R：比祝福快吗？

Question

Suppose that I have a matrix for 7 items with 5 options, and a vector for 10 people. What I need to calculate is to multiply each item value for each option with each person value. I was able to do it by using sapply , resulting in 5 different matrices. But, actually as my data is pretty huge, I was wondering if there is any other way faster than sapply . Below is the example.

set.seed(10)
person<-round(runif(10,-2,2),3)
[1]  0.030 -0.773 -0.292  0.772 -1.659 -1.098 -0.902 -0.911  0.463 -0.281
set.seed(10)
item<-round(matrix(runif(35,0,1),nrow=7),3)
      [,1]  [,2]  [,3]  [,4]  [,5]
[1,] 0.507 0.272 0.358 0.615 0.771
[2,] 0.307 0.616 0.429 0.775 0.356
[3,] 0.427 0.430 0.052 0.356 0.536
[4,] 0.693 0.652 0.264 0.406 0.093
[5,] 0.085 0.568 0.399 0.707 0.170
[6,] 0.225 0.114 0.836 0.838 0.900
[7,] 0.275 0.596 0.865 0.240 0.423

## multiplication
m1<-t(sapply(person, FUN=function(x) item[,1]* x))
m2<-t(sapply(person, FUN=function(x) item[,2]* x))
m3<-t(sapply(person, FUN=function(x) item[,3]* x))
m4<-t(sapply(person, FUN=function(x) item[,4]* x))
m5<-t(sapply(person, FUN=function(x) item[,5]* x))

Answer 1

You are doing this

A <- outer(person,item)

As shown

> all.equal(A[,,1],m1)
[1] TRUE
> all.equal(A[,,2],m2)
[1] TRUE
> all.equal(A[,,3],m3)
[1] TRUE
> all.equal(A[,,4],m4)
[1] TRUE
> all.equal(A[,,5],m5)
[1] TRUE