display only the top-right half of a correlation matrix?

Dear all
Is there an easy way to display only one half (top-right or
bottom-left) of a correlation matrix?

require(Hmisc)
rcorr(as.matrix(mtcars[ , 1:4]))

? ? ? mpg ? cyl ?disp ? ?hp
mpg ? 1.00 -0.85 -0.85 -0.78
cyl ?-0.85 ?1.00 ?0.90 ?0.83
disp -0.85 ?0.90 ?1.00 ?0.79
hp ? -0.78 ?0.83 ?0.79 ?1.00

Use as.dist: here's an example.

x = matrix(rnorm(5*100), 100, 5)
as.dist(cor(x))

1             2             3             4
2 -2.892981e-06
3  2.873711e-02  1.002969e-02
4 -5.803705e-02  4.022733e-02 -6.154211e-02
5  1.137083e-01 -8.065676e-02 -9.279316e-02 -8.201583e-02

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On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder

Use as.dist: here's an example.

Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T', but then as.dist() replaces the diagonal
values with zero. (See below.) Is there a way to prevent it from doing
that? Either keep the original values, or not display anything in the
diagonal (as for the upper part)?

Regards
Liviu

(xb <- rcorr(as.matrix(mtcars[ , 1:4])))

mpg   cyl  disp    hp
mpg   1.00 -0.85 -0.85 -0.78
cyl  -0.85  1.00  0.90  0.83
disp -0.85  0.90  1.00  0.79
hp   -0.78  0.83  0.79  1.00

n= 32


P
     mpg cyl disp hp
mpg       0   0    0
cyl   0       0    0
disp  0   0        0
hp    0   0   0

round(as.dist(xb$r, T), 2)

mpg   cyl  disp    hp
mpg   0.00
cyl  -0.85  0.00
disp -0.85  0.90  0.00
hp   -0.78  0.83  0.79  0.00

On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
<peter.langfelder at gmail.com> wrote:

Use as.dist: here's an example.

Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T', but then as.dist() replaces the diagonal
values with zero. (See below.) Is there a way to prevent it from doing
that? Either keep the original values, or not display anything in the
diagonal (as for the upper part)?

if as.dist doesn't work, use brute force:

x = matrix(rnorm(5*100), 100, 5)
mat = signif(cor(x), 2);
mat[lower.tri(mat)] = ""

data.frame(mat)

Peter

On Fri, Aug 19, 2011 at 9:38 PM, Peter Langfelder

if as.dist doesn't work, use brute force:

x = matrix(rnorm(5*100), 100, 5)
mat = signif(cor(x), 2);
mat[lower.tri(mat)] = ""

data.frame(mat)

Yes, brute force works. This isn't quite how I wanted to do this, but
the following seems to work for me.

Thanks all
Liviu

require(Hmisc)
print.rcorr <-
    function (x, upper=FALSE, ...)
{
    r <- format(round(x$r, 2))
    if(!is.null(upper)) r[if(!upper) upper.tri(r) else lower.tri(r)] <- ''
    print(data.frame(r))
    n <- x$n
    if (all(n == n[1, 1]))
        cat("\nn=", n[1, 1], "\n\n")
    else {
        cat("\nn\n")
        print(n)
    }
    cat("\nP\n")
    P <- x$P
    P <- ifelse(P < 0.0001, 0, P)
    p <- format(round(P, 4))
    p[is.na(P)] <- ""
    if(!is.null(upper)) p[if(!upper) upper.tri(p) else lower.tri(p)] <- ''
    print(p, quote = FALSE)
    invisible()
}

(xb <- rcorr(as.matrix(mtcars[ , 1:4])))

mpg   cyl  disp    hp
mpg   1.00
cyl  -0.85  1.00
disp -0.85  0.90  1.00
hp   -0.78  0.83  0.79  1.00

n= 32


P
     mpg cyl disp hp
mpg
cyl   0
disp  0   0
hp    0   0   0