如果是这样，为什么我们需要sapply？
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
最佳答案
除了运行lapply，sapply还会运行simplify2array尝试将输出拟合为数组。为了弄清楚这是否可行，该函数需要检查所有单个输出是否具有相同的长度：这是通过代价高昂的unique(lapply(..., length))来完成的，这占了您所看到的大部分时间差：
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
关于r - 为什么“unlist(lapply)”比“sapply”更快？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/18761541/