用rowmean替换所有行

Question

I have a matrix with 4 columns and I have sorted the matrix and calculated the rowMeans for each row. Now I want to replace all the values in the original matrix with the respective rowMeans . c5sub is the matrix with 4 columns

sortmat<-apply(c5sub, 2, sort) # sorted matrix by column
[1,]                     -7                     -6                    -17                     -6
[2,]                     -7                     -6                     -9                     -6
[3,]                     -6                     -5                     -8                     -6
[4,]                     -6                     -5                     -8                     -6
[5,]                     -6                     -5                     -7                     -6
[6,]                     -6                     -5                     -7                     -5

rwmeans<-apply(sortmat, 1, mean)# calculated rowmeans 
-9.00 -7.00 -6.25 -6.25 -6.00 -5.75

(a <- sweep(a,1,rwmeans,function(x,y) ifelse(x!=0,y,0)))
[1,]                  -9.00                  -9.00                  -9.00                  -9.00
[2,]                  -7.00                  -7.00                  -7.00                  -7.00
[3,]                  -6.25                  -6.25                  -6.25                  -6.25
[4,]                  -6.25                  -6.25                  -6.25                  -6.25
[5,]                  -6.00                  -6.00                  -6.00                  -6.00
[6,]                  -5.75                  -5.75                  -5.75                  -5.75

I have used above one to replace the original values with rowMeans but it does not replace the zeros because I have the ifelse . How do I modify it so that it replaces all the values?

Answer 1

To keep the original attributes of sortmat such as row and column names:

sortmat[] <- rowMeans(sortmat)

This works because 1) matrices in R are stored in column-major order, meaning all values in column 1, followed by all values in column 2, and so on; 2) vectors are recycled, so the vector of rowmeans gets replicated to the correct length for assignment; and 3) assignment with empty brackets on the LHS [] means "replace all elements in an object, but keep the object itself".

Answer 2

I think the problem here is in trying to literally replace all of the values, instead of trying to simply create the final matrix the OP desires.

Once the row means have been calculated, and with the size of the final matrix known, we can create the matrix a like so:

a <- matrix(rwmeans, nrow=length(rwmeans), ncol=Nc)

Where 'rwmeans' contains an array with the column means, and 'Nc' is the number of columns in the final matrix. The values in 'rwmeans' will be repeated for each column of the matrix.