元组的子元素

Question

我希望得到以下结论:

(('A',), ('B',), ('C',), ('D',))
(('A',), ('B',), ('C','D'))
(('A',), ('B','C'), ('D',))
(('A',), ('B','C','D'))
(('A','B'), ('C',), ('D',))
(('A','B'), ('C','D'))
(('A','B','C'), ('D',))
(('A','B','C','D'),)

在打电话时 sub_combinations(('A', 'B', 'C', 'D'))

这是我的尝试,但它不起作用:

def sub_combinations(segment):
   for i in range(1, len(segment)):
      for j in sub_combinations(segment[i:]):
         yield segment[:i]+j 
      yield segment 

但我认为我走在正确的轨道上.

另外,我想要一个名为limit的第二个参数来限制子元组的大小,例如它sub_combinations(('A', 'B', 'C', 'D'), 2)会给出:

(('A',), ('B',), ('C',), ('D',))
(('A',), ('B',), ('C','D'))
(('A',), ('B','C'), ('D',))
(('A','B'), ('C',), ('D',))
(('A','B'), ('C','D'))

我正在使用python 3.

Answer 1

处理基础案例 - 何时segment为空:

def sub_combinations(segment, size=0):
    if segment == ():
        yield ()
        return
    stop = min(size or len(segment), len(segment))
    for i in range(1, stop + 1):
        for j in sub_combinations(segment[i:], size):
            yield (segment[:i],) + j

用法示例:

>>> for x in sub_combinations(('A', 'B', 'C', 'D')):
...     print(x)
...
(('A',), ('B',), ('C',), ('D',))
(('A',), ('B',), ('C', 'D'))
(('A',), ('B', 'C'), ('D',))
(('A',), ('B', 'C', 'D'))
(('A', 'B'), ('C',), ('D',))
(('A', 'B'), ('C', 'D'))
(('A', 'B', 'C'), ('D',))
(('A', 'B', 'C', 'D'),)

>>> for x in sub_combinations(('A', 'B', 'C', 'D'), 2):
...     print(x)
...
(('A',), ('B',), ('C',), ('D',))
(('A',), ('B',), ('C', 'D'))
(('A',), ('B', 'C'), ('D',))
(('A', 'B'), ('C',), ('D',))
(('A', 'B'), ('C', 'D'))