Question

假设我们需要编写一个函数来给出集合中所有子集的列表。函数和doctest如下。我们需要完成函数的整个定义

def subsets(s):
   """Return a list of the subsets of s.

   >>> subsets({True, False})
   [{False, True}, {False}, {True}, set()]
   >>> counts = {x for x in range(10)} # A set comprehension
   >>> subs = subsets(counts)
   >>> len(subs)
   1024
   >>> counts in subs
   True
   >>> len(counts)
   10
   """
   assert type(s) == set, str(s) + ' is not a set.'
   if not s:
       return [set()]
   element = s.pop() 
   rest = subsets(s)
   s.add(element)

它必须不使用任何内置函数

我的方法是将“元素”添加到休息中并将它们全部返回，但我并不熟悉如何在Python中使用set，list。

Answer 1

查看powerset()中的itertools docs食谱。

from itertools import chain, combinations

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))

def subsets(s):
    return map(set, powerset(s))

Answer 2

>>> s=set(range(10))
>>> L=list(s)
>>> subs = [{L[j] for j in range(len(L)) if 1<<j&i} for i in range(1,1<<len(L))]
>>> s in subs
True
>>> set() in subs
False

Answer 3

>>> from itertools import combinations
>>> s=set(range(10))
>>> subs = [set(j) for i in range(len(s)) for j in combinations(s, i+1)]
>>> len(subs)
1023

Answer 4

list上关于powerset的通常实现是这样的：

def powerset(elements):
    if len(elements) > 0:
        head = elements[0]
        for tail in powerset(elements[1:]):
            yield [head] + tail
            yield tail
    else:
        yield []

只需要一点点适应来处理set。

Answer 5

稍微提高效率（比以前的答案更少复制）：

# Generate all subsets of the list v of length l.
def subsets(v, l):
  return _subsets(v, 0, l, [])

def _subsets(v, k, l, acc):
  if l == 0:
    return [acc]
  else:
    r = []
    for i in range(k, len(v)):
      # Take i-th position and continue with subsets of length l - 1:
      r.extend(_subsets(v, i + 1, l - 1, acc + [v[i]]))
    return r

Answer 6

如果您想不使用itertools或任何其他库来获取所有子集，则可以执行以下操作。

def generate_subsets(elementList):
    """Generate all subsets of a set""" 
    combination_count = 2**len(elementList)

    for i in range(0, combination_count):   
        tmp_str = str(bin(i)).replace("0b", "")
        tmp_lst = [int(x) for x in tmp_str]

        while (len(tmp_lst) < len(elementList)):
            tmp_lst = [0] + tmp_lst

        subset = list(filter(lambda x : tmp_lst[elementList.index(x)] == 1, elementList))
        print(subset)

Answer 7

>>> from itertools import combinations
>>> s=set([1,2,3])
>>> sum(map(lambda r: list(combinations(s, r)), range(1, len(s)+1)), [])
[(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]

产生元组，但它足够接近

在Python中获取集合的子集

7 个答案: