在一个键上连接两个字典列表 [英] join two lists of dictionaries on a single key

问题描述

给定以 m 字典作为元素的 n 列表，我想生成一个新列表，其中包含一组连接的字典.每个字典都保证有一个称为索引"的键，但可以有一组任意的键.非索引键永远不会跨列表重叠.例如，想象以下两个列表:

Given n lists with m dictionaries as their elements, I would like to produce a new list, with a joined set of dictionaries. Each dictionary is guaranteed to have a key called "index", but could have an arbitrary set of keys beyond that. The non-index keys will never overlap across lists. For example, imagine the following two lists:

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "green":"eggs"}]
l2 = [{"index":1, "c":4}, {"index":2, "c":5}]

("b" 永远不会出现在 l2 中，因为它出现在 l1 中，类似地，"c" 永远不会出现在 l1 中，因为它出现在 l2)

("b" would never appear in l2, since it appeared in l1, and similarly, "c" would never appear in l1, since it appeared in l2)

我想生成一个连接列表:

I would like to produce a joined list:

l3 = [{"index":1, "b":2, "c":4}, 
      {"index":2, "b":3, "c":5}, 
      {"index":3, "green":"eggs"}]

在 Python 中执行此操作的最有效方法是什么?

What is the most efficient way to do this in Python?

推荐答案

from collections import defaultdict

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "green":"eggs"}]
l2 = [{"index":1, "c":4}, {"index":2, "c":5}]

d = defaultdict(dict)
for l in (l1, l2):
    for elem in l:
        d[elem['index']].update(elem)
l3 = d.values()

# l3 is now:

[{'b': 2, 'c': 4, 'index': 1},
 {'b': 3, 'c': 5, 'index': 2},
 {'green': 'eggs', 'index': 3}]

EDIT:由于 l3 不能保证被排序(.values() 返回没有特定顺序的项目)，你可以这样做正如@user560833 建议的那样:

EDIT: Since l3 is not guaranteed to be sorted (.values() returns items in no specific order), you can do as @user560833 suggests:

from operator import itemgetter

...

l3 = sorted(d.values(), key=itemgetter("index"))

这篇关于在一个键上连接两个字典列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！