将矢量与矢量列表匹配

Question

I have a list of vectors lis which I need to match to another vector vec

lis <- list(c(2,0,0),c(1,1,0), c(1,0,1), c(0,2,0), c(0,1,1), c(0,0,2))
vec <- c(1,1,0)

So either I would get a logical output

 [1] FALSE  TRUE  FALSE  FALSE  FALSE  FALSE

Or just the position within lis of the match

I've been trying things along these lines:

unlist(lis) %in% vec

but the problem is the position of the number is important ie distinguish between c(1,1,0) and c(1,0,1) which I haven't been able to do. I would like to avoid for loops as this needs to be quite efficient (fast).

Answer 1

The answers by @agstudy and @Julius involve a loop over the (long) lis object; here's an alternative assuming that all elements of lis are the same length as vec to allow vector comparison of the unlisted reference

shortloop <- function(x, lst)
    colSums(matrix(unlist(lst) == x, length(x))) == length(x)

which is fast when lis is long compared to vec .

longloop <- function(x, lst)
    sapply(lst, identical, x)

l1 = rep(lis, 1000)
microbenchmark(shortloop(vec, l1), longloop(vec, l1))
## Unit: microseconds
##                expr       min         lq    median         uq      max neval
##  shortloop(vec, l1)   793.009   808.2175   896.299   905.8795  1058.79   100
##   longloop(vec, l1) 18732.338 21649.0770 21797.646 22107.7805 24788.86   100

Interestingly, using for is not that bad from a performance perspective compared to the implicit loop in lapply (though more complicated and error-prone)

longfor <- function(x, lst) {
    res <- logical(length(lst))
    for (i in seq_along(lst))
        res[[i]] = identical(x, lst[[i]])
    res
}
library(compiler)
longforc = cmpfun(longfor)
microbenchmark(longloop(vec, l1), longfor(vec, l1), longforc(vec, l1))
## Unit: milliseconds
##               expr      min       lq   median       uq      max neval
##  longloop(vec, l1) 18.92824 21.20457 21.71295 21.80009 23.23286   100
##   longfor(vec, l1) 23.64756 26.73481 27.43815 27.61699 28.33454   100
##  longforc(vec, l1) 17.40998 18.28686 20.47844 20.75303 21.49532   100

Answer 2

sapply(lis, identical, vec)
# [1] FALSE  TRUE FALSE FALSE FALSE FALSE

Benchmark:

l1 <- list(1:1000)[rep(1, 10000)]
v1 <- sample(1000)
AG <- function() sapply(l1,function(x)all(x==v1))
J <- function() sapply(l1, identical, v1)
microbenchmark(AG(), J())
# Unit: milliseconds
#  expr      min       lq    median        uq      max neval
#  AG() 76.42732 84.26958 103.99233 111.62671 148.2824   100
#   J() 32.14965 37.54198  47.34538  50.93195 104.4036   100

Answer 3

sapply(lis,function(x)all(x==vec))
[1] FALSE  TRUE FALSE FALSE FALSE FALSE