两个列表的所有可能替换？

Question

（我知道问题的标题可能具有误导性，但我找不到任何其他方式来表述它 - 请随时编辑它）

我有两个列表，长度相同：

a = [1,2,3]
b = [4,5,6]

我想用第二个列表获取第一个列表的所有可能替换。

output[0] = [1,2,3] # no replacements
output[1] = [4,2,3] # first item was replaced
output[2] = [1,5,3] # second item was replaced
output[3] = [1,2,6] # third item was replaced
output[4] = [4,5,3] # first and second items were replaced
output[5] = [4,2,6] # first and third items were replaced
output[6] = [1,5,6] # second and third items were replaced
output[7] = [4,5,6] # all items were replaced

请注意，以下问题无法回答此问题：

• How to get all possible combinations of a list’s elements?
• combinations between two lists?
• Python merging two lists with all possible permutations
• All combinations of a list of lists

涉及先前链接的答案的可能解决方案是创建多个列表，然后对它们使用 itertools.product 方法。例如，我可以创建 3 个包含 2 个元素的列表，而不是 2 个包含 3 个元素的列表；但是，这会使代码过于复杂，如果可以的话，我宁愿避免这种情况。

有没有简单快捷的方法？

Answer 1

创建 3 个包含两个元素的列表根本不会使代码过于复杂。 zip can "flip the axes" of multiple lists琐碎（把Y元素的X序列变成X元素的Y序列），方便使用itertools.product：

import itertools

a = [1,2,3]
b = [4,5,6]

# Unpacking result of zip(a, b) means you automatically pass
# (1, 4), (2, 5), (3, 6)
# as the arguments to itertools.product
output = list(itertools.product(*zip(a, b)))

print(*output, sep="\n")

哪些输出：

(1, 2, 3)
(1, 2, 6)
(1, 5, 3)
(1, 5, 6)
(4, 2, 3)
(4, 2, 6)
(4, 5, 3)
(4, 5, 6)

与您的示例输出的排序不同，但它是同一组可能的替换。

Answer 2

每个项目都可以单独更换或单独放置。这可以通过位为 1 或 0 来建模。如果您将每个项目视为一个单独的位，那么迭代所有可能性可以映射到迭代所有 n 位的组合.

换句话说，从 0 迭代到 2ⁿ-1 并查看位模式。

n = len(a)
for i in range(2**n):
    yield [a[j] if i & (1 << j) != 0 else b[j] for j in range(n)]

打破这一点，i & (1 << j) != 0 检查是否设置了 i 的第 j 位。如果是，则使用a[j]，否则使用b[j]。

结果：

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

Answer 3

好的，这与其他答案相似，但从两者中吸取了一点。您可以将问题建模为查找给定长度序列的所有可能位，并仅在有 1 时替换，否则不替换。

from itertools import product

a = [1,2,3]
b = [4,5,6]

## All binary combinations of length of a (or b)
combinations = product([0,1], repeat=len(a))

for combination in combinations:
    y = []
    for l, i in zip(zip(a,b),combination):
         y.append(l[i])
    print y

所有位组合为：

(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)

结果：

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