dapengde
搜了一下还是有好多答案的
https://stackoverflow.com/questions/8788050/how-to-define-a-piecewise-function-in-r
ifelse写法是简单，但有一次在so上有小伙伴提示ifelse比较慢之后都在避免用它.
刚benchmark了一下好像速度并没有可见的差异…
fun1 = function(){
f = function(x) {
sapply(x, function(x) {
y = numeric()
if (x >= 0 & x < 1)
y = 2
else if (x >= 1 & x < 2)
y = 0.1
else if (x >= 2 & x < 5)
y = -1
return(y)
})
}
x = seq(0, 500, 0.1)
y = f(x)
}
fun2 = function(){
x = seq(0, 500, 0.1)
y = ifelse(x > 0 & x < 1, 2, ifelse(x > 1 & x < 2, 1, ifelse(x > 2 & x < 5, -1, NA)))
}
fun3 = function(){
x = seq(0, 500, 0.1)
y <- (x > 0 & x < 1) * 2 +
(x > 1 & x < 2) * 1 +
(x > 2 & x < 5) * -1
}
microbenchmark::microbenchmark(list = c(fun1,fun2,fun3), times = 10000)
#> Unit: nanoseconds
#> expr
#> function () { f = function(x) { sapply(x, function(x) { y = numeric() if (x >= 0 & x < 1) y = 2 else if (x >= 1 & x < 2) y = 0.1 else if (x >= 2 & x < 5) y = -1 return(y) }) } x = seq(0, 500, 0.1) y = f(x) }
#> function () { x = seq(0, 500, 0.1) y = ifelse(x > 0 & x < 1, 2, ifelse(x > 1 & x < 2, 1, ifelse(x > 2 & x < 5, -1, NA))) }
#> function () { x = seq(0, 500, 0.1) y <- (x > 0 & x < 1) * 2 + (x > 1 & x < 2) * 1 + (x > 2 & x < 5) * -1 }
#> min lq mean median uq max neval cld
#> 21 28 89.1662 31 67 67965 10000 b
#> 21 28 75.9365 31 68 4953 10000 ab
#> 21 28 73.5522 31 67 4509 10000 a
<sup>Created on 2019-01-15 by the reprex package (v0.2.1)</sup>
