清单清单的所有组合

问题:清单清单的所有组合

我基本上是在寻找组合的 python版本List<List<int>>

给定一个列表列表,我需要一个新列表,该列表给出列表之间所有可能的项目组合。

[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]

列表的数量是未知的,因此我需要适用于所有情况的东西。优雅奖励积分!

I’m basically looking for a python version of Combination of List<List<int>>

Given a list of lists, I need a new list that gives all the possible combinations of items between the lists.

[[1,2,3],[4,5,6],[7,8,9,10]] -> [[1,4,7],[1,4,8],...,[3,6,10]]

The number of lists is unknown, so I need something that works for all cases. Bonus points for elegance!


回答 0

您需要itertools.product

>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

you need itertools.product:

>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

回答 1

最优雅的解决方案是在python 2.6中使用itertools.product

如果您不使用Python 2.6,itertools.product的文档实际上会显示等效的功能,以“手动”方式完成产品:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

The most elegant solution is to use itertools.product in python 2.6.

If you aren’t using Python 2.6, the docs for itertools.product actually show an equivalent function to do the product the “manual” way:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

回答 2

listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for list in itertools.product(*listOLists):
  print list;

希望您能像我初次接触时一样优雅。

listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for list in itertools.product(*listOLists):
  print list;

I hope you find that as elegant as I did when I first encountered it.


回答 3

Numpy可以做到:

 >>> import numpy
 >>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
 >>> [list(x) for x in numpy.array(numpy.meshgrid(*a)).T.reshape(-1,len(a))]
[[ 1, 4, 7], [1, 5, 7], [1, 6, 7], ....]

Numpy can do it:

 >>> import numpy
 >>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
 >>> [list(x) for x in numpy.array(numpy.meshgrid(*a)).T.reshape(-1,len(a))]
[[ 1, 4, 7], [1, 5, 7], [1, 6, 7], ....]

回答 4

直接执行此任务的递归没有什么问题,如果您需要使用字符串的版本,则可以满足您的需求:

combinations = []

def combine(terms, accum):
    last = (len(terms) == 1)
    n = len(terms[0])
    for i in range(n):
        item = accum + terms[0][i]
        if last:
            combinations.append(item)
        else:
            combine(terms[1:], item)


>>> a = [['ab','cd','ef'],['12','34','56']]
>>> combine(a, '')
>>> print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']

Nothing wrong with straight up recursion for this task, and if you need a version that works with strings, this might fit your needs:

combinations = []

def combine(terms, accum):
    last = (len(terms) == 1)
    n = len(terms[0])
    for i in range(n):
        item = accum + terms[0][i]
        if last:
            combinations.append(item)
        else:
            combine(terms[1:], item)


>>> a = [['ab','cd','ef'],['12','34','56']]
>>> combine(a, '')
>>> print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']

回答 5

一个人可以为此使用基本的python。该代码需要一个函数来拉平列表的列表:

def flatten(B):    # function needed for code below;
    A = []
    for i in B:
        if type(i) == list: A.extend(i)
        else: A.append(i)
    return A

然后可以运行:

L = [[1,2,3],[4,5,6],[7,8,9,10]]

outlist =[]; templist =[[]]
for sublist in L:
    outlist = templist; templist = [[]]
    for sitem in sublist:
        for oitem in outlist:
            newitem = [oitem]
            if newitem == [[]]: newitem = [sitem]
            else: newitem = [newitem[0], sitem]
            templist.append(flatten(newitem))

outlist = list(filter(lambda x: len(x)==len(L), templist))  # remove some partial lists that also creep in;
print(outlist)

输出:

[[1, 4, 7], [2, 4, 7], [3, 4, 7], 
[1, 5, 7], [2, 5, 7], [3, 5, 7], 
[1, 6, 7], [2, 6, 7], [3, 6, 7], 
[1, 4, 8], [2, 4, 8], [3, 4, 8], 
[1, 5, 8], [2, 5, 8], [3, 5, 8], 
[1, 6, 8], [2, 6, 8], [3, 6, 8], 
[1, 4, 9], [2, 4, 9], [3, 4, 9], 
[1, 5, 9], [2, 5, 9], [3, 5, 9], 
[1, 6, 9], [2, 6, 9], [3, 6, 9], 
[1, 4, 10], [2, 4, 10], [3, 4, 10], 
[1, 5, 10], [2, 5, 10], [3, 5, 10], 
[1, 6, 10], [2, 6, 10], [3, 6, 10]]

One can use base python for this. The code needs a function to flatten lists of lists:

def flatten(B):    # function needed for code below;
    A = []
    for i in B:
        if type(i) == list: A.extend(i)
        else: A.append(i)
    return A

Then one can run:

L = [[1,2,3],[4,5,6],[7,8,9,10]]

outlist =[]; templist =[[]]
for sublist in L:
    outlist = templist; templist = [[]]
    for sitem in sublist:
        for oitem in outlist:
            newitem = [oitem]
            if newitem == [[]]: newitem = [sitem]
            else: newitem = [newitem[0], sitem]
            templist.append(flatten(newitem))

outlist = list(filter(lambda x: len(x)==len(L), templist))  # remove some partial lists that also creep in;
print(outlist)

Output:

[[1, 4, 7], [2, 4, 7], [3, 4, 7], 
[1, 5, 7], [2, 5, 7], [3, 5, 7], 
[1, 6, 7], [2, 6, 7], [3, 6, 7], 
[1, 4, 8], [2, 4, 8], [3, 4, 8], 
[1, 5, 8], [2, 5, 8], [3, 5, 8], 
[1, 6, 8], [2, 6, 8], [3, 6, 8], 
[1, 4, 9], [2, 4, 9], [3, 4, 9], 
[1, 5, 9], [2, 5, 9], [3, 5, 9], 
[1, 6, 9], [2, 6, 9], [3, 6, 9], 
[1, 4, 10], [2, 4, 10], [3, 4, 10], 
[1, 5, 10], [2, 5, 10], [3, 5, 10], 
[1, 6, 10], [2, 6, 10], [3, 6, 10]]

回答 6

from itertools import product 
list_vals = [['Brand Acronym:CBIQ', 'Brand Acronym :KMEFIC'],['Brand Country:DXB','Brand Country:BH']]
list(product(*list_vals))

输出:

[(”品牌缩写:CBIQ’,’品牌国家:DXB’),
(’品牌缩写:CBIQ’,’品牌国家:BH’),
(’品牌缩写:KMEFIC’,’品牌国家:DXB’),
( ‘品牌缩写:KMEFIC’,’品牌国家:BH’)]

from itertools import product 
list_vals = [['Brand Acronym:CBIQ', 'Brand Acronym :KMEFIC'],['Brand Country:DXB','Brand Country:BH']]
list(product(*list_vals))

Output:

[(‘Brand Acronym:CBIQ’, ‘Brand Country :DXB’),
(‘Brand Acronym:CBIQ’, ‘Brand Country:BH’),
(‘Brand Acronym :KMEFIC’, ‘Brand Country :DXB’),
(‘Brand Acronym :KMEFIC’, ‘Brand Country:BH’)]