保持字典结构,同时减少嵌套字典
我有一个嵌套字典对的列表 dd
,并希望将结构维护为字典列表:
dd = [
[{'id': 'bla',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_1A', 'amount': '2'}]},
{'id': 'bla2',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_1B', 'amount': '1'}]}
],
[{'id': 'bla3',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_2A', 'amount': '3'}]},
{'id': 'bla4',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_2B', 'amount': '4'}]}
]
]
我想将其减少为配对字典列表,同时仅提取一些细节。例如,预期的输出可能如下所示:
[{'name': ['KEEP_PAIR_1A', 'KEEP_PAIR_1B'], 'amount': [2, 1]},
{'name': ['KEEP_PAIR_2A', 'KEEP_PAIR_2B'], 'amount': [3, 4]}]
我已经运行了我的代码:
pair=[]
for all_pairs in dd:
for output_pairs in all_pairs:
for d in output_pairs.get('detail'):
if d['name'] != 'discard':
pair.append(d)
output_pair = {
k: [d.get(k) for d in pair]
for k in set().union(*pair)
}
但它没有维护该结构:
{'name': ['KEEP_PAIR_1A', 'KEEP_PAIR_1B', 'KEEP_PAIR_2A', 'KEEP_PAIR_2B'],
'amount': ['2', '1', '3', '4']}
我假设我需要使用一些列表理解来解决这个问题,但是我应该在 for 循环中的哪个位置执行此操作维持结构。
I have a list of pairs of nested dict dd
and would like to maintain the structure to a list of dictionaries:
dd = [
[{'id': 'bla',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_1A', 'amount': '2'}]},
{'id': 'bla2',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_1B', 'amount': '1'}]}
],
[{'id': 'bla3',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_2A', 'amount': '3'}]},
{'id': 'bla4',
'detail': [{'name': 'discard', 'amount': '123'},
{'name': 'KEEP_PAIR_2B', 'amount': '4'}]}
]
]
I want to reduce this to a list of paired dictionaries while extracting only some detail. For example, an expected output may look like this:
[{'name': ['KEEP_PAIR_1A', 'KEEP_PAIR_1B'], 'amount': [2, 1]},
{'name': ['KEEP_PAIR_2A', 'KEEP_PAIR_2B'], 'amount': [3, 4]}]
I have run my code:
pair=[]
for all_pairs in dd:
for output_pairs in all_pairs:
for d in output_pairs.get('detail'):
if d['name'] != 'discard':
pair.append(d)
output_pair = {
k: [d.get(k) for d in pair]
for k in set().union(*pair)
}
But it didn't maintain that structure :
{'name': ['KEEP_PAIR_1A', 'KEEP_PAIR_1B', 'KEEP_PAIR_2A', 'KEEP_PAIR_2B'],
'amount': ['2', '1', '3', '4']}
I assume I would need to use some list comprehension to solve this but where in the for loop should I do that to maintain the structure.
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由于您想要组合列表中的字典,一种选择是使用 dict.setdefault:
输出:
Since you want to combine dictionaries in lists, one option is to use
dict.setdefault
:Output: