Question

所有我在网上被教导和阅读的都是随机播放和其他方法。我在想是否有另一种简单形式的方法，比如while或for。

如果有人可以帮助我，我是新来的

Answer 1

你应该看combinations。

from itertools import combinations

my_list = [1, 2, 3, 4]
list(combinations(my_list, 3))

# [(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]

Answer 2

你所要求的并不完全清楚。如果您只想快速了解列表的排列，请使用itertools.permutations：

for perm in itertools.permutations(sequence):
    # do something with permutation "perm"

如果你想了解如何自己产生排列，那么它会有点复杂。存在多种方式，并且通常选择不同的算法以便以特定顺序产生排列。这是描述in Wikipedia的算法，它以字典顺序生成序列的排列（与itertools相同）：

def permutations(seq):
    perm = sorted(seq) # the first permutation is the sequence in sorted order
    while True:
        yield perm

        # find largest index k such that perm[k] < perm[k+1]
        for k in range(len(perm)-2, -1, -1):
            if perm[k] < perm[k+1]:
                break
        else: # if none was found, we've already found the last permutation
            return

        # find the largest index l such that perm[k] < perm[l] (always exists)
        for l in range(len(perm)-1, -1, -1):
            if p[k] < p[l]:
                break

        # Build the next permutation. This is equivalent to copying perm,
        # swapping the values at indexes `k` and `l`, then reversing the values
        # from index `k+1` to the end.
        perm = perm[:k]+perm[l:l+1]+perm[-1:l:-1]+perm[k:k+1]+perm[l-1:k:-1]

我怎么做一个没有洗牌的列表的排列？

2 个答案: