xia*_*012 17 python merge dictionary list
给出两个词典列表:
>>> 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"}]
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有没有merge_lists_of_dicts什么方法可以实现基于字典项的键合并两个字典列表?
the*_*eye 14
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
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输出
[{'x': 'one', 'id': 1}, {'x': 'two', 'id': 2}, {'x': 'three', 'id': 3}]
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geo*_*org 11
也许是最简单的选择
result = {x['id']:x for x in lst1 + lst2}.values()
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这ids在列表中仅保留唯一,但不保留订单.
如果列表真的很大,那么更现实的解决方案就是对它们进行排序id并迭代合并.
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'}]
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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.
还要注意看似等效且更具吸引力的
set(lst1 + lst2)
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不会工作,因为dicts不可清洗.