在python中生成列表的所有组合
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Generating all combinations of a list in python
提问by Minas Abovyan
Here's the question:
这是问题：
Given a list of items in Python, how would I go by to get all the possible combinations of the items?
给定 Python 中的项目列表，我将如何获得项目的所有可能组合？
There are several similar questions on this site, that suggest using itertools.combine, but that returns only a subset of what I need:
该站点上有几个类似的问题，建议使用 itertools.combine，但这仅返回我需要的一个子集：
stuff = [1, 2, 3]
for L in range(0, len(stuff)+1):
for subset in itertools.combinations(stuff, L):
print(subset)
()
(1,)
(2,)
(3,)
(1, 2)
(1, 3)
(2, 3)
(1, 2, 3)
As you see, it returns only items in a strict order, not returning (2, 1), (3, 2), (3, 1), (2, 1, 3), (3, 1, 2), (2, 3, 1), and (3, 2, 1). Is there some workaround that? I can't seem to come up with anything.
如你所见，它只返回严格顺序的项，不返回 (2, 1), (3, 2), (3, 1), (2, 1, 3), (3, 1, 2), ( 2, 3, 1) 和 (3, 2, 1)。有什么解决方法吗？我似乎想不出什么。
采纳答案by Ashwini Chaudhary
>>> import itertools
>>> stuff = [1, 2, 3]
>>> for L in range(0, len(stuff)+1):
for subset in itertools.permutations(stuff, L):
print(subset)
...
()
(1,)
(2,)
(3,)
(1, 2)
(1, 3)
(2, 1)
(2, 3)
(3, 1)
....
help on itertools.permutations:
帮助itertools.permutations：
permutations(iterable[, r]) --> permutations object
Return successive r-length permutations of elements in the iterable.
permutations(range(3), 2) --> (0,1), (0,2), (1,0), (1,2), (2,0), (2,1)
>>>
回答by Brien
itertools.permutationsis going to be what you want. By mathematical definition, order does not matter for combinations, meaning (1,2)is considered identical to (2,1). Whereas with permutations, each distinct ordering counts as a unique permutation, so (1,2)and (2,1)are completely different.
itertools.permutations将是你想要的。根据数学定义，顺序与 无关combinations，含义(1,2)被认为与 相同(2,1)。而对于permutations，每个不同的排序都算作一个唯一的排列，因此(1,2)和(2,1)完全不同。
回答by Sukrit Kalra
Are you looking for itertools.permutationsinstead?
From help(itertools.permutations),
从help(itertools.permutations),
Help on class permutations in module itertools:
class permutations(__builtin__.object)
| permutations(iterable[, r]) --> permutations object
|
| Return successive r-length permutations of elements in the iterable.
|
| permutations(range(3), 2) --> (0,1), (0,2), (1,0), (1,2), (2,0), (2,1)
Sample Code :
示例代码：
>>> from itertools import permutations
>>> stuff = [1, 2, 3]
>>> for i in range(0, len(stuff)+1):
for subset in permutations(stuff, i):
print(subset)
()
(1,)
(2,)
(3,)
(1, 2)
(1, 3)
(2, 1)
(2, 3)
(3, 1)
(3, 2)
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
From Wikipedia, the difference between permutations and combinations :
从维基百科，排列和组合之间的区别：
Permutation :
排列：
Informally, a permutation of a set of objects is an arrangement of those objects into a particular order. For example, there are six permutations of the set {1,2,3}, namely (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1).
非正式地，一组对象的排列是将这些对象按特定顺序排列。例如集合{1,2,3}有六种排列，即(1,2,3),(1,3,2),(2,1,3),(2,3,1) , (3,1,2) 和 (3,2,1)。
Combination :
组合 ：
In mathematics a combination is a way of selecting several things out of a larger group, where (unlike permutations) order does not matter.
在数学中，组合是从更大的组中选择几个事物的一种方式，其中（与排列不同）顺序无关紧要。
回答by saimadhu.polamuri
You can generate all the combinations of a list in python using this simple code
您可以使用这个简单的代码在 python 中生成列表的所有组合
import itertools
a = [1,2,3,4]
for i in xrange(1,len(a)+1):
print list(itertools.combinations(a,i))
Result:
结果：
[(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by Shashank Singh
I assume you want all possible combinations as 'sets' of values. Here is a piece of code that I wrote that might help give you an idea:
我假设您希望将所有可能的组合作为值的“集合”。这是我写的一段代码，可能有助于给你一个想法：
def getAllCombinations(object_list):
uniq_objs = set(object_list)
combinations = []
for obj in uniq_objs:
for i in range(0,len(combinations)):
combinations.append(combinations[i].union([obj]))
combinations.append(set([obj]))
return combinations
Here is a sample:
这是一个示例：
combinations = getAllCombinations([20,10,30])
combinations.sort(key = lambda s: len(s))
print combinations
... [set([10]), set([20]), set([30]), set([10, 20]), set([10, 30]), set([20, 30]), set([10, 20, 30])]
I think this has n! time complexity, so be careful. This works but may not be most efficient
我认为这有n！时间复杂度，所以要小心。这有效但可能不是最有效的
回答by Uri Goren
Here is a solution without itertools
这是一个没有 itertools 的解决方案
First lets define a translation between an indicator vector of 0and 1s and a sub-list (1if the item is in the sublist)
首先让我们定义0和1s的指示向量和子列表（1如果项目在子列表中）之间的转换
def indicators2sublist(indicators,arr):
return [item for item,indicator in zip(arr,indicators) if int(indicator)==1]
Next, Well define a mapping from a number between 0and 2^n-1to the its binary vector representation (using string's formatfunction) :
接着，那么限定从之间的数的映射0和2^n-1（使用字符串的到其二进制向量表示format函数）：
def bin(n,sz):
return ('{d:0'+str(sz)+'b}').format(d=n)
All we have left to do, is to iterate all the possible numbers, and call indicators2sublist
我们剩下要做的就是迭代所有可能的数字，然后调用 indicators2sublist
def all_sublists(arr):
sz=len(arr)
for n in xrange(0,2**sz):
b=bin(n,sz)
yield indicators2sublist(b,arr)
回答by Matt Slap
just thought i'd put this out there since i couldn't fine EVERY possible outcome and keeping in mind i only have the rawest most basic of knowledge when it comes to python and there's probably a much more elegant solution...(also excuse the poor variable names
只是想我会把它放在那里，因为我无法确定所有可能的结果，请记住，当涉及到 python 时，我只有最原始的最基本的知识，并且可能有一个更优雅的解决方案......（也请原谅可怜的变量名
testing = [1, 2, 3]
测试 = [1, 2, 3]
testing2= [0]
测试2 = [0]
n = -1
n = -1
def testingSomethingElse(number):
def testingSomethingElse（数字）：
try:
testing2[0:len(testing2)] == testing[0]
n = -1
testing2[number] += 1
except IndexError:
testing2.append(testing[0])
while True:
为真：
n += 1
testing2[0] = testing[n]
print(testing2)
if testing2[0] == testing[-1]:
try:
n = -1
testing2[1] += 1
except IndexError:
testing2.append(testing[0])
for i in range(len(testing2)):
if testing2[i] == 4:
testingSomethingElse(i+1)
testing2[i] = testing[0]
i got away with == 4 because i'm working with integers but you may have to modify that accordingly...
我逃脱了 == 4 因为我正在使用整数，但您可能需要相应地修改它...
