如何在Python中为列表中的多个分组生成排列

我知道我们可以使用itertools.permutations来排列列表中的不同项目。但是，如果我有一个列表，其中很少有项目需要处于固定位置，很少有项目需要与一个或多个项目交换，很少有项目需要与两个或多个项目交换，该怎么办

例如：

test = [1, 6, 2, 12, 5, 13, 11, 14, 15]

如何使用Python itertools.permutation或其他方法生成具有以下约束的所有可能组合

更新：

1 and 5 have fixed positions
In position 2, I could have either 6 or 11
In position 3, I could have either 2 or 12
In position 4, I could have 2 or 12
In position 6, I could have either 13, 14, 15 and so on

因此，我的列表如下所示：

[1, (6, 11), (2, 12), (2,12), 5, (13, 14, 15), (6, 11), (13, 14, 15), (13, 14, 15)]

from itertools import permutations, product, chain

test = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
groups = [[1], [2, 3, 4], [5], [6, 7], [8, 9], [10], [11, 12, 13], [14], [15, 16]]

result = [list(chain.from_iterable(permutation)) for permutation in product(*map(permutations, groups))]

for e in result[:20]:
    print(e)

from functools import partial
from itertools import combinations, permutations, product, chain

choose_one = partial(lambda r, iterable: combinations(iterable, r), 1)
groups = [[[1]], combinations([6, 11], 1), permutations([2, 12]), [[5]], combinations([13, 14, 15], 1)]

for e in product(*groups, repeat=1):
    print(list(chain.from_iterable(e)))

我已经将数字分组，表示同一组中的数字可以相互交换

---

谢谢。

您可以这样做：

[1, (6, 11), (2, 12), (2,12), 5, (13, 14, 15), (6, 11), (13, 14, 15), (13, 14, 15)]

from itertools import permutations, product, chain

test = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
groups = [[1], [2, 3, 4], [5], [6, 7], [8, 9], [10], [11, 12, 13], [14], [15, 16]]

result = [list(chain.from_iterable(permutation)) for permutation in product(*map(permutations, groups))]

for e in result[:20]:
    print(e)

from functools import partial
from itertools import combinations, permutations, product, chain

choose_one = partial(lambda r, iterable: combinations(iterable, r), 1)
groups = [[[1]], combinations([6, 11], 1), permutations([2, 12]), [[5]], combinations([13, 14, 15], 1)]

for e in product(*groups, repeat=1):
    print(list(chain.from_iterable(e)))

输出

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 16, 15]

[1, 6, 2, 12, 5, 13]
[1, 6, 2, 12, 5, 14]
[1, 6, 2, 12, 5, 15]
[1, 6, 12, 2, 5, 13]
[1, 6, 12, 2, 5, 14]
[1, 6, 12, 2, 5, 15]
[1, 11, 2, 12, 5, 13]
[1, 11, 2, 12, 5, 14]
[1, 11, 2, 12, 5, 15]
[1, 11, 12, 2, 5, 13]
[1, 11, 12, 2, 5, 14]
[1, 11, 12, 2, 5, 15]

更新

根据新的约束条件，您可以执行以下操作：

[1, (6, 11), (2, 12), (2,12), 5, (13, 14, 15), (6, 11), (13, 14, 15), (13, 14, 15)]

from itertools import permutations, product, chain

test = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
groups = [[1], [2, 3, 4], [5], [6, 7], [8, 9], [10], [11, 12, 13], [14], [15, 16]]

result = [list(chain.from_iterable(permutation)) for permutation in product(*map(permutations, groups))]

for e in result[:20]:
    print(e)

from functools import partial
from itertools import combinations, permutations, product, chain

choose_one = partial(lambda r, iterable: combinations(iterable, r), 1)
groups = [[[1]], combinations([6, 11], 1), permutations([2, 12]), [[5]], combinations([13, 14, 15], 1)]

for e in product(*groups, repeat=1):
    print(list(chain.from_iterable(e)))

输出

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 16, 15]

[1, 6, 2, 12, 5, 13]
[1, 6, 2, 12, 5, 14]
[1, 6, 2, 12, 5, 15]
[1, 6, 12, 2, 5, 13]
[1, 6, 12, 2, 5, 14]
[1, 6, 12, 2, 5, 15]
[1, 11, 2, 12, 5, 13]
[1, 11, 2, 12, 5, 14]
[1, 11, 2, 12, 5, 15]
[1, 11, 12, 2, 5, 13]
[1, 11, 12, 2, 5, 14]
[1, 11, 12, 2, 5, 15]

---

是否有任何特定的逻辑说明为什么

测试的哪些元素是固定的或被认为是可交换的？如果没有，您可以只计算每个硬编码组的置换，并最终连接子结果以得到最终列表。如上所述，如果您知道可以置换的组，请将它们添加到自己的列表中

等。当你遇到一个列表时，循环并进行置换。@是的，这听起来也是个好主意。谢谢，如果我调用上面列表中的len方法result，我只得到288的值。我以为会有更多可能的组合？有没有办法从数学上知道组合的总数？：）@我相信数字是正确的，你有两组3个元素，3个2个元素和4个1个元素。一组3个元素的排列是6，2个元素的排列是2，一个元素的排列是1。然后你必须把它们全部乘以

6**2*2**3*1**4=288

谢谢，实际上我想我没有把我的问题说清楚。我现在更新了。你能检查一下吗？在该位置，这些可能的数字中只能有一个。例如，在位置2，我可以有6个或11个。“我认为这涉及到排列和组合。”NeonFlash更新了答案！正如你所猜测的，它包括排列和组合。