw, Andy [mailto:andy_liaw at merck.com]
> Sent: Monday, April 18, 2005 2:40 PM
> To: 'John Fox'; 'Dren Scott'
> Cc: 'R-Help'
> Subject: RE: [R] Pearson corelation and p-value for matrix
>
> I believe this will do:
>
> cor.pval2 <- function(x, alternative="two-sided") {
> corMat <- cor(x, use=if (any(is.na(x)))
> "pairwise.complete.obs" else
> "all")
> df <- crossprod(!is.na(x)) - 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
> }
>
> Some test:
>
> > set.seed(1)
> > x <- matrix(runif(2e3 * 1e2), 2e3)
> > system.time(res1 <- cor.pval(t(x)), gcFirst=TRUE)
> [1] 17.28 0.77 18.16 NA NA
> > system.time(res2 <- cor.pval2(t(x)), gcFirst=TRUE)
> [1] 19.51 1.05 20.70 NA NA
> > max(abs(res1 - res2))
> [1] 0
> > x[c(1, 3, 6), c(2, 4)] <- NA
> > x[30, 61] <- NA
> > system.time(res2 <- cor.pval2(t(x)), gcFirst=TRUE)
> [1] 24.48 0.71 25.28 NA NA
>
> This is a bit slower because of the extra computation for
> "df". One can try to save some computation by only computing
> with the lower (or upper) triangular part.
>
> Cheers,
> Andy
>
> > From: John Fox
> >
> > Dear Dren,
> >
> > 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
> > Canada L8S 4M4
> > 905-525-9140x23604
> > http://socserv.mcmaster.ca/jfox
> > --------------------------------
> >
> > > -----Original Message-----
> > > From: r-help-bounces at stat.math.ethz.ch
> > > [mailto:r-help-bounces at 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 at 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
> > > > Canada L8S 4M4
> > > > 905-525-9140x23604
> > > > http://socserv.mcmaster.ca/jfox
> > > > --------------------------------
> > > >
> > > > > -----Original Message-----
> > > > > From: Liaw, Andy [mailto:andy_liaw at merck.com]
> > > > > Sent: Friday, April 15, 2005 9:51 PM
> > > > > To: 'John Fox'; MSchwartz at 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]])$p.value), ncol =
> > > > > > ncol(mat)) }
> > > > > >
> > > > > >
> > > > > > My function is cor.pvalues and yours cor.pvals. This is
> > > > for a data
> > > > > > matrix with 1000 observations on 100 variables:
> > > > > >
> > > > > > > R <- diag(100)
> > > > > > > R[upper.tri(R)] <- R[lower.tri(R)] <- .5
> > > > > > > library(mvtnorm)
> > > > > > > X <- rmvnorm(1000, sigma=R)
> > > > > > > dim(X)
> > > > > > [1] 1000 100
> > > > > > >
> > > > > > > system.time(cor.pvalues(X))
> > > > > > [1] 5.53 0.00 5.53 NA NA
> > > > > > >
> > > > > > > system.time(cor.pvals(X))
> > > > > > [1] 12.66 0.00 12.66 NA NA
> > > > > > >
> > > > > >
> > > > > > I frankly didn't expect the advantage of my approach to be
> > > > > this large,
> > > > > > but there it is.
> > > > > >
> > > > > > Regards,
> > > > > > John
> > > > > >
> > > > > > -------------------------------- John Fox Department of
> > > > > > Sociology McMaster University Hamilton, Ontario
> Canada L8S 4M4
> > > > > > 905-525-9140x23604
> > > > > > http://socserv.mcmaster.ca/jfox
> > > > > > --------------------------------
> > > > > >
> > > > > > > -----Original Message-----
> > > > > > > From: Marc Schwartz [mailto:MSchwartz at MedAnalytics.com]
> > > > > > > Sent: Friday, April 15, 2005 7:08 PM
> > > > > > > To: John Fox
> > > > > > > Cc: 'Dren Scott'; R-Help
> > > > > > > Subject: RE: [R] Pearson corelation and p-value for matrix
> > > > > > >
> > > > > > > Here is what might be a slightly more efficient way to
> > > > > get to John's
> > > > > > > question:
> > > > > > >
> > > > > > > cor.pvals <- function(mat)
> > > > > > > {
> > > > > > > rows <- expand.grid(1:nrow(mat), 1:nrow(mat))
> > > matrix(apply(rows,
> > > > > > > 1,
> > > > > > > function(x) cor.test(mat[x[1], ], mat[x[2],
> > > ])$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 at stat.math.ethz.ch
> > > > > > > > > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of
> > > > > > Dren Scott
> > > > > > > > > Sent: Friday, April 15, 2005 4:33 PM
> > > > > > > > > To: r-help at 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
> > > > > > >
> > > > > > >
> > > > > >
> > > > > > ______________________________________________
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