如何从所有可能的组合中获得所有可能的独特变体

Question

我有一个积分列表

lst = [1,2,3,4,5]

使用 itertools 我得到所有可能的行组合 ((1,2)=(2,1))：itertools.combinations(lst, 2)

[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]

我想获取元组列表（具有唯一点），像这样

   [
    [(1,2),(3,4)],
    [(1,2),(3,5)],
    [(1,2),(4,5)],
    [(1,3),(2,4)],
    [(1,3),(2,5)],
    ...
    [(2,3),(4,5)]
   ]

Answer 1

好的，这很有趣 xD。您需要的一切都由 itertools 提供，您只需以正确的方式组合它。看看这个：

import itertools


lst = [1,2,3,4,5]

points = list(itertools.combinations(lst, 2))
# [(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]

f = lambda p: filter(lambda sub: not any(x in sub for x in p), points)

res = []
for p in points:
    res.extend(list(itertools.product([p], f(p))))

# corresponding list-comprehension solution
# res = list(itertools.chain.from_iterable(itertools.product([p], f(p)) for p in points))

返回：

res = [((1, 2), (3, 4)), ((1, 2), (3, 5)), ((1, 2), (4, 5)), ((1, 3), (2, 4)), ((1, 3), (2, 5)), ((1, 3), (4, 5)), ((1, 4), (2, 3)), ((1, 4), (2, 5)), ((1, 4), (3, 5)), ((1, 5), (2, 3)), ((1, 5), (2, 4)), ((1, 5), (3, 4)), ((2, 3), (1, 4)), ((2, 3), (1, 5)), ((2, 3), (4, 5)), ((2, 4), (1, 3)), ((2, 4), (1, 5)), ((2, 4), (3, 5)), ((2, 5), (1, 3)), ((2, 5), (1, 4)), ((2, 5), (3, 4)), ((3, 4), (1, 2)), ((3, 4), (1, 5)), ((3, 4), (2, 5)), ((3, 5), (1, 2)), ((3, 5), (1, 4)), ((3, 5), (2, 4)), ((4, 5), (1, 2)), ((4, 5), (1, 3)), ((4, 5), (2, 3))]

---

它基本上归结为itertools.combinations（你已经做过）和itertools.product。扭曲是在两者之间执行的过滤（请参阅f = lambda ...）。如果您在理解代码方面需要进一步帮助，请告诉我。