[英]Construct a dictionary from a list of dictionary keys, sub-keys and values
有一个字典列表:
list_dict = [['key1', {'subkey1': 0}],
['key1', {'subkey2': 2}],
['key1', {'subkey5': 5}],
['key2', {'subkey2': 4}],
['key2', {'subkey1': 8}],
['key1', {'sybkey5': 10}]]
如何将list_dict
转换为整洁的字典,如下所示?
{'key1': {'subkey1': 0, 'subkey2': 2, 'subkey5': 5},
'key2': {'subkey2': 4, 'subkey1': 8, 'sybkey5': 10}}
以下应该有效:
d = dict()
for item in list_dict:
d.setdefault(item[0], {}).update(item[1])
来自Python 3 文档:
设置默认(键[,默认])
如果键在字典中,则返回其值。 如果不是,则插入值为默认值的键并返回默认值。 默认默认为无。
你可以使用collections.defaultdict
:
from collections import defaultdict
list_dict = [['key1', {'subkey1': 0}],
['key1', {'subkey2': 2}],
['key1', {'subkey5': 5}],
['key2', {'subkey2': 4}],
['key2', {'subkey1': 8}],
['key1', {'sybkey5': 10}]]
d = defaultdict(dict)
for k, v in list_dict:
d[k].update(v)
d
output:
defaultdict(dict,
{'key1': {'subkey1': 0, 'subkey2': 2, 'subkey5': 5, 'sybkey5': 10},
'key2': {'subkey2': 4, 'subkey1': 8}})
或者您可以使用速度较慢的dict.setdefault
:
d = {}
for k, v in list_dict:
d.setdefault(k, {}).update(v)
output:
{'key1': {'subkey1': 0, 'subkey2': 2, 'subkey5': 5, 'sybkey5': 10},
'key2': {'subkey2': 4, 'subkey1': 8}}
这有效:
>>> from collections import defaultdict
>>> result = defaultdict(dict)
>>> for item in list_dict :
... result[item[0]].update(item[1])
...
>>> result
defaultdict(<type 'dict'>, {'key2': {'subkey2': 4, 'subkey1': 8}, 'key1': {'subkey5': 5, 'subkey2': 2, 'sybkey5': 10, 'subkey1': 0}})
>>>
常规方法
new_dict = {}
for i in list_dict:
if i[0] in new_dict.keys():
new_dict[i[0]].update(i[1])
else:
new_dict[i[0]]=i[1]
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