Question

假设我有两个列表；

a = ['A', 'B', 'C', 'D']
b  = ['1', '2', '3', '4']

我知道我可以像这样获得这两个列表的每个排列：

for r in itertools.product(a, b): print (r[0] + r[1])

但是我要寻找的是存储在元组中的每个成对组合。因此，例如，一些组合是：

[(A, 1), (B, 2), (C, 3), (D, 4)]
[(A, 1), (B, 3), (C, 2), (D, 4)]
[(A, 1), (B, 4), (C, 3), (D, 2)]
[(A, 1), (B, 3), (C, 4), (D, 2)]
[(A, 1), (B, 2), (C, 4), (D, 3)]

因此它将遍历所有可能的组合，以使字母都不具有相同的数字值。我对执行此操作的有效方法不知所措（特别是因为我需要在实际示例中将其缩放到三个列表）

Answer 1

我们可以对列表中的一个进行排列（例如此处的后一个），然后将每个排列与第一个列表一起压缩，例如：

from functools import partial
from itertools import permutations

def pairwise_comb(xs, ys):
    return map(partial(zip, xs), permutations(ys))

或者所有子元素都应该是列表（此处仍然是可迭代对象，当您“实现”这些元素时可以采用特定形状）：

from functools import partial
from itertools import permutations

def pairwise_comb(xs, ys):
    return map(list, map(partial(zip, xs), permutations(ys)))

对于给定的样本输入，我们获得：

>>> for el in pairwise_comb(a, b):
...     print(list(el))
... 
[('A', '1'), ('B', '2'), ('C', '3'), ('D', '4')]
[('A', '1'), ('B', '2'), ('C', '4'), ('D', '3')]
[('A', '1'), ('B', '3'), ('C', '2'), ('D', '4')]
[('A', '1'), ('B', '3'), ('C', '4'), ('D', '2')]
[('A', '1'), ('B', '4'), ('C', '2'), ('D', '3')]
[('A', '1'), ('B', '4'), ('C', '3'), ('D', '2')]
[('A', '2'), ('B', '1'), ('C', '3'), ('D', '4')]
[('A', '2'), ('B', '1'), ('C', '4'), ('D', '3')]
[('A', '2'), ('B', '3'), ('C', '1'), ('D', '4')]
[('A', '2'), ('B', '3'), ('C', '4'), ('D', '1')]
[('A', '2'), ('B', '4'), ('C', '1'), ('D', '3')]
[('A', '2'), ('B', '4'), ('C', '3'), ('D', '1')]
[('A', '3'), ('B', '1'), ('C', '2'), ('D', '4')]
[('A', '3'), ('B', '1'), ('C', '4'), ('D', '2')]
[('A', '3'), ('B', '2'), ('C', '1'), ('D', '4')]
[('A', '3'), ('B', '2'), ('C', '4'), ('D', '1')]
[('A', '3'), ('B', '4'), ('C', '1'), ('D', '2')]
[('A', '3'), ('B', '4'), ('C', '2'), ('D', '1')]
[('A', '4'), ('B', '1'), ('C', '2'), ('D', '3')]
[('A', '4'), ('B', '1'), ('C', '3'), ('D', '2')]
[('A', '4'), ('B', '2'), ('C', '1'), ('D', '3')]
[('A', '4'), ('B', '2'), ('C', '3'), ('D', '1')]
[('A', '4'), ('B', '3'), ('C', '1'), ('D', '2')]
[('A', '4'), ('B', '3'), ('C', '2'), ('D', '1')]

由于'A'，'B'，'C'和'D'的顺序保持固定，因此这有24种可能的组合方式，可以使用4个字符在4中分配！方式或 4！ = 4×3×2×1 = 24 。

Answer 2

这可能比您想象的容易得多。怎么样：

import itertools

a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']

for aperm in itertools.permutations(a):
    for bperm in itertools.permutations(b):
        print(list(zip(aperm, bperm)))

第一个输出：

[('A', '1'), ('B', '2'), ('C', '3'), ('D', '4')]
[('A', '1'), ('B', '2'), ('C', '4'), ('D', '3')]
[('A', '1'), ('B', '3'), ('C', '2'), ('D', '4')]
[('A', '1'), ('B', '3'), ('C', '4'), ('D', '2')]
[('A', '1'), ('B', '4'), ('C', '2'), ('D', '3')]
[('A', '1'), ('B', '4'), ('C', '3'), ('D', '2')]
[('A', '2'), ('B', '1'), ('C', '3'), ('D', '4')]
[('A', '2'), ('B', '1'), ('C', '4'), ('D', '3')]
[('A', '2'), ('B', '3'), ('C', '1'), ('D', '4')]
...

（这两个4元素列表印有576行）

编辑：如果要将其推广到更多可迭代项，则可以执行以下操作：

import itertools

a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']

gens = [itertools.permutations(lst) for lst in (a,b)]

for perms in itertools.product(*gens):
    print(list(zip(*perms)))

输出相同的内容，但可以轻松扩展，例如

import itertools

a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
c = ['W', 'X', 'Y', 'Z']

gens = [itertools.permutations(lst) for lst in (a,b,c)]   # add c

for perms in itertools.product(*gens):                    # no change
    print(list(zip(*perms)))                              # ''

Answer 3

您可以将yield的递归用于无导入解决方案：

a = ['A', 'B', 'C', 'D']
b  = ['1', '2', '3', '4']
def combinations(d, current = []):
  if len(current) == 4:
    yield current
  elif filter(None, d):
    for i in d[0]:
      _d0, _d1 = [c for c in d[0] if c != i], [c for c in d[1] if c != d[1][0]]
      yield from combinations([_d0, _d1] , current+[[i, d[1][0]]])

for i in combinations([a, b]):
  print(i)

输出：

[['A', '1'], ['B', '2'], ['C', '3'], ['D', '4']]
[['A', '1'], ['B', '2'], ['D', '3'], ['C', '4']]
[['A', '1'], ['C', '2'], ['B', '3'], ['D', '4']]
[['A', '1'], ['C', '2'], ['D', '3'], ['B', '4']]
[['A', '1'], ['D', '2'], ['B', '3'], ['C', '4']]
[['A', '1'], ['D', '2'], ['C', '3'], ['B', '4']]
[['B', '1'], ['A', '2'], ['C', '3'], ['D', '4']]
[['B', '1'], ['A', '2'], ['D', '3'], ['C', '4']]
[['B', '1'], ['C', '2'], ['A', '3'], ['D', '4']]
[['B', '1'], ['C', '2'], ['D', '3'], ['A', '4']]
[['B', '1'], ['D', '2'], ['A', '3'], ['C', '4']]
[['B', '1'], ['D', '2'], ['C', '3'], ['A', '4']]
[['C', '1'], ['A', '2'], ['B', '3'], ['D', '4']]
[['C', '1'], ['A', '2'], ['D', '3'], ['B', '4']]
[['C', '1'], ['B', '2'], ['A', '3'], ['D', '4']]
[['C', '1'], ['B', '2'], ['D', '3'], ['A', '4']]
[['C', '1'], ['D', '2'], ['A', '3'], ['B', '4']]
[['C', '1'], ['D', '2'], ['B', '3'], ['A', '4']]
[['D', '1'], ['A', '2'], ['B', '3'], ['C', '4']]
[['D', '1'], ['A', '2'], ['C', '3'], ['B', '4']]
[['D', '1'], ['B', '2'], ['A', '3'], ['C', '4']]
[['D', '1'], ['B', '2'], ['C', '3'], ['A', '4']]
[['D', '1'], ['C', '2'], ['A', '3'], ['B', '4']]
[['D', '1'], ['C', '2'], ['B', '3'], ['A', '4']]

获取两个列表之间的每对值的成对组合

3 个答案: