I have a matrix (named rating) with dim n x 140000 and another matrix (named trust) with dim nxn where n varying when I change the group and n might have value from 1-15000. I need to multiply each column of rating by trust. for example:

trust=                         rating=
a1 a2 a3 a4 a5                 1 2 3 4 5 6 7 8
b1 b2 b3 b4 b5                 2 5 7 8 9 2 1 6
c1 c2 c3 c4 c5                 3 5 3 6 8 1 2 5
d1 d2 d3 d4 d5                 4 7 8 2 4 5 6 7
e1 e2 e3 e4 e5                 5 2 5 7 8 9 1 4

answer1=                       answer2=
a1.1 a2.2 a3.3 a4.4 a5.5       a1.2 a2.5 a3.5 a4.7 a5.2
b1.1 b2.2 b3.3 b4.4 b5.5       b1.2 b2.5 b3.5 b4.7 b5.2
c1.1 c2.2 c3.3 c4.4 c5.5       c1.2 c2.5 c3.5 c4.7 c5.2
d1.1 d2.2 d3.3 d4.4 d5.5       d1.2 d2.5 d3.5 d4.7 d5.2
e1.1 e2.2 e3.3 e4.4 e5.5       e1.2 e2.5 e3.5 e4.7 e5.2

and answer3 must multiply by 3rd column and so on. Then add each rows of answer1, answer2, ... and store into a vector. Then store each vector into a list for future use.

 for (k in 1:ncol(rating)) {
   clmy <- as.matrix(rating[, k])
   answer <- sweep(trust, MARGIN = 2, clmy, '*')
   sumtrustbyrating <- rowSums(answer)
   LstsumRbyT[[k]] <- sumtrustbyrating
   sumtrustbyrating = NULL
 }

It is working perfectly if I change the ncol(rating) to a small value (about 100). But for the actual data, I have 140000 columns. It takes time and I couldn't get the final execution result. Please help me to enhance the performance of my code for a huge data set.

How about a matrix product? Or is that too slow?

rating <- matrix(c(1, 2, 3, 4, 5,2, 5, 5, 6, 3, 3, 4, 1, 2, 1), ncol=3)
trust <- matrix(rep(1:5, rep(5, 1)), 5, byrow=TRUE)

Running your code above yields

LstsumRbyT
[[1]]
[1] 55 55 55 55 55

[[2]]
[1] 66 66 66 66 66

[[3]]
[1] 27 27 27 27 27

which is the same as

 trust %*% rating
     [,1] [,2] [,3]
[1,]   55   66   27
[2,]   55   66   27
[3,]   55   66   27
[4,]   55   66   27
[5,]   55   66   27

If this isn't enough then this could be improved a bit in RCppArmadillo I guess.

To add to the benchmarking discussion. If your for loop above is renamed f() then I get

microbenchmark(trust %*% rating, f())
Unit: microseconds
             expr     min       lq      mean   median       uq      max neval cld
 trust %*% rating   1.418   1.7010   2.97663   2.7215   3.5965   14.452   100  a
              f() 593.890 700.9775 764.00515 766.5535 792.6375 1511.104   100   b

which is quite a substantial speedup with the normal matrix product.