问题：如何获得列表元素的所有可能组合？
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32,768个组合。
我已经找到了一些代码（通过Googling），这些代码显然可以满足我的需求，但是我发现代码相当不透明并且对使用它很谨慎。另外，我觉得必须有一个更优雅的解决方案。
对我而言，唯一发生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形式作为筛选器来选择适当的数字。
有谁知道更好的方法吗？使用map()，也许？
I have a list with 15 numbers in, and I need to write some code that produces all 32,768 combinations of those numbers.
I’ve found some code (by Googling) that apparently does what I’m looking for, but I found the code fairly opaque and am wary of using it. Plus I have a feeling there must be a more elegant solution.
The only thing that occurs to me would be to just loop through the decimal integers 1–32768 and convert those to binary, and use the binary representation as a filter to pick out the appropriate numbers.
Does anyone know of a better way? Using map(), maybe?
回答 0
看看itertools.combinations：
itertools.combinations(iterable, r)
从可迭代的输入中返回元素的r长度子序列。
组合按字典顺序排序。因此，如果将输入的iterable排序，则将按排序顺序生成组合元组。
从2.6开始，包括电池！
Have a look at itertools.combinations:
itertools.combinations(iterable, r)
Return r length subsequences of elements from
the input iterable.
Combinations are emitted in lexicographic sort order. So, if the
input iterable is sorted, the
combination tuples will be produced in
sorted order.
Since 2.6, batteries are included!
回答 1
这个答案错过了一个方面：OP要求所有组合……而不仅仅是长度“ r”的组合。
因此，您要么必须遍历所有长度“ L”：
import itertools
stuff = [1, 2, 3]
for L in range(0, len(stuff)+1):
for subset in itertools.combinations(stuff, L):
print(subset)
或者-如果您想变得眼花or乱（或者想让以后阅读代码的人都屈服）-您可以生成“ combinations（）”生成器链，然后进行迭代：
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)
This answer missed one aspect: the OP asked for ALL combinations… not just combinations of length “r”.
So you’d either have to loop through all lengths “L”:
import itertools
stuff = [1, 2, 3]
for L in range(0, len(stuff)+1):
for subset in itertools.combinations(stuff, L):
print(subset)
Or — if you want to get snazzy (or bend the brain of whoever reads your code after you) — you can generate the chain of “combinations()” generators, and iterate through that:
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)
回答 2
这是一个懒惰的单行代码，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)) )
答案背后的主要思想是：有2 ^ N个组合-与长度为N的二进制字符串的数目相同。对于每个二进制字符串，请选择与“ 1”相对应的所有元素。
items=abc * mask=###
|
V
000 ->
001 -> c
010 -> b
011 -> bc
100 -> a
101 -> a c
110 -> ab
111 -> abc
注意事项：
- 这就需要你可以调用
len(...)的items（解决方法：如果items是这样的迭代就像一台生成器，与第一把它变成一个列表items=list(_itemsArg)）
- 这要求迭代的顺序
items不是随机的（解决方法：不要疯了）
- 这就要求项目是独一无二的，要不然
{2,2,1}并{2,1,1}都将崩溃{2,1}（解决方法：使用collections.Counter作为一个下拉更换set;它基本上是一个多集…尽管你可能需要以后使用tuple(sorted(Counter(...).elements()))，如果你需要它是可哈希）
演示版
>>> 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'}]
Here’s a lazy one-liner, also using 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)) )
Main idea behind this answer: there are 2^N combinations — same as the number of binary strings of length N. For each binary string, you pick all elements corresponding to a “1”.
items=abc * mask=###
|
V
000 ->
001 -> c
010 -> b
011 -> bc
100 -> a
101 -> a c
110 -> ab
111 -> abc
Things to consider:
- This requires that you can call
len(...) on items (workaround: if items is something like an iterable like a generator, turn it into a list first with items=list(_itemsArg))
- This requires that the order of iteration on
items is not random (workaround: don’t be insane)
- This requires that the items are unique, or else
{2,2,1} and {2,1,1} will both collapse to {2,1} (workaround: use collections.Counter as a drop-in replacement for set; it’s basically a multiset… though you may need to later use tuple(sorted(Counter(...).elements())) if you need it to be hashable)
Demo
>>> 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'}]
回答 3
在@Dan H 极力支持的答案的评论powerset()中，itertools文档中提到了该配方—包括Dan本人的配方。但是，到目前为止，还没有人将其发布为答案。由于它可能是解决问题的最佳方法，即使不是最好的方法之一，并且在另一位评论者的鼓励下，它显示如下。该函数产生的所有列表中的元件的独特组合每长度可能的（包括那些含有零和所有的元素）。
注意：如果，微妙的不同，目标是获得唯一的独特元素的组合，改线s = list(iterable)，以s = list(set(iterable))消除任何重复的元素。无论如何，iterable最终会变成的事实list意味着它将与生成器一起使用（与其他几个答案不同）。
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) # allows duplicate elements
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
stuff = [1, 2, 3]
for i, combo in enumerate(powerset(stuff), 1):
print('combo #{}: {}'.format(i, combo))
输出：
combo #1: ()
combo #2: (1,)
combo #3: (2,)
combo #4: (3,)
combo #5: (1, 2)
combo #6: (1, 3)
combo #7: (2, 3)
combo #8: (1, 2, 3)
In comments under the highly upvoted answer by @Dan H, mention is made of the powerset() recipe in the itertools documentation—including one by Dan himself. However, so far no one has posted it as an answer. Since it’s probably one of the better if not the best approach to the problem—and given a little encouragement from another commenter, it’s shown below. The function produces all unique combinations of the list elements of every length possible (including those containing zero and all the elements).
Note: If the, subtly different, goal is to obtain only combinations of unique elements, change the line s = list(iterable) to s = list(set(iterable)) to eliminate any duplicate elements. Regardless, the fact that the iterable is ultimately turned into a list means it will work with generators (unlike several of the other answers).
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) # allows duplicate elements
return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
stuff = [1, 2, 3]
for i, combo in enumerate(powerset(stuff), 1):
print('combo #{}: {}'.format(i, combo))
Output:
combo #1: ()
combo #2: (1,)
combo #3: (2,)
combo #4: (3,)
combo #5: (1, 2)
combo #6: (1, 3)
combo #7: (2, 3)
combo #8: (1, 2, 3)
回答 4
这是使用递归的一种：
>>> 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']
Here is one using recursion:
>>> 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']
回答 5
这种单行代码为您提供了所有组合（如果原始列表/集合包含不同的元素，则在0和n项之间n），并使用本机方法itertools.combinations：
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']]
在线尝试：
http://ideone.com/COghfX
This one-liner gives you all the combinations (between 0 and n items if the original list/set contains n distinct elements) and uses the native method itertools.combinations:
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)], [])
The output will be:
[[],
['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']]
Try it online:
http://ideone.com/COghfX
回答 6
我同意Dan H的观点，Ben确实要求所有组合。itertools.combinations()没有给出所有组合。
另一个问题是，如果可迭代的输入很大，则最好返回一个生成器而不是列表中的所有内容：
iterable = range(10)
for s in xrange(len(iterable)+1):
for comb in itertools.combinations(iterable, s):
yield comb
I agree with Dan H that Ben indeed asked for all combinations. itertools.combinations() does not give all combinations.
Another issue is, if the input iterable is big, it is perhaps better to return a generator instead of everything in a list:
iterable = range(10)
for s in xrange(len(iterable)+1):
for comb in itertools.combinations(iterable, s):
yield comb
回答 7
这是一种可以轻松转移到支持递归的所有编程语言的方法（没有itertools，没有yield，没有列表理解）：
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]]
This is an approach that can be easily transfered to all programming languages supporting recursion (no itertools, no yield, no list comprehension):
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]]
回答 8
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))
结果将是：
[()]
[(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)]
You can generating all combinations of a list in python using this simple code
import itertools
a = [1,2,3,4]
for i in xrange(0,len(a)+1):
print list(itertools.combinations(a,i))
Result would be :
[()]
[(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)]
回答 9
我认为我可以为那些寻求答案的人添加此功能，而无需导入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
简单良率生成器用法：
for i in powerSet([1,2,3,4]):
print (i, ", ", end="")
上面用法示例的输出：
[]，[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]，
I thought I would add this function for those seeking an answer without importing itertools or any other extra libraries.
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
Simple Yield Generator Usage:
for i in powerSet([1,2,3,4]):
print (i, ", ", end="")
Output from Usage example above:
[] , [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] ,
回答 10
这是涉及使用itertools.combinations函数的另一种解决方案（单行），但是这里我们使用了双重列表理解（与for循环或求和相对）：
def combs(x):
return [c for i in range(len(x)+1) for c in combinations(x,i)]
演示：
>>> combs([1,2,3,4])
[(),
(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)]
Here is yet another solution (one-liner), involving using the itertools.combinations function, but here we use a double list comprehension (as opposed to a for loop or sum):
def combs(x):
return [c for i in range(len(x)+1) for c in combinations(x,i)]
Demo:
>>> combs([1,2,3,4])
[(),
(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)]
回答 11
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))
输出
[
['A'],
['B'],
['C'],
['A', 'B'],
['A', 'C'],
['B', 'C'],
['A', 'B', 'C']
]
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))
output
[
['A'],
['B'],
['C'],
['A', 'B'],
['A', 'C'],
['B', 'C'],
['A', 'B', 'C']
]
回答 12
下面是一个“标准递归答案”，类似于其他类似的答案https://stackoverflow.com/a/23743696/711085。（我们实际上不必担心会耗尽堆栈空间，因为我们不可能处理所有N！个置换。）
它依次访问每个元素，然后接受或离开它（我们可以从该算法直接看到2 ^ N基数）。
def combs(xs, i=0):
if i==len(xs):
yield ()
return
for c in combs(xs,i+1):
yield c
yield c+(xs[i],)
演示：
>>> list( combs(range(5)) )
[(), (0,), (1,), (1, 0), (2,), (2, 0), (2, 1), (2, 1, 0), (3,), (3, 0), (3, 1), (3, 1, 0), (3, 2), (3, 2, 0), (3, 2, 1), (3, 2, 1, 0), (4,), (4, 0), (4, 1), (4, 1, 0), (4, 2), (4, 2, 0), (4, 2, 1), (4, 2, 1, 0), (4, 3), (4, 3, 0), (4, 3, 1), (4, 3, 1, 0), (4, 3, 2), (4, 3, 2, 0), (4, 3, 2, 1), (4, 3, 2, 1, 0)]
>>> list(sorted( combs(range(5)), key=len))
[(),
(0,), (1,), (2,), (3,), (4,),
(1, 0), (2, 0), (2, 1), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (4, 3),
(2, 1, 0), (3, 1, 0), (3, 2, 0), (3, 2, 1), (4, 1, 0), (4, 2, 0), (4, 2, 1), (4, 3, 0), (4, 3, 1), (4, 3, 2),
(3, 2, 1, 0), (4, 2, 1, 0), (4, 3, 1, 0), (4, 3, 2, 0), (4, 3, 2, 1),
(4, 3, 2, 1, 0)]
>>> len(set(combs(range(5))))
32
Below is a “standard recursive answer”, similar to the other similar answer https://stackoverflow.com/a/23743696/711085 . (We don’t realistically have to worry about running out of stack space since there’s no way we could process all N! permutations.)
It visits every element in turn, and either takes it or leaves it (we can directly see the 2^N cardinality from this algorithm).
def combs(xs, i=0):
if i==len(xs):
yield ()
return
for c in combs(xs,i+1):
yield c
yield c+(xs[i],)
Demo:
>>> list( combs(range(5)) )
[(), (0,), (1,), (1, 0), (2,), (2, 0), (2, 1), (2, 1, 0), (3,), (3, 0), (3, 1), (3, 1, 0), (3, 2), (3, 2, 0), (3, 2, 1), (3, 2, 1, 0), (4,), (4, 0), (4, 1), (4, 1, 0), (4, 2), (4, 2, 0), (4, 2, 1), (4, 2, 1, 0), (4, 3), (4, 3, 0), (4, 3, 1), (4, 3, 1, 0), (4, 3, 2), (4, 3, 2, 0), (4, 3, 2, 1), (4, 3, 2, 1, 0)]
>>> list(sorted( combs(range(5)), key=len))
[(),
(0,), (1,), (2,), (3,), (4,),
(1, 0), (2, 0), (2, 1), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (4, 3),
(2, 1, 0), (3, 1, 0), (3, 2, 0), (3, 2, 1), (4, 1, 0), (4, 2, 0), (4, 2, 1), (4, 3, 0), (4, 3, 1), (4, 3, 2),
(3, 2, 1, 0), (4, 2, 1, 0), (4, 3, 1, 0), (4, 3, 2, 0), (4, 3, 2, 1),
(4, 3, 2, 1, 0)]
>>> len(set(combs(range(5))))
32
回答 13
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 )
输出为：
['A', 'A']
['A', 'B']
['A', 'C']
['B', 'B']
['B', 'C']
['C', 'C']
Using list comprehension:
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 )
Output would be:
['A', 'A']
['A', 'B']
['A', 'C']
['B', 'B']
['B', 'C']
['C', 'C']
回答 14
该代码采用带有嵌套列表的简单算法…
# 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()
This code employs a simple algorithm with nested lists…
# 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()
回答 15
我知道这是更为实际使用itertools得到所有的组合，但你可以想代码实现这个部分，只有列表中理解，如果你这么碰巧的愿望，给予了很多
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，就这么简单：
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.combinations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)]
I know it’s far more practical to use itertools to get the all the combinations, but you can achieve this partly with only list comprehension if you so happen to desire, granted you want to code a lot
For combinations of two pairs:
lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]
And, for combinations of three pairs, it’s as easy as this:
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:]]
The result is identical to using itertools.combinations:
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)]
回答 16
不使用itertools：
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']))
Without using itertools:
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']))
回答 17
这是两个实现 itertools.combinations
返回列表的一个
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
一个还生成器
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
请注意，建议为这些函数提供帮助函数，因为prepend参数是静态的，并且不会在每次调用时更改
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]]
这是一个非常肤浅的案例，但总比后悔要安全
Here are two implementations of itertools.combinations
One that returns a list
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
One returns a generator
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
Please note that providing a helper function to those is advised because the prepend argument is static and is not changing with every call
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]]
This is a very superficial case but better be safe than sorry
回答 18
怎么样..使用字符串而不是列表，但是同样的事情..字符串可以像Python中的列表一样对待：
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)
How about this.. used a string instead of list, but same thing.. string can be treated like a list in Python:
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)
回答 19
来自itertools的组合
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))
谢谢
Combination from itertools
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))
Thanks
回答 20
没有 itertools Python 3，您可以执行以下操作：
def combinations(arr, carry):
for i in range(len(arr)):
yield carry + arr[i]
yield from combinations(arr[i + 1:], carry + arr[i])
最初在哪里 carry = "".
Without itertools in Python 3 you could do something like this:
def combinations(arr, carry):
for i in range(len(arr)):
yield carry + arr[i]
yield from combinations(arr[i + 1:], carry + arr[i])
where initially carry = "".
回答 21
3个功能：
- n个元素的所有组合列表
- n个元素的所有组合列出了顺序不同的地方
- 所有排列
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)
3 functions:
- all combinations of n elements list
- all combinations of n elements list where order is not distinct
- all permutations
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)
回答 22
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
This is my implementation
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
回答 23
您也可以使用优质包装中的Powerset功能more_itertools。
from more_itertools import powerset
l = [1,2,3]
list(powerset(l))
# [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
我们还可以验证它是否符合OP的要求
from more_itertools import ilen
assert ilen(powerset(range(15))) == 32_768
You can also use the powerset function from the excellent more_itertools package.
from more_itertools import powerset
l = [1,2,3]
list(powerset(l))
# [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
We can also verify, that it meets OP’s requirement
from more_itertools import ilen
assert ilen(powerset(range(15))) == 32_768
回答 24
def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = range(r)
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
x = [2, 3, 4, 5, 1, 6, 4, 7, 8, 3, 9]
for i in combinations(x, 2):
print i
def combinations(iterable, r):
# combinations('ABCD', 2) --> AB AC AD BC BD CD
# combinations(range(4), 3) --> 012 013 023 123
pool = tuple(iterable)
n = len(pool)
if r > n:
return
indices = range(r)
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != i + n - r:
break
else:
return
indices[i] += 1
for j in range(i+1, r):
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
x = [2, 3, 4, 5, 1, 6, 4, 7, 8, 3, 9]
for i in combinations(x, 2):
print i
回答 25
如果有人正在寻找反向列表，就像我曾经那样：
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)
If someone is looking for a reversed list, like I was:
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)
回答 26
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')
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')
