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

Question

I have a site by species matrix. The dimensions are 375 x 360. Each value represents the frequency of a species in samples of that site.
I am trying to convert this matrix from frequencies to relative abundances at each site.
I've tried a few ways to achieve this and the only one that has worked is using a for loop. However, this takes an incredibly long time or simply never finishes.
Is there a function or a vectorised method of achieving this? I've included my for-loop as an example of what I am trying to do.
relative_abundance <- matrix(0, nrow= nrow(data_wide),
ncol=ncol(data), dimnames = dimnames(data))
i=0
j=0
for(i in 1:nrow(relative_abundance)){
for(j in 1:ncol(relative_abundance)){
species_freq <- data[i,j]
row_sum <- sum(data[i,])
relative_abundance[i,j] <- species_freq/row_sum
}
}

Answer 1

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.

Answer 2

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))

Answer 3

Using some simple linear algebra we can produce faster results. Simply multiply on the left by a diagonal matrix with the scaling factors you need, like this:
library(Matrix)
set.seed(0)
relative_abundance <- matrix(sample(1:10, 360*375, TRUE), nrow= 375)
Diagonal_Matrix <- diag(1/rowSums(relative_abundance))
And then we multiply from the left:
row_normalized_matrix <- Diagonal_Matrix %*% relative_abundance
If you want to normalize columnwise simply make:
Diagonal_Matrix <- diag(1/colSums(relative_abundance))
and multiply from the right.

Answer 4

You can do something like this
relative_abundance <- matrix(sample(1:10, 360*375, TRUE), nrow= 375)
datnorm <- relative_abundance/rowSums(relative_abundance)
this will be faster if relative_abundance is a matrix rather than a data.frame