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
R = corrcoef( A ) returns the matrix of correlation coefficients for A , where the columns of A represent random variables and the rows represent observations. R = corrcoef( A , B ) returns coefficients between two random variables A and B .
Initialize two variables, col1 and col2, and assign them the columns that you want to find the correlation of. Find the correlation between col1 and col2 by using df[col1]. corr(df[col2]) and save the correlation value in a variable, corr. Print the correlation value, corr.
The correlation coefficient is determined by dividing the covariance by the product of the two variables' standard deviations. Standard deviation is a measure of the dispersion of data from its average. Covariance is a measure of how two variables change together.
By using corr() function we can get the correlation between two columns in the dataframe.
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
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
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.
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