如何在不更换逐步样？

问题描述

Python有 my_sample = random.sample（范围（100），10）来随机抽取无需更换从 [0，100）。

Python has my_sample = random.sample(range(100), 10) to randomly sample without replacement from [0, 100).

假设我已经取样 N 这样的数字，现在我想品尝多一个不用更换（不包括任何的previously采样的 ñ），如何做到这一点的超级效率？

Suppose I have sampled n such numbers and now I want to sample one more without replacement (without including any of the previously sampled n), how to do so super efficiently?

更新：由「合理有效地到超级高效的（但忽略常数因子）

update: changed from "reasonably efficiently" to "super efficiently" (but ignoring constant factors)

推荐答案

注意从OP读者：请考虑在看的来理解其中的逻辑，然后明白这个答案。

Note to readers from OP: Please consider looking at the originally accepted answer to understand the logic, and then understand this answer.

Aaaaaand的完整性的缘故：这是 no_answer_not_upvoted的回答的概念，但适应所以需要禁止的号码清单作为输入。这是一样的code在我的previous答案的，但我们从<$构建一个状态C $ C>禁止，之前我们生成的数字。

Aaaaaand for completeness sake: This is the concept of no_answer_not_upvoted’s answer, but adapted so it takes a list of forbidden numbers as input. This is just the same code as in my previous answer, but we build a state from forbid, before we generate numbers.

• 这是时间 O（F + K）和内存 O（F + K）。显然，这是最快的东西可能不朝禁止（排序/套）的格式要求。我想，这使得这个以某种方式^^赢家。
• 如果禁止是一组，反复猜测的方法是更快 O（k⋅n/（N-（F + K）） ），这是非常接近 O（K）为 F + K 不是非常的接近 N 。
• 如果禁止进行排序，我可笑的算法是快搭配：
  

• This is time O(f+k) and memory O(f+k). Obviously this is the fastest thing possible without requirements towards the format of forbid (sorted/set). I think this makes this a winner in some way ^^.
• If forbid is a set, the repeated guessing method is faster with O(k⋅n/(n-(f+k))), which is very close to O(k) for f+k not very close to n.
• If forbid is sorted, my ridiculous algorithm is faster with:
  

import random
def sample_gen(n, forbid):
    state = dict()
    track = dict()
    for (i, o) in enumerate(forbid):
        x = track.get(o, o)
        t = state.get(n-i-1, n-i-1)
        state[x] = t
        track[t] = x
        state.pop(n-i-1, None)
        track.pop(o, None)
    del track
    for remaining in xrange(n-len(forbid), 0, -1):
        i = random.randrange(remaining)
        yield state.get(i, i)
        state[i] = state.get(remaining - 1, remaining - 1)
        state.pop(remaining - 1, None)

用法：

gen = sample_gen(10, [1, 2, 4, 8])
print gen.next()
print gen.next()
print gen.next()
print gen.next()

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