如何按行和列对R中的稀疏矩阵进行归一化？

Question

I have created a sparse matrix using the R package "Matrix". The matrix is not square, and its dimensions are 4561 by 68825.

I'm looking to standardize this matrix so that each value x is equal to x / row sum + column sum. I've found a solution on stack which I could alter to solve this problem here . However, in the solution seen in the linked question, the problem uses a square matrix, so Diaganal can be used.In my case, my matrix is not square so I can't make this solution work.

How can I normalize a sparse matrix in R by both rows and columns?

Answer 1

Hope this helps!

m_final <- t(t(m/rowSums(m)) + rowSums(t(m)))
m_final

Output is:

           [,1]     [,2]       [,3]
 [1,] 0.9748283 3.326324 -0.8274075
 [2,] 1.4574957 2.776025 -0.7597753
 [3,] 1.9265464 2.937874 -1.3906749
 [4,] 0.7105211 3.337394 -0.5741696
 [5,] 1.4808831 3.030777 -1.0379153
 [6,] 2.2123599 2.537209 -1.2758243
 [7,] 2.8672471 2.437124 -1.8306263
 [8,] 4.8144351 6.952963 -8.2936531
 [9,] 1.9486587 3.382196 -1.8571098
[10,] 0.8897446 3.329129 -0.7451281

#sample data:
set.seed(1)
m <- replicate(3,rnorm(10))
> m
            [,1]        [,2]        [,3]
 [1,] -0.6264538  1.51178117  0.91897737
 [2,]  0.1836433  0.38984324  0.78213630
 [3,] -0.8356286 -0.62124058  0.07456498
 [4,]  1.5952808 -2.21469989 -1.98935170
 [5,]  0.3295078  1.12493092  0.61982575
 [6,] -0.8204684 -0.04493361 -0.05612874
 [7,]  0.4874291 -0.01619026 -0.15579551
 [8,]  0.7383247  0.94383621 -1.47075238
 [9,]  0.5757814  0.82122120 -0.47815006
[10,] -0.3053884  0.59390132  0.41794156

Edit:
In case you want to have below calculation then you can try

m/(row_sum + col_sum)

m/outer(rowSums(m), colSums(m), FUN = "+")

Answer 2

If you simply want to divide each cell by sum of row sum and col sum, here is a simple way to do so:

test = matrix(1:20, 4, 5)
test
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    5    9   13   17
[2,]    2    6   10   14   18
[3,]    3    7   11   15   19
[4,]    4    8   12   16   20

rs = rowSums(test)
cs = colSums(test)

for(j in 1:ncol(test)){
  for(i in 1:nrow(test)){
    test[i,j] = test[i,j]/(rs[i] + cs[j])
  }
}

test
           [,1]       [,2]      [,3]      [,4]      [,5]
[1,] 0.01818182 0.07042254 0.1034483 0.1262136 0.1428571
[2,] 0.03333333 0.07894737 0.1086957 0.1296296 0.1451613
[3,] 0.04615385 0.08641975 0.1134021 0.1327434 0.1472868
[4,] 0.05714286 0.09302326 0.1176471 0.1355932 0.1492537