Divide each each cell of large matrix by sum of its row

You could do this using apply, but scale in this case makes things even simplier. Assuming you want to divide columns by their sums:

set.seed(0)
relative_abundance <- matrix(sample(1:10, 360*375, TRUE), nrow= 375)

freqs <- scale(relative_abundance, center = FALSE, 
               scale = colSums(relative_abundance))

The matrix is too big to output here, but here's how it shoud look like:

> head(freqs[, 1:5])
            [,1]         [,2]        [,3]        [,4]         [,5]
[1,] 0.004409603 0.0014231499 0.003439803 0.004052685 0.0024026910
[2,] 0.001469868 0.0023719165 0.002457002 0.005065856 0.0004805382
[3,] 0.001959824 0.0018975332 0.004914005 0.001519757 0.0043248438
[4,] 0.002939735 0.0042694497 0.002948403 0.002532928 0.0009610764
[5,] 0.004899559 0.0009487666 0.000982801 0.001519757 0.0028832292
[6,] 0.001469868 0.0023719165 0.002457002 0.002026342 0.0009610764

And a sanity check:

> head(colSums(freqs))
[1] 1 1 1 1 1 1

Using apply:

freqs2 <- apply(relative_abundance, 2, function(i) i/sum(i))

This has the advatange of being easly changed to run by rows, but the results will be joined as columns anyway, so you'd have to transpose it.

Firstly, you could just do

relative_abundance[i,j] <- data[i,j]/sum(data[i,])

so you dont create the variables...

But to vectorise it, I suggest: compute the row sums with rowsum function(fast) and then you can just use apply by columns and each of that divide by the rowsums:

 relative_freq<-apply(data,2,function(x) data[,x]/rowsum(data))