Question

我想从区间[1，N]中随机采样两个整数x和y，使得| x-y | ＆gt; = D，对于某些D＆lt; N.下面的代码（用R编写）是我一直在使用的，但效率非常低。这种抽样有更好的方法吗？谢谢你。

N <- 100; D <- 10;

i <- sample(1:N, 2)

while ( abs( i[1] - i[2] ) < D ){
   i <- sort(sample(1:N, 2))
}

Answer 1

我想关键是要意识到y依赖于x（或者相反）。这里的算法应该最多可以在三个步骤中起作用：

1. sample x from [1:N]
2. sample y from [1:(x-D)] if (x-D) >= 1
   sample y from [x + D:N] if (x+D) <= N
3. If both conditions for y are met, choose one of the generated y  uniform at random

这个想法是，一旦x被采样，y需要在[1：（x-D）]或[x + D：N]的范围内，以满足| x-y |。＆gt; = D.

示例：

N = 100; d = 10

a) x is close to N

1. x is sampled from 1:N as 95
2. to satisfy |x-y| >= D, y can be at most 85, so the range to sample y is [1:85]

b) x is close to 1

1. x is sampled from 1:N as 9
2. y must be at least 19, so the range to sample y is [19:N]

c) x is close to 50

1. x is sampled from 1:N as 45
2. y must be either at most 35, or at least 55, so the ranges to sample from are [1:35] and [55:N]

Answer 2

我会先通过随机抽样得出这个数字之间的差异大于或等于D。换句话说，我们希望在D和N-1之间抽样替换。

difference <- sample(D:(N-1), 20, replace = TRUE)

现在我们需要做的就是选择1和N - difference之间的数字来选择较低的数字。我们可以使用vapply执行此操作。

lowerval <- vapply(N - difference, sample, numeric(1), 1)

最后，我们通过将差值添加到较低值来获得上限值。

upperval <- lowerval + difference

随机采样与距离条件

2 个答案: