python reduce找到集合的并集 [英] python reduce to find the union of sets

问题描述

我正在尝试找到集合的并集.具体来说，我希望对networkx图的字典中称为periodic_gs的每个键的节点列表进行并集.我想使用reduce函数，因为采用所有periodic_gs[x].nodes()的并集似乎是合理的，其中x是periodic_gs的键.

I am trying to find the union of set of sets. Specifically I want the union of the list of nodes for each key in the dictionary of networkx graphs called periodic_gs. I would like to use the reduce function as it seems reasonable to take the union of all periodic_gs[x].nodes() where x is a key of periodic_gs.

这是我的尝试:

reduce(lambda x,y: set(periodic_gs[x].nodes()).union(set(periodic_gs[y].nodes())), periodic_gs.keys(), {})

对我来说，这就是说将字典中每个图的节点并集.出于某种原因，python告诉我:TypeError: unhashable type: 'dict'我看不到此TypeError，因为periodic_gs.keys()是键的列表(它们是字符串，但我看不出这有什么用)，以及在替换时lambda函数的参数将起作用.

To me, this says take the union of the nodes across each graph in the dictionary. For some reason, python tells me: TypeError: unhashable type: 'dict' I do not see this TypeError, because periodic_gs.keys() is a list of the keys (they are strings but I do not see how this would matter), and when substituted in for the arguments to the lambda function will work.

什么是导致类型错误的原因，我该如何解决?

What is causing the type error and how do I fix it?

推荐答案

您可以使用

You can use set.union like this:

>>> lis = [{1, 2, 3, 4}, {3, 4, 5}, {7, 3, 6}]
>>> set().union(*lis)
set([1, 2, 3, 4, 5, 6, 7])

可以使用 reduce 进行此操作，但是不要:

>>> reduce(set.union, lis)
set([1, 2, 3, 4, 5, 6, 7])

因为此reduce由于需要构建和丢弃所有中间集，因此需要花费二次时间:

because this reduce takes quadratic time due to all the intermediate sets it builds and discards:

In [1]: from functools import reduce

In [2]: sets = [{x} for x in range(1000)]

In [3]: %timeit set().union(*sets)
40.9 µs ± 1.43 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [4]: %timeit reduce(set.union, sets)
4.09 ms ± 587 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

此测试用例的速度降低了100倍，而且很容易变得更糟.

That's a 100x slowdown on this test case, and it can easily be even worse.

对于您的代码，这应该做到:

For your code, this should do it:

set().union(*(x.nodes() for x in periodic_gs.values()))

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