配对字典中元素的组合而无需重复 [英] Pair combinations of elements in dictionary without repetition

问题描述

在python中，我有一个这样的字典...

In python, I have a dictionary like this...

pleio = {'firstLine': {'enf1': ['54', 'set'], 
                      'enf2': ['48', 'free'], 
                      'enf3': ['34', 'set'], 
                      'enf4': ['12', 'free']}

        'secondLine':{'enf5': ['56','bgb']
                      'enf6': ['67','kiol']
                      'enf7': ['11','dewd']
                      'enf8': ['464','cona']}}

我想在不重复内部字典中元素的情况下进行配对组合，以得到这样的结果。 ..

I would like to make paired combinations with no repetition of the elements in the inner dictionary, to end up with a result like this...

{'enf3': ['34', 'set'], 'enf2': ['48', 'free']}
{'enf3': ['34', 'set'], 'enf1': ['54', 'set']}
{'enf3': ['34', 'set'], 'enf4': ['12', 'free']}
{'enf2': ['48', 'free'], 'enf1': ['54', 'set']}
{'enf2': ['48', 'free'], 'enf4': ['12', 'free']}
{'enf1': ['54', 'set'], 'enf4': ['12', 'free']}

我构建了一个函数，可以让我做...

I built a function which lets me do it...

import itertools

def pairwise():
    '''
    '''
    leti=[]
    for snp, enfs in pleio.items():        
        for x in itertools.combinations(enfs, 2 ):
            leti.append(x)    
    pleopairs=[]
    for i in leti:
        pipi={}
        for c in i:
            pipi[c]= enfs[c]
        pleopairs.append(pipi)

..但是我想知道是否有更有效的方法，例如itertools的另一个特定功能或任何其他来源。顺便说一下，我在 itertools 文档中找到了一个称为成对的函数。但是我不知道如何适应它，或者我是否会尝试改进。有帮助吗？

..but i was wondering if there's a more efficient way, like another specific function from itertools, or any other source. By the way, I found a function called "pairwise" in the itertools documentation. But I don't know how to adapt it, if would be possible in my case, or improve my attempt. Any help?

推荐答案

您的组合方法是正确的，您只需要再次将每个组合的结果转换为字典：

Your combinations approach was correct, you just need to turn the results of each combination into a dict again:

import itertools

def pairwise(input):
    for values in input.itervalues():
        for pair in itertools.combinations(values.iteritems(), 2):
            yield dict(pair)

此版本是一个生成器，可以高效地生成对，没有什么比绝对必要的存储在内存中了。如果需要列表，只需在生成器上调用 list（）：

This version is a generator, yielding pairs efficiently, nothing is held in memory any longer than absolutely necessary. If you need a list, just call list() on the generator:

list(pairwise(pleio))

输出：

>>> from pprint import pprint
>>> pprint(list(pairwise(pleio)))
[{'enf2': ['48', 'free'], 'enf3': ['34', 'set']},
 {'enf1': ['54', 'set'], 'enf3': ['34', 'set']},
 {'enf3': ['34', 'set'], 'enf4': ['12', 'free']},
 {'enf1': ['54', 'set'], 'enf2': ['48', 'free']},
 {'enf2': ['48', 'free'], 'enf4': ['12', 'free']},
 {'enf1': ['54', 'set'], 'enf4': ['12', 'free']}]

您甚至可以合并整个将东西放到单层生成器中：

You can even combine the whole thing into a one-liner generator:

from itertools import combinations

for paired in (dict(p) for v in pleio.itervalues() for p in combinations(v.iteritems(), 2)):
    print paired

哪些输出：

>>> for paired in (dict(p) for v in pleio.itervalues() for p in combinations(v.iteritems(), 2)):
...     print paired
... 
{'enf3': ['34', 'set'], 'enf2': ['48', 'free']}
{'enf3': ['34', 'set'], 'enf1': ['54', 'set']}
{'enf3': ['34', 'set'], 'enf4': ['12', 'free']}
{'enf2': ['48', 'free'], 'enf1': ['54', 'set']}
{'enf2': ['48', 'free'], 'enf4': ['12', 'free']}
{'enf1': ['54', 'set'], 'enf4': ['12', 'free']}

如果您使用的是Python 3，请替换 .itervalues（）并 .iteritems（）由 .values（）和 .items（）。

If you are on Python 3, replace .itervalues() and .iteritems() by .values() and .items() respectively.

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