blocks|key|1572610|text|您需要itertools.product|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1572611|>>>+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)]|code-block|syntax|javascript|1572612|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/itertools.html#itertools.product^0|3|H|3|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

you need <a href="https://docs.python.org/2/library/itertools.html#itertools.product" rel="noreferrer"><code>itertools.product</code></a>:

<pre><code>&gt;&gt;&gt; import itertools
&gt;&gt;&gt; a = [[1,2,3],[4,5,6],[7,8,9,10]]
&gt;&gt;&gt; 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)]
</code></pre>

blocks|key|1572658|text|最优雅的解决方案是在Python2.6中使用itertools.product。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1572659|如果你使用的不是Python2.6，itertools.product的文档实际上展示了一个等效的功能，以“手动”的方式完成产品：|1572660|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%2B[y]+for+x+in+result+for+y+in+pool]
++++for+prod+in+result:
++++++++yield+tuple(prod)|code-block|syntax|javascript|1572661|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/itertools.html#itertools.product^0|M|H|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

The most elegant solution is to use <a href="https://docs.python.org/2/library/itertools.html#itertools.product" rel="noreferrer">itertools.product</a> 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:

<pre><code>def product(*args, **kwds):
 # product('ABCD', 'xy') --&gt; Ax Ay Bx By Cx Cy Dx Dy
 # product(range(2), repeat=3) --&gt; 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)
</code></pre>

blocks|key|1357526|text|listOLists+=+[[1,2,3],[4,5,6],[7,8,9,10]]
for+list+in+itertools.product(*listOLists):
++print+list;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1357527|我希望你能发现它和我第一次遇到它时一样优雅。|unstyled|1357528|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|1357586|text|对于这个任务，直接递归没有错，如果你需要一个可以处理字符串的版本，这可能满足你的需要：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1357587|combinations+=+[]

def+combine(terms,+accum):
++++last+=+(len(terms)+==+1)
++++n+=+len(terms[0])
++++for+i+in+range(n):
++++++++item+=+accum+%2B+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']|code-block|syntax|javascript|1357588|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>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)


&gt;&gt;&gt; a = [['ab','cd','ef'],['12','34','56']]
&gt;&gt;&gt; combine(a, '')
&gt;&gt;&gt; print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']
</code></pre>

blocks|key|1357537|text|Numpy可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1357538|+>>>+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],+....]|code-block|syntax|javascript|1357539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Numpy can do it:

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

blocks|key|1357612|text|可以使用base+python来实现这一点。代码需要一个函数来展平列表列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1357613|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|code-block|syntax|javascript|1357614|然后就可以运行了：|1357615|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)|1357616|输出：|1357617|[[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]]|1357618|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

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

<pre><code>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
</code></pre>

Then one can run: 

<pre><code>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)
</code></pre>

Output: 

<pre><code>[[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]]
</code></pre>

blocks|key|1357627|text|这个答案不像使用itertools那样清晰，但其中的想法可能是有用的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1357628|从zip()+here的构造中得到灵感，我们可以做以下事情。|offset|length|1357629|>>>+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+%2B+[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]]|code-block|syntax|javascript|1357630|我们使用a作为迭代器，以便连续获得它的下一项，而不需要知道有多少先验的。当我们用完next中的列表时，sentinel命令将输出here+(这是一个专门创建用于进行此比较的对象，请参阅here以获得一些解释)，从而导致if语句触发，因此我们跳出了循环。|style|CODE|1357631|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#zip|1|https://python-patterns.guide/python/sentinel-object/|2^0|0|7|4|0|0|0|4|1|15|4|1F|8|31|2|1S|4|1|2K|4|2|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@]|9|@$D|11|E|12|1|13]]|A|$]]|$1|F|3|G|5|H|7|14|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|15|8|@$D|16|E|17|M|N]|$D|18|E|19|M|N]|$D|1A|E|1B|M|N]|$D|1C|E|1D|M|N]]|9|@$D|1E|E|1F|1|1G]|$D|1H|E|1I|1|1J]]|A|$]]|$1|O|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]|Y|$5|R|S|T|A|$U|X]]]]

This answer isn't as clean as using itertools but the ideas could be useful.
Drawing inspiration from the construction of zip() <a href="https://docs.python.org/3/library/functions.html#zip" rel="nofollow noreferrer">here</a>, we could do the following.
<pre><code>&gt;&gt;&gt; a = iter([[1,2,3],[4,5,6],[7,8,9,10]])
&gt;&gt;&gt; sentinel = object()
&gt;&gt;&gt; result = [[]]
&gt;&gt;&gt; while True:
&gt;&gt;&gt; l = next(a,sentinel)
&gt;&gt;&gt; if l == sentinel:
&gt;&gt;&gt; break
&gt;&gt;&gt; result = [ r + [digit] for r in result for digit in l]
&gt;&gt;&gt; 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]]
</code></pre>
We use <code>a</code> as an iterator in order to successively get the next item of it without needing to know how many there are a priori. The <code>next</code> command will output <code>sentinel</code> (which is an object created solely to make this comparison, see <a href="https://python-patterns.guide/python/sentinel-object/" rel="nofollow noreferrer">here</a> for some explanation) when we run out of lists in <code>a</code>, causing the <code>if</code> statement to trigger so we break out of the loop.

I'm basically looking for a python version of <a href="https://stackoverflow.com/questions/545703/combination-of-listlistint">Combination of <code>List&lt;List&lt;int&gt;&gt;</code></a>

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

<pre><code>[[1,2,3],[4,5,6],[7,8,9,10]] -&gt; [[1,4,7],[1,4,8],...,[3,6,10]]
</code></pre>

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

All combinations of a list of lists

我基本上是在寻找的python版本给定一个列表列表，我需要一个新的列表，它给出列表之间所有可能的项组合。[[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，所以我需要一个适用于所有情况的东西。优雅的加分！

问列表列表的所有组合
EN

回答 7

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