Question

list_a = [("A","<",1), ("A","==",5)]
list_b = [("B","<",5), ("B","==",7), ("B",">=",8)]
list_c = [("C","<",10),("C","<=",6),("C",">",4),("C","<=",6)]

我想列出每个可能的组合，每个列表只允许一个约束。

我可以看到itertools.product有点接近我想要的东西，我知道我可以做类似的事情

new_list = []
for a in list_a:
  for b in list_b:
    for c in list_c
      new_list.append(list(itertools.combinations([a,b,c],2)))

但n**3这里的[("A","<",1)] [("A","<",1),("B","<",5)] [("A","<",1),("B","==",7)] [("A","<",1),("B",">=",8)] [("A","<",1),("B",">=",8),("C","<",10)]复杂性似乎是一个令人难以置信的糟糕解决方案，因为我希望最终使用大小为30+的9个列表（即list_c，list_d，list_e等等）以下是一些可接受的可能输出：

{{1}}

基本上我理解当你有一组数字时使用itertools，例如itertools.product（（'ABCD'），3）会给出AAA，AAB，AAC，AAD，BAA，BAB的输出， BAC等，但我似乎无法弄清楚如何使用stdlib将“每个列表唯一一个”约束应用到最大延伸而不会破解一些非常低效的解决方案。

Answer 1

怎么样：

import itertools

list_a = [("A","<",1), ("A","==",5)]
list_b = [("B","<",5), ("B","==",7), ("B",">=",8)]
list_c = [("C","<",10),("C","<=",6),("C",">",4),("C","<=",6)]

lists = [list_a, list_b, list_c]


for l in lists: l.insert(0, None)

for x in itertools.product(*lists):
    print list(filter(None, x))

对于这些列表，我得到60个元素，包括一个空元素。

作为参考，下面列出了示例元素的索引：

[("A","<",1)]                           # 20
[("A","<",1),("B","<",5)]               # 25
[("A","<",1),("B","==",7)]              # 30
[("A","<",1),("B",">=",8)]              # 35
[("A","<",1),("B",">=",8),("C","<",10)] # 36

多列表的每种可能组合

1 个答案: