Question

我很难理解如何组合列表中的元素。

我有：

a = [[1, 2], [3, 4, 5], [6], [7, 8, 9, 10]]

我想通过从每个列表成员中获取元素来创建三元组但总是包括第一个元素。这段代码完成了一半作业：

r=[[]]
for x in a[0:3]:
    t = []
    for y in x:
        for i in r:
            t.append(i+[y])
r = t

[R

[[1, 3, 6], [2, 3, 6], [1, 4, 6], [2, 4, 6], [1, 5, 6], [2, 5, 6]]

但我也想要：

[[1,6,7], [1,6,8], [1,6,9] etc.]

有人可以建议一个好的方法吗？

Answer 1

使用itertools：

import itertools

a = [[1, 2], [3, 4, 5], [6], [7, 8, 9, 10]]
for xss in itertools.combinations(a, 3): # pick 3 lists.
    # xss => ([1, 2], [3, 4, 5], [6])
    #        ([1, 2], [3, 4, 5], [7, 8, 9, 10])
    #        ...

    indexed = [enumerate(x) for x in xss]
    # indexed => [[(0, 1), (1, 2)], [(0, 3), (1, 4), (2, 5)], [(0, 6)]]
    #             ^^^^^^^^^^^^^^^^ list(enumerate([1, 2]))

    for xs in itertools.product(*indexed):
        # xs => ((0, 1), (0, 3), (0, 6))
        #       ((0, 1), (0, 3), (0, 7))
        #       ((0, 1), (0, 6), (0, 7))
        #        ...
        if all(i > 0 for i, x in xs): # exclude no first item is selected.
            continue
        print [x for i, x in xs]

更新对评论的回复。

import itertools

a = [[1, 2], [3, 4, 5], [6], [7, 8, 9, 10]]
for xss in itertools.combinations(a[1:], 2):
    xss = (a[0],) + xss
    indexed = [enumerate(x) for x in xss]
    for xs in itertools.product(*indexed):
        if all(i > 0 for i, x in xs):
            continue
        print [x for i, x in xs]

输出：

[1, 3, 6]
[1, 4, 6]
[1, 5, 6]
[2, 3, 6]
[2, 4, 6]
[2, 5, 6]
[1, 3, 7]
[1, 3, 8]
[1, 3, 9]
[1, 3, 10]
[1, 4, 7]
[1, 4, 8]
[1, 4, 9]
[1, 4, 10]
[1, 5, 7]
[1, 5, 8]
[1, 5, 9]
[1, 5, 10]
[2, 3, 7]
[2, 3, 8]
[2, 3, 9]
[2, 3, 10]
[2, 4, 7]
[2, 5, 7]
[1, 6, 7]
[1, 6, 8]
[1, 6, 9]
[1, 6, 10]
[2, 6, 7]
[2, 6, 8]
[2, 6, 9]
[2, 6, 10]

当时组合列表3的元素

1 个答案: