Multiplying Combinations of a list of lists in R

Question

Given a list of two lists, I am trying to obtain, without using for loops, a list of all element-wise products of the first list with the second. For example:

Answer 1

Have no idea if this is fast or memory intensive just that it works, Joris Meys's answer is more eloquent:

x <- expand.grid(1:length(a), 1:length(b))
x <- x[order(x$Var1), ]    #gives the order you asked for
FUN <- function(i)  diag(outer(a[[x[i, 1]]], b[[x[i, 2]]], "*"))
sapply(1:nrow(x), FUN)     #I like this out put
lapply(1:nrow(x), FUN)     #This one matches what you asked for

EDIT: Now that Brian introduced benchmarking (which I love (LINK)) I have to respond. I actually have a faster answer using what I call expand.grid2 that's a lighter weight version of the original that I stole from HERE. I was going to throw it up before but when I saw how fast Joris's is I figured why bother, both short and sweet but also fast. But now that Diggs has dug I figured I'd throw up here the expand.grid2 for educational purposes.

expand.grid2 <-function(seq1,seq2) {
    cbind(Var1 = rep.int(seq1, length(seq2)), 
    Var2 = rep.int(seq2, rep.int(length(seq1),length(seq2))))
}

x <- expand.grid2(1:length(a), 1:length(b))
x <- x[order(x[,'Var1']), ]    #gives the order you asked for
FUN <- function(i)  diag(outer(a[[x[i, 1]]], b[[x[i, 2]]], "*"))
lapply(1:nrow(x), FUN)

Here's the results (same labeling as Bryan's except TylerEG2 is using the expand.grid2):

Unit: microseconds
            expr      min       lq   median       uq      max
1   DiggsL(a, b) 5102.296 5307.816 5471.578 5887.516 70965.58
2   DiggsM(a, b)  384.912  428.769  443.466  461.428 36213.89
3    Joris(a, b)   91.446  105.210  123.172  130.171 16833.47
4 TylerEG2(a, b)  392.377  425.503  438.100  453.263 32208.94
5   TylerL(a, b) 1752.398 1808.852 1847.577 1975.880 49214.10
6   TylerM(a, b) 1827.515 1888.867 1925.959 2090.421 75766.01
7 Wojciech(a, b) 1719.740 1771.760 1807.686 1924.325 81666.12

And if I take the ordering step out I can squeak out even more but it still isn't close to Joris's answer.