R 性能：具有多个调用的全局和本地声明

为什么在R中无论在何处声明函数（两者的性能几乎相同）：

库（微基准）
f1定义函数的性能成本可以忽略不计。仅当调用函数时，才会计算函数体

microbenchmark(fun <- function() 1:10000, 
               fun <- function() 1:100000, times = 1000)

#Unit: nanoseconds
#                      expr min  lq    mean median  uq   max neval cld
# fun <- function() 1:10000 198 506 568.462  511.5 548 54620  1000   a
# fun <- function() 1:1e+05 199 504 570.826  511.0 551 18620  1000   a

microbenchmark（乐趣
library(microbenchmark)

f1 <- function() {
    lapply(1:100000, function(x) {
        var <- 1:10000
        var
    })
}

f2 <- function() {
    var <- 1:10000
    lapply(1:100000, function(x) {
        var
    })
}

microbenchmark(f1(), f2(), times = 10)

# Unit: milliseconds
# expr       min        lq      mean    median        uq      max neval
# f1() 516.07492 567.71822 611.44760 630.57550 642.47586 701.3975    10
# f2()  49.30975  50.12807  72.44492  52.53448  58.85256 159.2140    10

microbenchmark(fun <- function() 1:10000, 
               fun <- function() 1:100000, times = 1000)

#Unit: nanoseconds
#                      expr min  lq    mean median  uq   max neval cld
# fun <- function() 1:10000 198 506 568.462  511.5 548 54620  1000   a
# fun <- function() 1:1e+05 199 504 570.826  511.0 551 18620  1000   a

microbenchmark(var <- 1:10000, times = 100)
#Unit: microseconds
#           expr   min     lq    mean median    uq    max neval
# var <- 1:10000 4.183 4.3305 4.92081 4.4135 4.538 15.283   100