我有两个嵌套的字典数据。我想将它们合并以在python中创建一个字典。字典数据:
dict1 = {'employee':{'dev1': 'Roy'}}
dict2 = {'employee':{'dev2': 'Biswas'}}
现在,我尝试从中创建像波纹管一样的字典。必需输出
dict_output = {'employee':{
'dev1': 'Roy',
'dev2': 'Biswas'
}
}
我的尝试:
import json
dict1 = {'employee':{'dev1': 'Roy'}}
dict2 = {'employee':{'dev2': 'Biswas'}}
dict1.update(dict2)
print(json.dumps(dict1, indent=2))
输出:
{
"employee": {
"dev2": "Biswas"
}
}
我无法合并两个字典。需要帮助来合并它们
您可以只更新内部字典。
>>> dict1 = {'employee':{'dev1': 'Roy'}}
>>> dict2 = {'employee':{'dev2': 'Biswas'}}
>>>
>>> for key in dict1:
... if key in dict2:
... dict1[key].update(dict2[key])
...
>>> dict1
{'employee': {'dev2': 'Biswas', 'dev1': 'Roy'}}
即使两个字典具有不同的键,并且您希望保留所有键,这也是一种可行的解决方案。
from collections import defaultdict
dict1 = {'employee': {'dev1': 'Roy'}, 'aKeyNotInDict2': {}}
dict2 = {'employee': {'dev2': 'Biswas'}, 'aKeyNotInDict1': {}}
merged_dict = defaultdict(dict)
merged_dict.update(dict1)
for key, nested_dict in dict2.items():
merged_dict[key].update(nested_dict)
print(dict(merged_dict))
输出:
{
'employee': {'dev2': 'Biswas', 'dev1': 'Roy'},
'aKeyNotInDict2': {},
'aKeyNotInDict1': {}
}
这是我对嵌套字典的尝试。它适用于使用递归的嵌套数据。而且它也有一些限制。例如,如果dict1和dict2具有不属于字典的相同值,则dict2具有优先权。另一方面,如果dict1包含字典,而dict2包含具有相同键的值,则优先级位于dict1上,而dict2被忽略。其他限制将要求更改代码。
def merge_dict(dict1, dict2):
for key, val in dict1.items():
if type(val) == dict:
if key in dict2 and type(dict2[key] == dict):
merge_dict(dict1[key], dict2[key])
else:
if key in dict2:
dict1[key] = dict2[key]
for key, val in dict2.items():
if not key in dict1:
dict1[key] = val
return dict1
dict1 = merge_dict(dict1, dict2)
#use a dict comprehension. Adding {} in get() is to set a default return value if the key doesn't exist in dict1
{k:dict(dict1.get(k,{}).items() + v.items()) for k,v in dict2.items()}
Out[689]: {'employee': {'dev1': 'Roy', 'dev2': 'Biswas'}}
#Alternatively, a less readable way to merge the dicts using the dict constructor.
{k:dict(dict1.get(k,{}), **v) for k,v in dict2.items()}
Out[690]: {'employee': {'dev1': 'Roy', 'dev2': 'Biswas'}}
[key1 == key2 and dict1.get(key1).update(dict2.get(key2))
for key1, key2 in zip(dict1, dict2)]
print dict1