由网友(昔日暖阳)分享简介：我需要一个号码列表的唯一随机排列而不更换，高效。我目前的做法：I need a number of unique random permutations of a list without replacement, efficiently. My current approach:total_permutation...

我需要一个号码列表的唯一随机排列而不更换，高效。我目前的做法：

I need a number of unique random permutations of a list without replacement, efficiently. My current approach:

total_permutations = math.factorial(len(population))
permutation_indices = random.sample(xrange(total_permutations), k)
k_permutations = [get_nth_permutation(population, x) for x in permutation_indices]

其中， get_nth_permutation 不正是它听起来像，有效（意为O（N））。但是，这仅适用于 len个（人口）LT = 20 ，仅仅是因为21！如此mindblowingly久，的xrange（math.factorial（21））将无法工作：

where get_nth_permutation does exactly what it sounds like, efficiently (meaning O(N)). However, this only works for len(population) <= 20, simply because 21! is so mindblowingly long that xrange(math.factorial(21)) won't work:

OverflowError: Python int too large to convert to C long

有没有更好的算法，以样本k独特的排列没有更换O（N）？

Is there a better algorithm to sample k unique permutations without replacement in O(N)?

推荐答案

而不是使用的xrange 只是继续产生随机数，直到你有很多，你所需要的。使用设置确保它们都是唯一的。

Instead of using xrange simply keep generating random numbers until you have as many as you need. Using a set makes sure they're all unique.

permutation_indices = set()
while len(permutation_indices) < k:
    permutation_indices.add(random.randrange(total_permutations))