I was reading that R uses column-major storage in matrices, which means that elements in nearby columns are stored in contiguous blocks or something of that sort. This made me wonder: Is it faster to fill a matrix by row (using byrow=TRUE in the base R function matrix()) or is it faster to fill the matrix by column first (using the default byrow=FALSE) and then transpose it using t()?

I tried benchmarking it.

Filling a Matrix by Row

> microbenchmark(matrix(1, n, n, byrow=TRUE))
Unit: seconds
                          expr      min       lq     mean   median       uq      max neval
 matrix(1, n, n, byrow = TRUE) 1.047379 1.071353 1.105468 1.081795 1.112995 1.628675   100

Filling a Matrix by Column and then Transpose it

> microbenchmark(t(matrix(1, n, n)))
Unit: seconds
               expr     min       lq     mean  median       uq      max neval
 t(matrix(1, n, n)) 1.43931 1.536333 1.692572 1.61793 1.726244 3.070821   100

Conclusion

It seems that it's faster to fill the matrix by row! Am I missing something? I would have thought that R would just do some relabelling with t() but it's actually slower than filling in the matrix by row!

Is there an explanation for this? I'm quite baffled.

Observation

After ThomasIsCoding's answer and after benchmarking myself a few times it looks like it depends on the number of rows and number of columns.

• Number of Rows < Number of Columns: t() is faster.
• Number of Rows = Number of Columns: byrow=TRUE is faster.
• Number of Rows > Number of Columns: byrow=TRUE is faster.