Pearson corelation and p-value for matrix

-----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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