There probably are more elegant ways, but this works:
m <- matrix(1:9, ncol=3, byrow=TRUE)
# m[rev(seq_len(nrow(m))), ] # Initial answer
m[nrow(m):1, ]
[,1] [,2] [,3]
[1,] 7 8 9
[2,] 4 5 6
[3,] 1 2 3
This works because you are indexing the matrix with a reversed sequence of integers as the row index. nrow(m):1
results in 3 2 1
.
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