Question

我目前在Python的土地上。这就是我需要做的。我已经查看了itertools库，但它似乎只做了排列。

我想要一个输入列表，比如['yahoo'，'wikipedia'，'freebase']，并生成一个项目与零个或多个其他项目的每个独特组合......

['yahoo', 'wikipedia', 'freebase']
['yahoo', 'wikipedia']
['yahoo', 'freebase']
['wikipedia', 'freebase']
['yahoo']
['freebase']
['wikipedia']

一些笔记。顺序无关紧要，我正在尝试设计方法来获取任何大小的列表。另外，这种组合有名称吗？

感谢您的帮助！

Answer 1

>>> l = ['yahoo', 'wikipedia', 'freebase']
>>> import itertools
>>> for i in range(1, len(l) +1):
    print(list(itertools.combinations(l, r=i)))


[('yahoo',), ('wikipedia',), ('freebase',)]
[('yahoo', 'wikipedia'), ('yahoo', 'freebase'), ('wikipedia', 'freebase')]
[('yahoo', 'wikipedia', 'freebase')]

P.S。为什么这个维基？

Answer 2

它被称为powerset。这是itertools docs：

的实现

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

Answer 3

这称为动力装置。只需按照algorithm进行操作即可。这是一个简单的实现：

def powerset(seq):
  if len(seq):
    head = powerset(seq[:-1])
    return head + [item + [seq[-1]] for item in head]
  else:
    return [[]]

>>> powerset(['yahoo', 'wikipedia', 'freebase'])
[[], ['yahoo'], ['wikipedia'], ['yahoo', 'wikipedia'], ['freebase'], ['yahoo', 'freebase'], ['wikipedia', 'freebase'], ['yahoo', 'wikipedia', 'freebase']]

另一个：

def powerset(s):
  sets = []
  indicator = lambda x: x & 1
  for element in xrange(2**len(s)):
    n = element
    subset = []
    for x in s:
        if indicator(n):
            subset.append(x)
        n >>= 1
    sets.append(subset)
  return sets

Answer 4

你基本上在二进制中从1到2 ⁿ -1 计数：

0 0 1 ['freebase'] 0 1 0 ['wikipedia'] 0 1 1 ['wikipedia', 'freebase'] 1 0 0 ['yahoo'] 1 0 1 ['yahoo', 'freebase'] 1 1 0 ['yahoo', 'wikipedia'] 1 1 1 ['yahoo', 'wikipedia', 'freebase']

给定列表中的项目组合

4 个答案: