Reputation: 740
I have two large matrices P and Q around (10k x 50k dim in both, but to test this yourself a random 10x10 matrix for P and Q is sufficient). I have a list of indices, e.g.
i j
1 4
1 625
1 9207
2 827
... ...
etc. This means that I need to find the dot product of column 1 in P and column 4 in Q, then column 1 in P and column 625 in Q and so on. I could easily solve this with a for loop but I know they are not very efficient in R. Anyone got any ideas?
edit: asked for a reproducible example
P <- matrix(c(1,0,1,0,0,1,0,1,0), nrow = 3, ncol = 3)
Q <- matrix(c(0,0,1,0,1,0,1,0,1), nrow = 3, ncol = 3)
i <- c(1,1,2)
j <- c(2,1,3)
gives output (if in dot product form)
1: 0
2: 1
3: 1
Upvotes: 1
Views: 37
Reputation: 38520
Using matrix multiplication, you can do
diag(t(P[, i]) %*% Q[, j])
[1] 0 1 1
Here is second a solution with apply
.
apply(cbind(i, j), 1, function(x) t(P[, x[1]]) %*% Q[, x[2]])
[1] 0 1 1
To verify these agree in a second example:
set.seed(1234)
A <- matrix(sample(0:10, 100, replace=TRUE), 10, 10)
B <- matrix(sample(0:10, 100, replace=TRUE), 10, 10)
inds <- matrix(sample(10, 10, replace=TRUE), 5)
matrix multiplication
diag(t(A[, inds[,1]]) %*% B[, inds[,2]])
[1] 215 260 306 237 317
and apply
apply(inds, 1, function(x) t(A[, x[1]]) %*% B[, x[2]])
[1] 215 260 306 237 317
Upvotes: 2
Reputation: 3650
P <- matrix(1:50, nrow = 5,ncol = 10)
Q <- matrix(1:50, nrow = 5, ncol = 10)
i <- c(1,2,4,7)
j <- c(5,3,7,2)
P
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 1 6 11 16 21 26 31 36 41 46
# [2,] 2 7 12 17 22 27 32 37 42 47
# [3,] 3 8 13 18 23 28 33 38 43 48
# [4,] 4 9 14 19 24 29 34 39 44 49
# [5,] 5 10 15 20 25 30 35 40 45 50
Q
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 1 6 11 16 21 26 31 36 41 46
# [2,] 2 7 12 17 22 27 32 37 42 47
# [3,] 3 8 13 18 23 28 33 38 43 48
# [4,] 4 9 14 19 24 29 34 39 44 49
# [5,] 5 10 15 20 25 30 35 40 45 50
P[,i] * Q[, j]
# [,1] [,2] [,3] [,4]
# [1,] 21 66 496 186
# [2,] 44 84 544 224
# [3,] 69 104 594 264
# [4,] 96 126 646 306
# [5,] 125 150 700 350
Upvotes: 3