Question

我想获得列表列表的组合，例如

[['a0', 'a1'], ['b0', 'b1'], ['c0', 'c1', 'c2']]

我可以使用list(itertools.product(*listOfTuples))并获取

[('a0', 'b0', 'c0'), ('a0', 'b0', 'c1'), ('a0', 'b0', 'c2')... ('a1', 'b1', 'c2')]

如何获得长度为2的可能组合的列表？例如：

[('a0', 'b0'), ('a0', 'c0'), ('b0', 'c0')...

Answer 1

您可以展平列表，并创建一个简单的递归函数（对于在子列表中没有重复的结果）：

d = [['a0', 'a1'], ['b0', 'b1'], ['c0', 'c1', 'c2']]
def combinations(d, current = []):
  if len(current) == 2:
    yield current
  else:
    for i in d:
      if i not in current:
        yield from combinations(d, current+[i])

print(list(combinations([i for b in d for i in b]))

输出：

[['a0', 'a1'], ['a0', 'b0'], ['a0', 'b1'], ['a0', 'c0'], ['a0', 'c1'], ['a0', 'c2'], ['a1', 'a0'], ['a1', 'b0'], ['a1', 'b1'], ['a1', 'c0'], ['a1', 'c1'], ['a1', 'c2'], ['b0', 'a0'], ['b0', 'a1'], ['b0', 'b1'], ['b0', 'c0'], ['b0', 'c1'], ['b0', 'c2'], ['b1', 'a0'], ['b1', 'a1'], ['b1', 'b0'], ['b1', 'c0'], ['b1', 'c1'], ['b1', 'c2'], ['c0', 'a0'], ['c0', 'a1'], ['c0', 'b0'], ['c0', 'b1'], ['c0', 'c1'], ['c0', 'c2'], ['c1', 'a0'], ['c1', 'a1'], ['c1', 'b0'], ['c1', 'b1'], ['c1', 'c0'], ['c1', 'c2'], ['c2', 'a0'], ['c2', 'a1'], ['c2', 'b0'], ['c2', 'b1'], ['c2', 'c0'], ['c2', 'c1']]

Answer 2

您可以使用itertools.combinations()拨打r=2以获取每个三元组的所有对，然后展平，例如：

In []:
import itertools as it
[t for x in it.product(*s) for t in it.combinations(x, r=2)]

Out[]:
[('a0', 'b0'), ('a0', 'c0'), ('b0', 'c0'), ('a0', 'b0'), ('a0', 'c1'), ('b0', 'c1'),
 ('a0', 'b0'), ('a0', 'c2'), ('b0', 'c2'), ('a0', 'b1'), ('a0', 'c0'), ('b1', 'c0'), 
 ('a0', 'b1'), ('a0', 'c1'), ('b1', 'c1'), ('a0', 'b1'), ('a0', 'c2'), ('b1', 'c2'), 
 ('a1', 'b0'), ('a1', 'c0'), ('b0', 'c0'), ('a1', 'b0'), ('a1', 'c1'), ('b0', 'c1'), 
 ('a1', 'b0'), ('a1', 'c2'), ('b0', 'c2'), ('a1', 'b1'), ('a1', 'c0'), ('b1', 'c0'), 
 ('a1', 'b1'), ('a1', 'c1'), ('b1', 'c1'), ('a1', 'b1'), ('a1', 'c2'), ('b1', 'c2')]

但是，这会有重复，因为('a0', 'b0', 'c0')，('a0', 'b0', 'c1')和('a0', 'b0', 'c2')都会生成('a0', 'b0')，所以如果你想删除，那么你可以使用{{ 1}}，但你失去了秩序：

set

对它进行排序确实将其按顺序排列：

In []:
{t for x in it.product(*s) for t in it.combinations(x, r=2)}

Out[]:
{('a1', 'c0'), ('b0', 'c1'), ('b0', 'c2'), ('a0', 'c1'), ('a0', 'c2'), ('b1', 'c1'), 
 ('b1', 'c2'), ('a1', 'b1'), ('a1', 'c1'), ('a1', 'c2'), ('b0', 'c0'), ('a0', 'c0'), 
 ('a1', 'b0'), ('b1', 'c0'), ('a0', 'b1'), ('a0', 'b0')}

从列表python中生成值的组合

2 个答案: