19

もしそうなら、なぜ必要なのsapplyですか？

x <- list(a=1, b=1)
y <- list(a=1)
JSON <- rep(list(x,y),10000)
microbenchmark(sapply(JSON, function(x) x$a),
               unlist(lapply(JSON, function(x) x$a)),
               sapply(JSON, "[[", "a"),
               unlist(lapply(JSON, "[[", "a"))
               )

Unit: milliseconds
                                  expr      min       lq   median       uq      max neval
         sapply(JSON, function(x) x$a) 25.22623 28.55634 29.71373 31.76492 88.26514   100
 unlist(lapply(JSON, function(x) x$a)) 17.85278 20.25889 21.61575 22.67390 78.54801   100
               sapply(JSON, "[[", "a") 18.85529 20.06115 21.53790 23.42480 38.56610   100
       unlist(lapply(JSON, "[[", "a")) 11.33859 11.69198 12.25329 13.37008 27.81361   100
4

2 に答える 2

19

の実行に加えてlapply、をsapply実行simplify2arrayして、出力を配列に適合させようとします。それが可能かどうかを判断するには、関数は個々の出力がすべて同じ長さであるかどうかを確認する必要がunique(lapply(..., length))あります。

b <- lapply(JSON, "[[", "a")

microbenchmark(lapply(JSON, "[[", "a"),
               unlist(b),
               unique(lapply(b, length)),
               sapply(JSON, "[[", "a"),
               sapply(JSON, "[[", "a", simplify = FALSE),
               unlist(lapply(JSON, "[[", "a"))
)

# Unit: microseconds
#                                       expr       min        lq   median        uq       max neval
#                    lapply(JSON, "[[", "a") 14809.151 15384.358 15774.26 16905.226 24944.863   100
#                                  unlist(b)   920.047  1043.719  1158.62  1223.091  8056.231   100
#                  unique(lapply(b, length)) 10778.065 11060.452 11456.11 12581.414 19717.740   100
#                    sapply(JSON, "[[", "a") 24827.206 25685.535 26656.88 30519.556 93195.751   100
#  sapply(JSON, "[[", "a", simplify = FALSE) 14283.541 14922.780 15526.42 16654.058 26865.022   100
#            unlist(lapply(JSON, "[[", "a")) 15334.026 16133.146 16607.12 18476.182 30080.544   100
于 2013-09-12T11:32:51.890 に答える