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: [EMAIL PROTECTED]
> > [mailto:[EMAIL PROTECTED] 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" <[EMAIL PROTECTED]> 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:[EMAIL PROTECTED]
> > > > Sent: Friday, April 15, 2005 9:51 PM
> > > > To: 'John Fox'; [EMAIL PROTECTED]
> > > > 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]])$p.value), ncol =
> > > > > ncol(mat)) }
> > > > >
> > > > >
> > > > > My function is cor.pvalues and yours cor.pvals. This is
> > > for a data
> > > > > matrix with 1000 observations on 100 variables:
> > > > >
> > > > > > R <- diag(100)
> > > > > > R[upper.tri(R)] <- R[lower.tri(R)] <- .5
> > > > > > library(mvtnorm)
> > > > > > X <- rmvnorm(1000, sigma=R)
> > > > > > dim(X)
> > > > > [1] 1000 100
> > > > > >
> > > > > > system.time(cor.pvalues(X))
> > > > > [1] 5.53 0.00 5.53 NA NA
> > > > > >
> > > > > > system.time(cor.pvals(X))
> > > > > [1] 12.66 0.00 12.66 NA NA
> > > > > >
> > > > >
> > > > > I frankly didn't expect the advantage of my approach to be
> > > > this large,
> > > > > but there it is.
> > > > >
> > > > > Regards,
> > > > > John
> > > > >
> > > > > --------------------------------
> > > > > John Fox
> > > > > Department of Sociology
> > > > > McMaster University
> > > > > Hamilton, Ontario
> > > > > Canada L8S 4M4
> > > > > 905-525-9140x23604
> > > > > http://socserv.mcmaster.ca/jfox
> > > > > --------------------------------
> > > > >
> > > > > > -----Original Message-----
> > > > > > From: Marc Schwartz [mailto:[EMAIL PROTECTED]
> > > > > > Sent: Friday, April 15, 2005 7:08 PM
> > > > > > To: John Fox
> > > > > > Cc: 'Dren Scott'; R-Help
> > > > > > Subject: RE: [R] Pearson corelation and p-value for matrix
> > > > > >
> > > > > > Here is what might be a slightly more efficient way to
> > > > get to John's
> > > > > > question:
> > > > > >
> > > > > > cor.pvals <- function(mat)
> > > > > > {
> > > > > > rows <- expand.grid(1:nrow(mat), 1:nrow(mat))
> > matrix(apply(rows,
> > > > > > 1,
> > > > > > function(x) cor.test(mat[x[1], ], mat[x[2],
> > ])$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
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