Python 如何获得列表的所有可能组合’；s元素？
我有一个包含15个数字的列表，我需要编写一些代码来生成这些数字的所有32768个组合 我（通过谷歌搜索）发现，这显然符合我的要求，但我发现代码相当不透明，并对使用它持谨慎态度。另外，我觉得一定有更优雅的解决方案 我唯一想到的就是循环遍历十进制整数1–32768并将其转换为二进制，然后使用二进制表示作为过滤器来选择适当的数字 有人知道更好的方法吗？使用Python 如何获得列表的所有可能组合’；s元素？,python,combinations,Python,Combinations,我有一个包含15个数字的列表，我需要编写一些代码来生成这些数字的所有32768个组合 我（通过谷歌搜索）发现，这显然符合我的要求，但我发现代码相当不透明，并对使用它持谨慎态度。另外，我觉得一定有更优雅的解决方案 我唯一想到的就是循环遍历十进制整数1–32768并将其转换为二进制，然后使用二进制表示作为过滤器来选择适当的数字 有人知道更好的方法吗？使用map（），也许可以看看： 从返回元素的r长度子序列 输入不可编辑 组合按字典排序顺序发出。那么，如果 如果对输入iterable进行排序，则 组合
map（）from itertools import chain, combinations
def all_subsets(ss):
return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))
for subset in all_subsets(stuff):
print(subset)
这是一个懒惰的单行程序，也使用itertools：
from itertools import compress, product
def combinations(items):
return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) )
# alternative: ...in product([0,1], repeat=len(items)) )
def combine(inp):
return combine_helper(inp, [], [])
def combine_helper(inp, temp, ans):
for i in range(len(inp)):
current = inp[i]
remaining = inp[i + 1:]
temp.append(current)
ans.append(tuple(temp))
combine_helper(remaining, temp, ans)
temp.pop()
return ans
print(combine(['a', 'b', 'c', 'd']))
- 这要求您可以调用
上的items
（解决方法：如果len（…）
类似于生成器，请首先使用items
将其转换为列表）items=list（_itemsArg） - 这要求
项的迭代顺序不是随机的（解决方法：不要疯狂） - 这要求项目是唯一的，否则
和{2,2,1}
都将折叠为{2,1,1}
（解决方法：使用{2,1}集合。计数器作为集合的替代品；它基本上是一个多集…尽管以后可能需要使用
如果您需要它是可散列的）元组（排序（计数器（…）.elements（））
演示
>>> list(combinations(range(4)))
[set(), {3}, {2}, {2, 3}, {1}, {1, 3}, {1, 2}, {1, 2, 3}, {0}, {0, 3}, {0, 2}, {0, 2, 3}, {0, 1}, {0, 1, 3}, {0, 1, 2}, {0, 1, 2, 3}]
>>> list(combinations('abcd'))
[set(), {'d'}, {'c'}, {'c', 'd'}, {'b'}, {'b', 'd'}, {'c', 'b'}, {'c', 'b', 'd'}, {'a'}, {'a', 'd'}, {'a', 'c'}, {'a', 'c', 'd'}, {'a', 'b'}, {'a', 'b', 'd'}, {'a', 'c', 'b'}, {'a', 'c', 'b', 'd'}]
使用列表理解：
def selfCombine( list2Combine, length ):
listCombined = str( ['list2Combine[i' + str( i ) + ']' for i in range( length )] ).replace( "'", '' ) \
+ 'for i0 in range(len( list2Combine ) )'
if length > 1:
listCombined += str( [' for i' + str( i ) + ' in range( i' + str( i - 1 ) + ', len( list2Combine ) )' for i in range( 1, length )] )\
.replace( "', '", ' ' )\
.replace( "['", '' )\
.replace( "']", '' )
listCombined = '[' + listCombined + ']'
listCombined = eval( listCombined )
return listCombined
list2Combine = ['A', 'B', 'C']
listCombined = selfCombine( list2Combine, 2 )
['A', 'A']
['A', 'B']
['A', 'C']
['B', 'B']
['B', 'C']
['C', 'C']
[()]
[(1,), (2,), (3,), (4,)]
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
[(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
[(1, 2, 3, 4)]
我同意Dan H的观点，Ben确实要求所有的组合
itertools.combinations（）iterable = range(10)
for s in xrange(len(iterable)+1):
for comb in itertools.combinations(iterable, s):
yield comb
下面是一个使用递归的例子：
>>> import copy
>>> def combinations(target,data):
... for i in range(len(data)):
... new_target = copy.copy(target)
... new_data = copy.copy(data)
... new_target.append(data[i])
... new_data = data[i+1:]
... print new_target
... combinations(new_target,
... new_data)
...
...
>>> target = []
>>> data = ['a','b','c','d']
>>>
>>> combinations(target,data)
['a']
['a', 'b']
['a', 'b', 'c']
['a', 'b', 'c', 'd']
['a', 'b', 'd']
['a', 'c']
['a', 'c', 'd']
['a', 'd']
['b']
['b', 'c']
['b', 'c', 'd']
['b', 'd']
['c']
['c', 'd']
['d']
此一行代码提供了所有组合（如果原始列表/集合包含
n0n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']]
在线试用：
您可以使用以下简单代码在Python中生成列表的所有组合：
import itertools
a = [1,2,3,4]
for i in xrange(0,len(a)+1):
print list(itertools.combinations(a,i))
['A', 'A']
['A', 'B']
['A', 'C']
['B', 'B']
['B', 'C']
['C', 'C']
[()]
[(1,), (2,), (3,), (4,)]
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
[(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
[(1, 2, 3, 4)]
这段代码使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()
演示：
这里还有另一个解决方案（一行代码），涉及使用
itertools.combines演示： 如中所述
在@Dan H高度投票的评论中，提到了中的
powerset（）s=list（iterable）s=list（set（iterable））iterable列表，这意味着它将与生成器一起工作（与其他几个答案不同）
输出：
组合#1:（）
组合#2:（1，）
组合#3:（2，）
组合#4:（3，）
组合#5:（1,2）
组合#6:（1,3）
组合#7:（2,3）
组合#8:（1,2,3）
我想我会为那些寻求答案的人添加此函数，而不必导入itertools或任何其他额外的库
def powerSet(items):
"""
Power set generator: get all possible combinations of a list’s elements
Input:
items is a list
Output:
returns 2**n combination lists one at a time using a generator
Reference: edx.org 6.00.2x Lecture 2 - Decision Trees and dynamic programming
"""
N = len(items)
# enumerate the 2**N possible combinations
for i in range(2**N):
combo = []
for j in range(N):
# test bit jth of integer i
if (i >> j) % 2 == 1:
combo.append(items[j])
yield combo
简单产量生成器用法：
for i in powerSet([1,2,3,4]):
print (i, ", ", end="")
上述使用示例的输出：
[]，[1]，[2]，[1,2]，[3]，[1,3]，[2,3]，[1,2,3]，[4]，
[1, 4] , [2, 4] , [1, 2, 4] , [3, 4] , [1, 3, 4] , [2, 3, 4] , [1, 2,
3,4],
这是我的实现
def get_combinations(list_of_things):
"""gets every combination of things in a list returned as a list of lists
Should be read : add all combinations of a certain size to the end of a list for every possible size in the
the list_of_things.
"""
list_of_combinations = [list(combinations_of_a_certain_size)
for possible_size_of_combinations in range(1, len(list_of_things))
for combinations_of_a_certain_size in itertools.combinations(list_of_things,
possible_size_of_combinations)]
return list_of_combinations
我知道使用itertools获取所有组合要实际得多，但如果你想编写大量代码，那么你可以通过列表理解部分实现这一点
对于两对的组合：
lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]
而且，对于三对的组合，很容易做到：
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:]]
结果与使用itertools.compositions相同：
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)]
不使用itertools：
from itertools import compress, product
def combinations(items):
return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) )
# alternative: ...in product([0,1], repeat=len(items)) )
def combine(inp):
return combine_helper(inp, [], [])
def combine_helper(inp, temp, ans):
for i in range(len(inp)):
current = inp[i]
remaining = inp[i + 1:]
temp.append(current)
ans.append(tuple(temp))
combine_helper(remaining, temp, ans)
temp.pop()
return ans
print(combine(['a', 'b', 'c', 'd']))
下面是两种itertools.compositions的实现
返回列表的人
def combinations(lst, depth, start=0, items=[]):
if depth <= 0:
return [items]
out = []
for i in range(start, len(lst)):
out += combinations(lst, depth - 1, i + 1, items + [lst[i]])
return out
这是一个非常肤浅的案例，但如果有人正在寻找一个相反的列表，最好是安全的，比如我：
stuff = [1, 2, 3, 4]
def reverse(bla, y):
for subset in itertools.combinations(bla, len(bla)-y):
print list(subset)
if y != len(bla):
y += 1
reverse(bla, y)
reverse(stuff, 1)
这个怎么样。。使用了一个st
def get_combinations(list_of_things):
"""gets every combination of things in a list returned as a list of lists
Should be read : add all combinations of a certain size to the end of a list for every possible size in the
the list_of_things.
"""
list_of_combinations = [list(combinations_of_a_certain_size)
for possible_size_of_combinations in range(1, len(list_of_things))
for combinations_of_a_certain_size in itertools.combinations(list_of_things,
possible_size_of_combinations)]
return list_of_combinations
lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]
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:]]
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)]
def combine(inp):
return combine_helper(inp, [], [])
def combine_helper(inp, temp, ans):
for i in range(len(inp)):
current = inp[i]
remaining = inp[i + 1:]
temp.append(current)
ans.append(tuple(temp))
combine_helper(remaining, temp, ans)
temp.pop()
return ans
print(combine(['a', 'b', 'c', 'd']))
def combinations(lst, depth, start=0, items=[]):
if depth <= 0:
return [items]
out = []
for i in range(start, len(lst)):
out += combinations(lst, depth - 1, i + 1, items + [lst[i]])
return out
def combinations(lst, depth, start=0, prepend=[]):
if depth <= 0:
yield prepend
else:
for i in range(start, len(lst)):
for c in combinations(lst, depth - 1, i + 1, prepend + [lst[i]]):
yield c
print([c for c in combinations([1, 2, 3, 4], 3)])
# [[1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]
# get a hold of prepend
prepend = [c for c in combinations([], -1)][0]
prepend.append(None)
print([c for c in combinations([1, 2, 3, 4], 3)])
# [[None, 1, 2, 3], [None, 1, 2, 4], [None, 1, 3, 4], [None, 2, 3, 4]]
stuff = [1, 2, 3, 4]
def reverse(bla, y):
for subset in itertools.combinations(bla, len(bla)-y):
print list(subset)
if y != len(bla):
y += 1
reverse(bla, y)
reverse(stuff, 1)
def comb(s, res):
if not s: return
res.add(s)
for i in range(0, len(s)):
t = s[0:i] + s[i + 1:]
comb(t, res)
res = set()
comb('game', res)
print(res)
import itertools
col_names = ["aa","bb", "cc", "dd"]
all_combinations = itertools.chain(*[itertools.combinations(col_names,i+1) for i,_ in enumerate(col_names)])
print(list(all_combinations))
def combinations(arr, carry):
for i in range(len(arr)):
yield carry + arr[i]
yield from combinations(arr[i + 1:], carry + arr[i])
def combs(a):
if len(a) == 0:
return [[]]
cs = []
for c in combs(a[1:]):
cs += [c, c+[a[0]]]
return cs
>>> combs([1,2,3,4,5])
[[], [1], [2], [2, 1], [3], [3, 1], [3, 2], ..., [5, 4, 3, 2, 1]]
flag = 0
requiredCals =12
from itertools import chain, combinations
def powerset(iterable):
s = list(iterable) # allows duplicate elements
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
stuff = [2,9,5,1,6]
for i, combo in enumerate(powerset(stuff), 1):
if(len(combo)>0):
#print(combo , sum(combo))
if(sum(combo)== requiredCals):
flag = 1
break
if(flag==1):
print('True')
else:
print('else')
from itertools import permutations, 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))
[
['A'],
['B'],
['C'],
['A', 'B'],
['A', 'C'],
['B', 'C'],
['A', 'B', 'C']
]
lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
for i in lst:
lstCombos.append(lst[lst.index(i):lst.index(i)+Length])
lst = [S, T, A, R]
lstCombos = []
for Length in range(0,len(lst)+1):
for i in lst:
if not lst[lst.index(i):lst.index(i)+Length]) in lstCombos:
lstCombos.append(lst[lst.index(i):lst.index(i)+Length])
for subList in lstCombos:
if subList = '':
lstCombos.remove(subList)