Index values from a matrix using row, col indices

Almost. Needs to be offered to "[" as a two column matrix:

dat$matval <- mat[ cbind(dat$I, dat$J) ] # should do it.

There is a caveat: Although this also works for dataframes, they are first coerced to matrix-class and if any are non-numeric, the entire matrix becomes the "lowest denominator" class.

Using a matrix to index as DWin suggests is of course much cleaner, but for some strange reason doing it manually using 1-D indices is actually slightly faster:

# Huge sample data
mat <- matrix(sin(1:1e7), ncol=1000)
dat <- data.frame(I=sample.int(nrow(mat), 1e7, rep=T), 
                  J=sample.int(ncol(mat), 1e7, rep=T))

system.time( x <- mat[cbind(dat$I, dat$J)] )     # 0.51 seconds
system.time( mat[dat$I + (dat$J-1L)*nrow(mat)] ) # 0.44 seconds

The dat$I + (dat$J-1L)*nrow(m) part turns the 2-D indices into 1-D ones. The 1L is the way to specify an integer instead of a double value. This avoids some coercions.

...I also tried gsk3's apply-based solution. It's almost 500x slower though:

system.time( apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat ) ) # 212

Here's a one-liner using apply's row-based operations

> dat <- as.data.frame(matrix(rep(seq(4),4),ncol=2))
> colnames(dat) <- c('I','J')
> dat
   I  J
1  1  1
2  2  2
3  3  3
4  4  4
5  1  1
6  2  2
7  3  3
8  4  4
> mat <- matrix(seq(16),ncol=4)
> mat
     [,1] [,2] [,3] [,4]
[1,]    1    5    9   13
[2,]    2    6   10   14
[3,]    3    7   11   15
[4,]    4    8   12   16

> dat$K <- apply( dat, 1, function(x,mat) mat[ x[1], x[2] ], mat=mat )
> dat
  I J  K
1 1 1  1
2 2 2  6
3 3 3 11
4 4 4 16
5 1 1  1
6 2 2  6
7 3 3 11
8 4 4 16