Reputation: 8042
I have vector:
v1 = c(1,2,3)
From this vector I want to create matrix where element on i,j
position will be sum
of vector members on i,j
positions:
[,1] [,2] [,3]
[1,] 2 3 4
[2,] 3 4 5
[3,] 4 5 6
Questions:
i,j
and j,i
is the same, so there is no reason to compute it 2x
for better performance. How to achieve this?i == j
and simply returns NA
instead? I'm not asking for diag(m) <- NA
command, I'm asking how to prevent computing those elements.PS: This is reduced version of my problem
Upvotes: 2
Views: 1186
Reputation: 2718
There is an approach that is much faster than a straightforward calculation with 2 nested loops. It's not optimized in terms that you described in the question 1, but it's pretty fast because it's vectorized. Maybe, it will be enough for your purpose. Vectorized (or even matrix) approach itself:
f1 <- function(x){
n <- length(x)
m <- matrix(rep(x,n),n)
m + t(m)
}
> f1(1:3)
[,1] [,2] [,3]
[1,] 2 3 4
[2,] 3 4 5
[3,] 4 5 6
We can also create a function for straightforward approach to perform benchmark. This function does even less than needed: it calculates only upper triangle, but we will see that it's much slower.
f2 <- function(x){
n <- length(x)
m <- matrix(rep(NA,n^2),n)
for(i in 1:(n-1)){
for(j in (i+1):n) m[i,j] <- x[[i]] + x[[j]]
}
m
}
> f2(1:3)
[,1] [,2] [,3]
[1,] NA 3 4
[2,] NA NA 5
[3,] NA NA NA
Benchmark:
library(microbenchmark)
> microbenchmark(f1(1:100), f2(1:100))
Unit: microseconds
expr min lq mean median uq max neval
f1(1:100) 124.775 138.6175 181.6401 187.731 196.454 294.301 100
f2(1:100) 10227.337 10465.1285 11000.1493 10616.830 10907.148 15826.259 100
Upvotes: 1