All is in the title. I would expect that order uses sort to find the order of the values in a vector. Thus sort should be quicker than order to sort a vector, but this is not the case:
library(microbenchmark)
ss=sample(100,10000,replace=T)
microbenchmark(sort(ss))
microbenchmark(ss[order(ss)])
result:
> microbenchmark(sort(ss))
Unit: microseconds
expr min lq mean median uq max neval
sort(ss) 141.535 144.6415 173.6581 146.358 150.2295 2531.762 100
> microbenchmark(ss[order(ss)])
Unit: microseconds
expr min lq mean median uq max neval
ss[order(ss)] 109.198 110.9865 115.6275 111.901 115.3655 197.204 100
Example with a larger vector:
ss=sample(100,1e8,replace=T)
microbenchmark(sort(ss), ss[order(ss)], times = 5)
# Unit: seconds
# expr min lq mean median uq max neval
# sort(ss) 5.427966 5.431971 5.892629 6.049515 6.207060 6.346633 5
# ss[order(ss)] 3.381253 3.500134 3.562048 3.518079 3.625778 3.784997 5
The treatment of NA values under the default arguments is different. In sort, the entire vector must be scanned for NA values, which are then removed; in order, they are simply put last. When the argument sort.last = TRUE is used in both, the performance is basically identical.
ss=sample(100,1e8,replace=T)
bench::mark(sort(ss), ss[order(ss)], sort(ss, na.last = TRUE))
# A tibble: 3 x 14
expression min mean median max `itr/sec` mem_alloc n_gc n_itr total_time result
<chr> <bch:> <bch:> <bch:> <bch:> <dbl> <bch:byt> <dbl> <int> <bch:tm> <list>
1 sort(ss) 2.610s 2.610s 2.610s 2.610s 0.383 762.940MB 0 1 2.610s <int ~
2 ss[order(~ 1.597s 1.597s 1.597s 1.597s 0.626 762.940MB 0 1 1.597s <int ~
3 sort(ss, ~ 1.592s 1.592s 1.592s 1.592s 0.628 762.940MB 0 1 1.592s <int ~
# ... with 3 more variables: memory <list>, time <list>, gc <list>