Speed up of the calculation of the sum the point-wise difference in R

Question

Suppose I have two datasets. The first one is:

t1<-sample(1:10,10,replace = T)
t2<-sample(1:10,10,replace = T)
t3<-sample(1:10,10,replace = T)
t4<-sample(11:20,10,replace = T)
t5<-sample(11:20,10,replace = T)
xtrain<-rbind(t1,t2,t3,t4,t5)
xtrain
   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
t1    7    3    9   10    4    9    2    1    6     9
t2    5    1    1    6    5    3   10    2    6     3
t3    8    6    9    7    9    2    3    5    1     8
t4   16   18   14   17   19   20   15   15   20    19
t5   13   14   18   13   11   19   13   17   16    14

The second one is:

t6<-sample(1:10,10,replace = T)
t7<-sample(11:20,10,replace = T)
xtest<-rbind(t6,t7)
xtest
   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
t6    1    5    8    2   10    2    3    4    8     5
t7   14   18   15   12   17   20   17   13   16    17

What I did like to do is to calculate the sum of the distance between each row of xtest and each row of xtrain. For example:

sum((7-1)^2+(3-5)^2+(9-8)^2+.....(9-5)^2)
sum((5-1)^2+(1-5)^2+(1-8)^2+.....(4-5)^2)
...
sum((14-13)^2+(18-14)^2+(15-18)^2+.....(17-14)^2)

What I currently have is to use two for-loops (see below), which I don't think can handle large data sets:

sumPD<-function(vector1,vector2){
  sumPD1<-sum((vector1-vector2)^2)
  return(sumPD1)
}
loc<-matrix(NA,nrow=dim(xtrain)[1],ncol=dim(xtest)[1])
for(j in 1:dim(xtest)[1]){    
  for(i in 1:dim(xtrain)[1]){
     loc[i,j]<-sumPD(xtrain[i,],xtest[j,])
   }
 }

I'd like to ask for suggestions on how to modify the code to make it efficient. Thank you in advance! Hope to have a good discussion!

score 3 · Accepted Answer · answered Oct 03 '19 at 23:07

3

The rdist package has functions for quickly calculating these kinds of pairwise distances:

rdist::cdist(xtrain, xtest)^2

Output:

     [,1] [,2]
[1,]   65 1029
[2,]   94 1324
[3,]  165 1103
[4,] 1189  213
[5,] 1271  191

answered Oct 03 '19 at 23:07

Marius

58,213
16
107
105

score 2 · Answer 2 · answered Oct 03 '19 at 22:51

2

One option would be outer

f1 <- Vectorize(function(i, j) sumPD(xtrain[i,], xtest[j,]))
loc2 <- outer(seq_len(nrow(xtrain)), seq_len(nrow(xtest)), f1)
identical(loc, loc2)
#[1] TRUE

answered Oct 03 '19 at 22:51

akrun

874,273
37
540
662

score 1 · Answer 3 · answered Oct 03 '19 at 23:11

1

You could transpose your matrix, use vector difference and a single loop:

ftrain <- t(xtrain)
ftest <- t(xtest)


sapply(1:(dim(ftest)[2]),function(i){
  colSums((ftrain - ftest[,i])^2)
})


   [,1] [,2]
t1  103 1182
t2  125 1262
t3  130 1121
t4 1478  159
t5 1329  142

colSums is quite efficient, but have a look there if you want more speed

answered Oct 03 '19 at 23:11

denis

5,580
1
13
40

1

I think you can avoid the loop with `ntrain = nrow(xtrain); ntest = nrow(xtest); matrix(rowSums( (xtrain[rep(1:ntrain, ntest), ] - xtest[rep(1:ntest, each=ntrain),])^2), nc=ntest)` although your way the intent is clearer – user20650 Oct 03 '19 at 23:34

score 0 · Answer 4 · answered Oct 03 '19 at 23:10

Here are two simple ways.

Using dist - will calculate more distances than needed:

dists <- as.matrix(dist(rbind(xtrain, xtest))^2)
dists <- dists[rownames(xtrain), rownames(xtest)]
dists
     t6   t7
t1  140 1179
t2  134  693
t3  119  974
t4 1028   91
t5 1085   44

Using a simple custom functions that works on X matrix and y vector:

euclid <- function(X,y) colSums((X-y)^2)
dists  <- mapply(euclid, list(t(xtrain)), split(xtest, row(xtest)))
dists
   [,1] [,2]
t1  140 1179
t2  134  693
t3  119  974
t4 1028   91
t5 1085   44

Speed up of the calculation of the sum the point-wise difference in R

4 Answers4