I have a matrix like
A= [ 1 2 4 2 3 1 3 1 2 ] and I would like to calculate its cumulative sum by row and by column, that is, I want the result to be
B = [ 1 3 7 3 8 13 6 12 19 ] Any ideas of how to make this in R in a fast way? (Possibly using the function cumsum) (I have huge matrices)
Thanks!
A one-liner:
t(apply(apply(A, 2, cumsum)), 1, cumsum)) The underlying observation is that you can first compute the cumulative sums over the columns and then the cumulative sum of this matrix over the rows.
Note: When doing the rows, you have to transpose the resulting matrix.
Your example:
> apply(A, 2, cumsum) [,1] [,2] [,3] [1,] 1 2 4 [2,] 3 5 5 [3,] 6 6 7 > t(apply(apply(A, 2, cumsum), 1, cumsum)) [,1] [,2] [,3] [1,] 1 3 7 [2,] 3 8 13 [3,] 6 12 19 About performance: I have now idea how good this approach scales to big matrices. Complexity-wise, this should be close to optimal. Usually, apply is not that bad in performance as well.
Now I was getting curious - what approach is the better one? A short benchmark:
> A <- matrix(runif(1000*1000, 1, 500), 1000) > > system.time( + B <- t(apply(apply(A, 2, cumsum), 1, cumsum)) + ) User System elapsed 0.082 0.011 0.093 > > system.time( + C <- lower.tri(diag(nrow(A)), diag = TRUE) %*% A %*% upper.tri(diag(ncol(A)), diag = TRUE) + ) User System elapsed 1.519 0.016 1.530 Thus: Apply outperforms matrix multiplication by a factor of 15. (Just for comparision: MATLAB needed 0.10719 seconds.) The results do not really surprise, as the apply-version can be done in O(n^2), while the matrix multiplication will need approx. O(n^2.7) computations. Thus, all optimizations that matrix multiplication offers should be lost if n is big enough.
Here is a more efficient implementation using the matrixStats package and a larger example matrix:
library(matrixStats) A <- matrix(runif(10000*10000, 1, 500), 10000) # Thilo's answer system.time(B <- t(apply(apply(A, 2, cumsum), 1, cumsum))) user system elapsed 3.684 0.504 4.201 # using matrixStats system.time(C <- colCumsums(rowCumsums(A))) user system elapsed 0.164 0.068 0.233 all.equal(B, C) [1] TRUE