我如何计算两个矩阵的对应列之间的相关性，而不是像 output 那样获得其他相关性

Question

I have these data

> a
     a    b    c
1    1   -1    4
2    2   -2    6
3    3   -3    9
4    4   -4   12
5    5   -5    6

> b
     d    e    f
1    6   -5    7
2    7   -4    4
3    8   -3    3
4    9   -2    3
5   10   -1    9

> cor(a,b)
           d            e             f
a  1.0000000    1.0000000     0.1767767
b -1.0000000    -1.000000    -0.1767767
c  0.5050763    0.5050763    -0.6964286

The result I want is just:

cor(a,d) = 1
cor(b,e) = -1
cor(c,f) = -0.6964286

Answer 1

The first answer above calculates all pairwise correlations, which is fine unless the matrices are large, and the second one doesn't work. As far as I can tell, efficient computation must be done directly, such as this code borrowed from borrowed from the arrayMagic Bioconductor package, works efficiently for large matrices:

> colCors = function(x, y) { 
+   sqr = function(x) x*x
+   if(!is.matrix(x)||!is.matrix(y)||any(dim(x)!=dim(y)))
+     stop("Please supply two matrices of equal size.")
+   x   = sweep(x, 2, colMeans(x))
+   y   = sweep(y, 2, colMeans(y))
+   cor = colSums(x*y) /  sqrt(colSums(sqr(x))*colSums(sqr(y)))
+   return(cor)
+ }

> set.seed(1)
> a=matrix(rnorm(15),nrow=5)
> b=matrix(rnorm(15),nrow=5)
> diag(cor(a,b))
[1]  0.2491625 -0.5313192  0.5594564
> mapply(cor,a,b)
 [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> colCors(a,b)
[1]  0.2491625 -0.5313192  0.5594564

Answer 2

I would probably personally just use diag :

> diag(cor(a,b))
[1]  1.0000000 -1.0000000 -0.6964286

But you could also use mapply :

> mapply(cor,a,b)
         a          b          c 
 1.0000000 -1.0000000 -0.6964286

Answer 3

mapply works with data frames but not matrices. That is because in data frames each column is an element, while in matrices each entry is an element.

In the answer above mapply(cor,as.data.frame(a),as.data.frame(b)) works just fine.

set.seed(1)
a=matrix(rnorm(15),nrow=5)
b=matrix(rnorm(15),nrow=5)
diag(cor(a,b))
[1]  0.2491625 -0.5313192  0.5594564
mapply(cor,as.data.frame(a),as.data.frame(b))
    V1         V2         V3 
 0.2491625 -0.5313192  0.5594564 

This is much more efficient for large matrices.