问题描述
我想创建一个函数，该函数根据其输入生成一个随机数并将其应用于布尔向量.此函数将用于生成包含约 5 亿个元素的测试数据.
I want to create a function that generates a random number based on its input and apply it to a boolean vector. This function will be used to generate test data with approx 500M elements.
f <- function(x, p) ifelse(x, runif(1)^p, runif(1)^(1/p))
f(c(T,T,T,F,F,F), 2)
我得到的不是我想要的.
What I get is not what I wanted.
[1] 0.0054 0.0054 0.0054 0.8278 0.8278 0.8278
我希望输入向量的每个元素都有一个新的随机数，而不是两个重复的随机数.为什么我会得到这个结果，我怎样才能得到与
I'd expect a new random number for every element of my input vector, not two random numbers repeated. Why do I get this result, and how can I get the desired result which would be the same as
c(runif(3)^2, runif(3)^(1/2))
为每个元素产生一个新的随机数
which yields a new random number for every element
0.0774 0.7071 0.2184 0.8719 0.9990 0.8819
推荐答案
您需要制作两个不同的随机数向量，其长度与 x-vector 的长度相同.
You would need to make two different vectors of random numbers of the same length as the x-vector.
f <- function(x, p) ifelse(x, runif(6)^p, runif(6)^(1/p))
f(c(T,T,T,F,F,F), 2)
[1] 0.3040201 0.5543376 0.7291466 0.5205014 0.3563542 0.8697398
或更一般地说:
f <- function(x, p) ifelse(x, runif( length(x) )^p, runif( length(x) )^(1/p))
ifelse 函数并没有真正进行循环.第二个和第三个参数各计算一次.
The ifelse-function is not really doing looping. The second and third arguments get evaluated once each.
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