在公共密钥中合并具有相同值的多个字典列表 [英] Merge multiple list of dict with same value in common key

问题描述

我有4份字典

list1 = [{'a':0,'b':23}, {'a':3,'b':77},{'a':1,'b':99}]

list2 = [{'a':1,'c':666},{'a':4,'c':546}]

list3 = [{'d':33,'a':3},{'d':1111,'a':4},{'d':76,'a':1},{'d':775,'a':0}]

list4 = [{'a':2,'e':12},{'a':4,'e':76}]

列表中的每个字典都有一个公用键"a".我的要求是应合并所有列表中dict中具有相同值的"a"键，如果合并时dict中不存在特定键，则为这些键分配0或忽略这些键.例如对于具有值1的键"a"，在上面的示例中，我们有2个字典，一个来自list1，即{'a':0，'b':23}，一个是list3，最后一个字典，即{'d':775，a':0}，因此我们首先确定了具有相同'a'值的字典，现在需要合并这些字典即{'a':0，'b':23，'c':0，'d':775，'e':0}，因为两个字典都没有'c'，所以分配了'c'此处为0

Every dict in the list has a common key 'a'. My requirement is 'a' key with same value in dict from all the list should be merged and if a particular key does not exist in the dicts while merging, then assign 0 for those keys or just omit those keys. for eg. for key 'a' with value 1, from above example we have 2 dicts, one from list1 i.e {'a':0,'b':23} and one is list3, last dict i.e {'d':775,'a':0}, so 1st we have identified the dicts with same 'a' value, now need to merge these dicts i.e {'a':0, 'b':23, 'c':0, 'd':775, 'e':0}, since both the dict didn't have 'c', 'c' is assigned as 0 here

我的输出应为:

[{'a':0,'b':23,'c':0,'d':775, 'e':0},{'a':1,'b':99,'c':666,'d':76,'e':0},{'a':2,'b':0,'c':0,'d':0,'e':12},{'a':3,'b':77,'c':0,'d':33,'e':0}, {'a':4,'b':0,'c':546,'d':1111,'e':76}]

使用最小循环或列表理解

usings minimum loops or list comprehension

推荐答案

如果您想使用更Python化的方式:

If you want a more pythonic way:

from itertools import groupby
from pprint import pprint
from collections import ChainMap

a = [{'a':0,'b':23}, {'a':3,'b':77}, {'a':1,'b':99}]
b = [{'a':1,'c':666}, {'a':4,'c':546}]
c = [{'d':33,'a':3}, {'d':1111,'a':4}, {'d':76,'a':1}, {'d':775,'a':0}]
d = [{'a':2,'e':12}, {'a':4,'e':76}]

dict_list = a + b + c + d

# You just need to specify the key you want to use in the lambda function
# There's no need to declare the different key values previously
res = map(lambda dict_tuple: dict(ChainMap(*dict_tuple[1])),
          groupby(sorted(dict_list,
                         key=lambda sub_dict: sub_dict["a"]),
                  key=lambda sub_dict: sub_dict["a"]))

pprint(list(res))

输出:

[{'a': 0, 'b': 23, 'd': 775},
 {'a': 1, 'b': 99, 'c': 666, 'd': 76},
 {'a': 2, 'e': 12},
 {'a': 3, 'b': 77, 'd': 33},
 {'a': 4, 'c': 546, 'd': 1111, 'e': 76}]

编辑(改进):

您也可以使用

from _operator import itemgetter
key=itemgetter("a")

代替

key=lambda sub_dict: sub_dict["a"]

带有itemgetter的版本要快得多.使用您提供的示例:

The version with itemgetter is much faster. Using the example you provided:

- Lambda: 0.037109375ms
- Itemgetter: 0.009033203125ms

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