就地改组列表的一部分

Question

我有一个list，我想就地洗牌它的一部分。

我知道random.shuffle()并且它可以就地工作，但是如果我对列表进行切片，它会打乱原始输入的切片副本，而原始输入list保持不变：

import random

l = list(range(20))

print(l)

# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

random.shuffle(l[:10])  # I wish it was shuffling the first half

print(l)  # but does nothing to `l`

# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

我有什么选择？

Answer 1

不是真的到位，但具有预期的结果：import randoml = list(range(20))lpart = l[:10]random.shuffle(lpart)l[:10] = lpartprint(l)

Answer 2

修改列表 n-place 不会仅对列表的一部分起作用。您可以改为使用random.sample不替换的随机样本，然后切片分配：k = 10l[:k] = random.sample(l[:k], k=k)print(l)# [1, 7, 6, 0, 2, 3, 4, 9, 8, 5, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

Answer 3

可以使用Fisher-Yates shuffle从根本上重新实现random.shuffle()以接受 afirst和lastindex 作为参数，例如：import randomdef valid_index(i, n):&nbsp; &nbsp; assert(-n <= i < n)&nbsp; &nbsp; return i % ndef shuffle(seq, first=0, last=-1, rand_int_gen=None):&nbsp; &nbsp; n = len(seq)&nbsp; &nbsp; first = valid_index(first, n)&nbsp; &nbsp; last = valid_index(last, n)&nbsp; &nbsp; # use Fisher-Yates shuffle (Durstenfeld method)&nbsp; &nbsp; if callable(rand_int_gen):&nbsp; &nbsp; &nbsp; &nbsp; for i in range(first, last):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; j = rand_int_gen(i, last)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; seq[i], seq[j] = seq[j], seq[i]&nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; getrandbits = random.getrandbits&nbsp; &nbsp; &nbsp; &nbsp; for i in range(first, last + 1):&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; size = last - i + 1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; j = getrandbits(size.bit_length()) % size + i&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; seq[i], seq[j] = seq[j], seq[i]&nbsp; &nbsp; return seq像这样使用：l = list(range(20))print(l)# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]random.seed(0)&nbsp; # just to show reproducible resultsshuffle(l, 0, 9)print(l)# [6, 7, 2, 5, 8, 4, 9, 3, 0, 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]从时间上看，这实际上比random.shuffle()改组整个序列还要快几个百分点。这本质上更快，因为它直接从中获取随机值，random.getrandbits()这是用于随机整数生成的最接近的方法random，其他方法，例如randint()并randrange()最终减少到这个。最后两个最终在内部使用_getrandbelow()，这可能会getrandbits()更频繁地调用必要的。for k in range(1, 7):&nbsp;&nbsp; &nbsp; n = 10 ** k&nbsp;&nbsp; &nbsp; print(n)&nbsp;&nbsp; &nbsp; %timeit l = list(range(n)); random.shuffle(l)&nbsp;&nbsp; &nbsp; %timeit l = list(range(n)); shuffle(l)&nbsp;&nbsp; &nbsp; print()&nbsp;10100000 loops, best of 3: 6.16 µs per loop100000 loops, best of 3: 3.85 µs per loop10010000 loops, best of 3: 54.3 µs per loop10000 loops, best of 3: 28 µs per loop10001000 loops, best of 3: 585 µs per loop1000 loops, best of 3: 341 µs per loop10000100 loops, best of 3: 6.01 ms per loop100 loops, best of 3: 3.56 ms per loop10000010 loops, best of 3: 71.7 ms per loop10 loops, best of 3: 44.1 ms per loop10000001 loop, best of 3: 815 ms per loop1 loop, best of 3: 582 ms per loop正如@usr2564301 所指出的，这里也建议了这种方法。不幸的是，我认为没有更好的方法可以就地执行此操作。