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Python cartesian-product

列表列表的所有组合

发布于 2024-07-17 21:31:41 字数 332 浏览 6 评论 0 原文

我基本上正在寻找 List> 的 python 版本

给定一个列表列表，我需要一个新列表来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]]

列表的数量未知，所以我需要适用于所有情况的东西。优雅加分！

原文

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!

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小…红帽 2024-07-24 21:31:41

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)]

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撩心不撩汉 2024-07-24 21:31:41

只需使用itertools.product：

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

Simply use itertools.product:

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

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伴梦长久 2024-07-24 21:31:41

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

如果您没有使用 Python 2.6，itertools.product 的文档实际上显示了以“手动”方式执行产品的等效函数：

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)

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如痴如狂 2024-07-24 21:31:41

对于此任务，直接递归没有任何问题，不需要外部依赖项，如果您需要一个使用字符串的版本，这可能适合您的需求：

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, no need for external dependencies, 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']

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沫雨熙 2024-07-24 21:31:41

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], ....]

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暖心男生 2024-07-24 21:31:41

可以使用基础 python 来实现这一点。该代码需要一个函数来展平列表列表：

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]]

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不知在何时 2024-07-24 21:31:41

这主要模仿 Jarret Hardie 的回答使用 itertools.product，但有以下区别：

它将参数内联传递给 itertools.product，而不是通过变量 a - 因此，
如果您的 mypy type-linter 的作用与我的类似，您可以使用内联 *args 语法让您的代码以其他方式“工作” >product 参数（如 product(*[[1,2,3],[4,5,6],[7,8,9,10]])），< code>mypy 可能仍然会失败（类似于 error: No overridevariant of "product" matches argument type "List[object]"）
所以解决方案 mypy< /code>，就是不使用 *args 语法，如下所示：

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

This mostly mimics solutions like Answer by Jarret Hardie using itertools.product, but has these distinctions:

this passes parameters to itertools.product in-line, instead of via variable a - so no *args syntax needed on the inline parameters
if your mypy type-linter acts like mine, and you can get your code to otherwise "work" with the *args syntax with inline product parameters (like product(*[[1,2,3],[4,5,6],[7,8,9,10]])), mypy might still fail it (with something like error: No overload variant of "product" matches argument type "List[object]")
So solution to that mypy, is to not use *args syntax, like this:

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

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绿光 2024-07-24 21:31:41

这个答案并不像使用 itertools 那么干净，但这些想法可能有用。

从 zip() 此处的构造中汲取灵感，我们可以这样做下列。

>>> a = iter([[1,2,3],[4,5,6],[7,8,9,10]])
>>> sentinel = object()
>>> result = [[]]
>>> while True:
>>>     l = next(a,sentinel)
>>>     if l == sentinel:
>>>         break
>>>     result = [ r + [digit] for r in result for digit in l]
>>> print(result)
[[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]]

我们使用a作为迭代器，以便连续获取它的下一项，而不需要知道先验有多少。 next 命令将输出 sentinel （这是一个专门为了进行比较而创建的对象，请参阅此处进行一些解释）当我们用完 a 中的列表时，导致 if 语句触发所以我们跳出了循环。

This answer isn't as clean as using itertools but the ideas could be useful.

Drawing inspiration from the construction of zip() here, we could do the following.

>>> a = iter([[1,2,3],[4,5,6],[7,8,9,10]])
>>> sentinel = object()
>>> result = [[]]
>>> while True:
>>>     l = next(a,sentinel)
>>>     if l == sentinel:
>>>         break
>>>     result = [ r + [digit] for r in result for digit in l]
>>> print(result)
[[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]]

We use a as an iterator in order to successively get the next item of it without needing to know how many there are a priori. The next command will output sentinel (which is an object created solely to make this comparison, see here for some explanation) when we run out of lists in a, causing the if statement to trigger so we break out of the loop.

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能否归途做我良人 2024-07-24 21:31:41

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/地区：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')]