Python：合并两个字典列表
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Python: Merge two lists of dictionaries
提问by xiaohan2012
Given two lists of dictionaries:
给定两个字典列表：
>>> lst1 = [{id: 1, x: "one"},{id: 2, x: "two"}]
>>> lst2 = [{id: 2, x: "two"}, {id: 3, x: "three"}]
>>> merge_lists_of_dicts(lst1, lst2) #merge two lists of dictionary items by the "id" key
[{id: 1, x: "one"}, {id: 2, x: "two"}, {id: 3, x: "three"}]
Any way to implement merge_lists_of_dictswhat merges two lists of dictionary based on the dictionary items' keys?
有merge_lists_of_dicts什么方法可以根据字典项的键来实现合并两个字典列表的内容吗？
回答by roippi
One possible way to define it:
一种可能的定义方式：
lst1 + [x for x in lst2 if x not in lst1]
Out[24]: [{'id': 1, 'x': 'one'}, {'id': 2, 'x': 'two'}, {'id': 3, 'x': 'three'}]
Note that this will keep both{'id': 2, 'x': 'three'}and {'id': 2, 'x': 'two'}as you did not define what should happen in that case.
请注意，这将保留两者{'id': 2, 'x': 'three'}，{'id': 2, 'x': 'two'}因为您没有定义在这种情况下应该发生什么。
Also note that the seemingly-equivalent and more appealing
另请注意，看似等效且更具吸引力的
set(lst1 + lst2)
will NOT work since dicts are not hashable.
将不起作用，因为dicts 不可散列。
回答by thefourtheye
lst1 = [{"id": 1, "x": "one"}, {"id": 2, "x": "two"}]
lst2 = [{"id": 2, "x": "two"}, {"id": 3, "x": "three"}]
result = []
lst1.extend(lst2)
for myDict in lst1:
if myDict not in result:
result.append(myDict)
print result
Output
输出
[{'x': 'one', 'id': 1}, {'x': 'two', 'id': 2}, {'x': 'three', 'id': 3}]
回答by georg
Perhaps the simplest option
也许是最简单的选择
result = {x['id']:x for x in lst1 + lst2}.values()
This keeps only unique idsin the list, not preserving the order though.
这仅ids在列表中保持唯一，但不保留顺序。
If the lists are really big, a more realistic solution would be to sort them by idand merge iteratively.
如果列表真的很大，更现实的解决方案是对它们进行排序id并迭代合并。
