我知道使用itertools来获得所有的组合要实际得多，但是如果你碰巧想要，假设你想要编写很多代码，你可以只使用列表理解来部分实现这一点

对于两对组合:

lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]

而且，对于三对组合，它是这样简单的:

lambda l: [(a, b, c) for i, a in enumerate(l) for ii, b in enumerate(l[i+1:]) for c in l[i+ii+2:]]

结果和使用itertools.combination是一样的:

import itertools
combs_3 = lambda l: [
    (a, b, c) for i, a in enumerate(l) 
    for ii, b in enumerate(l[i+1:]) 
    for c in l[i+ii+2:]
]
data = ((1, 2), 5, "a", None)
print("A:", list(itertools.combinations(data, 3)))
print("B:", combs_3(data))
# A: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]
# B: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]

这段代码采用了一个简单的嵌套列表算法…

# FUNCTION getCombos: To generate all combos of an input list, consider the following sets of nested lists...
#
#           [ [ [] ] ]
#           [ [ [] ], [ [A] ] ]
#           [ [ [] ], [ [A],[B] ],         [ [A,B] ] ]
#           [ [ [] ], [ [A],[B],[C] ],     [ [A,B],[A,C],[B,C] ],                   [ [A,B,C] ] ]
#           [ [ [] ], [ [A],[B],[C],[D] ], [ [A,B],[A,C],[B,C],[A,D],[B,D],[C,D] ], [ [A,B,C],[A,B,D],[A,C,D],[B,C,D] ], [ [A,B,C,D] ] ]
#
#  There is a set of lists for each number of items that will occur in a combo (including an empty set).
#  For each additional item, begin at the back of the list by adding an empty list, then taking the set of
#  lists in the previous column (e.g., in the last list, for sets of 3 items you take the existing set of
#  3-item lists and append to it additional lists created by appending the item (4) to the lists in the
#  next smallest item count set. In this case, for the three sets of 2-items in the previous list. Repeat
#  for each set of lists back to the initial list containing just the empty list.
#

def getCombos(listIn = ['A','B','C','D','E','F'] ):
    listCombos = [ [ [] ] ]     # list of lists of combos, seeded with a list containing only the empty list
    listSimple = []             # list to contain the final returned list of items (e.g., characters)

    for item in listIn:
        listCombos.append([])   # append an emtpy list to the end for each new item added
        for index in xrange(len(listCombos)-1, 0, -1):  # set the index range to work through the list
            for listPrev in listCombos[index-1]:        # retrieve the lists from the previous column
                listCur = listPrev[:]                   # create a new temporary list object to update
                listCur.append(item)                    # add the item to the previous list to make it current
                listCombos[index].append(listCur)       # list length and append it to the current list

                itemCombo = ''                          # Create a str to concatenate list items into a str
                for item in listCur:                    # concatenate the members of the lists to create
                    itemCombo += item                   # create a string of items
                listSimple.append(itemCombo)            # add to the final output list

    return [listSimple, listCombos]
# END getCombos()

这一行代码给出了所有的组合(如果原始列表/set包含n个不同的元素，则在0到n个元素之间)，并使用本机方法itertools.combination:

Python 2

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([map(list, combinations(input, i)) for i in range(len(input) + 1)], [])

Python 3

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([list(map(list, combinations(input, i))) for i in range(len(input) + 1)], [])

输出将是:

[[],
 ['a'],
 ['b'],
 ['c'],
 ['d'],
 ['a', 'b'],
 ['a', 'c'],
 ['a', 'd'],
 ['b', 'c'],
 ['b', 'd'],
 ['c', 'd'],
 ['a', 'b', 'c'],
 ['a', 'b', 'd'],
 ['a', 'c', 'd'],
 ['b', 'c', 'd'],
 ['a', 'b', 'c', 'd']]

在网上试试吧:

http://ideone.com/COghfX

我知道使用itertools来获得所有的组合要实际得多，但是如果你碰巧想要，假设你想要编写很多代码，你可以只使用列表理解来部分实现这一点

对于两对组合:

lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]

而且，对于三对组合，它是这样简单的:

lambda l: [(a, b, c) for i, a in enumerate(l) for ii, b in enumerate(l[i+1:]) for c in l[i+ii+2:]]

结果和使用itertools.combination是一样的:

import itertools
combs_3 = lambda l: [
    (a, b, c) for i, a in enumerate(l) 
    for ii, b in enumerate(l[i+1:]) 
    for c in l[i+ii+2:]
]
data = ((1, 2), 5, "a", None)
print("A:", list(itertools.combinations(data, 3)))
print("B:", combs_3(data))
# A: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]
# B: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]

3个功能:

列出n个元素的所有组合 列出n个元素的所有组合，其中顺序不明确 所有的排列

import sys

def permutations(a):
    return combinations(a, len(a))

def combinations(a, n):
    if n == 1:
        for x in a:
            yield [x]
    else:
        for i in range(len(a)):
            for x in combinations(a[:i] + a[i+1:], n-1):
                yield [a[i]] + x

def combinationsNoOrder(a, n):
    if n == 1:
        for x in a:
            yield [x]
    else:
        for i in range(len(a)):
            for x in combinationsNoOrder(a[:i], n-1):
                yield [a[i]] + x
    
if __name__ == "__main__":
    for s in combinations(list(map(int, sys.argv[2:])), int(sys.argv[1])):
        print(s)

from itertools import combinations


features = ['A', 'B', 'C']
tmp = []
for i in range(len(features)):
    oc = combinations(features, i + 1)
    for c in oc:
        tmp.append(list(c))

输出

[
 ['A'],
 ['B'],
 ['C'],
 ['A', 'B'],
 ['A', 'C'],
 ['B', 'C'],
 ['A', 'B', 'C']
]