CODEWITHSUNDEEP
r
Guillaume Poussel
Guillaume Poussel

Reputation: 9822

Multiply each row of a matrix by the matrix

I am a new R-user and I have a kind of algorithm problem. I made some research on the web and on Stackoverflow, but can't find my answer.

I have a squared matrix, for example :

  A B C D
A 0 0 0 1
B 0 1 1 0
C 1 0 0 0
D 0 1 1 1

This matrix represents links between keywords (A, B, C and D here). A '1' (or a TRUE) means keywords are in relation. For example, the '1' on the first row means A is linked to D.

I need to find the two most linked keywords on the matrix. I know I need to compute the scalar product between each row and the initial matrix. Then I take the sum of the rows and get the maximum. But, what is the R program which put in a new matrix the product between each row of my matrix, and the matrix itself ?

Thanks!

Upvotes: 0

Views: 664

Answers (2)

Ben Bolker
Ben Bolker

Reputation: 226801

I thought I had a cleverer answer but it turns out to be slower ...

tmp1 <- function(a) {
  n <- nrow(a)
  aa <- apply(array(apply(a,1,"*",a),
                    rep(n,3)),3,rowSums)
  apply(aa,2,which.max)
}

Previous solution:

tmp2 <- function(a) {
  n <- nrow(a)
  r <- numeric(n)
  for(i in seq(n)) {
    b <- rowSums(a[i,]*a)
    r[i] <- which.max(b)
  }
  r
}

Test this on something reasonably large:

n <- 50
a <- matrix(0,nrow=n,ncol=n)
a[sample(length(a),size=n^2/5,replace=TRUE)] <- 1 

all(tmp1(a)==tmp2(a)) ## TRUE

library(rbenchmark)
benchmark(tmp1(a),tmp2(a))
> benchmark(tmp1(a),tmp2(a))
     test replications elapsed relative user.self sys.self 
1 tmp1(a)          100   4.030 9.264368     2.052     1.96  
2 tmp2(a)          100   0.435 1.000000     0.232     0.20

You will presumably do even better if you can do it in terms of sparse matrices.

Upvotes: 3

Benjamin
Benjamin

Reputation: 11860

Like this?

a=matrix(c(0,0,0,1,0,1,1,0,1,0,0,0,0,1,1,1), ncol=4, byrow=T)
for(i in 1:4){
 b = rowSums(a[i,]*a)
 print(which(b==max(b)))
}

Upvotes: 1

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