I'm trying to perform the dot product on all possible combinations of vectors. I am able to find all the possible combinations. I just can't quite figure out how the FUN argument in combn() works. Below is my code, thanks for any help!
def=c("Normal.def","Fire.def","Water.def","Electric.def","Grass.def","Ice.def",
"Fighting.def","Poison.def","Ground.def","Flying.def","Pyschic.def","Bug.def",
"Rock.def","Ghost.def","Dragon.def","Null.def")
combn(def,2,FUN=def%*%def,simplify=TRUE)
Using @BrodieG's sample data, you can just use the crossprod function:
set.seed(1)
vec1 <- sample(1:10)
vec2 <- sample(1:10)
vec3 <- sample(1:10)
crossprod(cbind(vec1, vec2, vec3))
# vec1 vec2 vec3
# vec1 385 298 284
# vec2 298 385 296
# vec3 284 296 385
Some benchmarks, out of curiosity:
The functions to run:
fun1 <- function() {
A <- crossprod(do.call(cbind, lst))
A[upper.tri(A)]
}
fun2 <- function() {
A <- do.call(rbind, lst) %*% do.call(cbind, lst)
A[upper.tri(A)]
}
fun3 <- function() {
combn(
seq_along(lst), 2,
FUN=function(idx) c(lst[[idx[[1]]]] %*% lst[[idx[[2]]]])
)
}
Benchmarking on "small number of large vectors".
library(microbenchmark)
set.seed(1)
n <- 5
lst <- setNames(replicate(n, sample(1:100000), simplify = FALSE),
paste0("V", sequence(n)))
microbenchmark(fun1(), fun2(), fun3())
# Unit: milliseconds
# expr min lq median uq max neval
# fun1() 6.909651 6.992031 8.432346 8.520301 74.12263 100
# fun2() 17.290101 18.811134 19.144601 21.292544 88.10602 100
# fun3() 22.841209 24.283113 24.427876 25.820158 91.14007 100
Not being patient enough to benchmark on medium numbers of medium vectors:
set.seed(1)
n <- 1000
lst <- setNames(replicate(n, sample(1:1000), simplify = FALSE),
paste0("V", sequence(n)))
system.time(fun1())
# user system elapsed
# 0.245 0.004 0.251
system.time(fun2())
# user system elapsed
# 0.407 0.016 0.425
system.time(fun3())
# user system elapsed
# 14.216 0.004 14.339
Why don't you just matrix multiply the whole thing. For example:
set.seed(1)
vec1 <- sample(1:10)
vec2 <- sample(1:10)
vec3 <- sample(1:10)
rbind(vec1, vec2, vec3) %*% cbind(vec1, vec2, vec3)
produces:
vec1 vec2 vec3
vec1 385 298 284
vec2 298 385 296
vec3 284 296 385
Where each cell of a matrix is the dot product of the two vectors in the col and row labels. Alternatively, if you really want to do it with combn:
vec.lst <- list(vec1, vec2, vec3)
combn(
seq_along(vec.lst), 2,
FUN=function(idx) c(vec.lst[[idx[[1]]]] %*% vec.lst[[idx[[2]]]])
)
Which produces:
[1] 298 284 296
Notice how those numbers correspond to the upper triangle of the matrix. For small data sets the matrix multiply approach is much faster. For large ones, particularly ones were the vectors are very large but there aren't that many of them, the combn approach might be faster since it doesn't run as many computations (only the upper triangle basically).
