如何比较不同键的两个词典,例如下面的词典?
dictionary={"name":"abc","age":23,"male":True}
new_dictionary={"my_name":"abc","my_age":23,"male":1}
比较示例中的两个词典时,比较应返回true
。
答案 0 :(得分:2)
>>> dictionary={"name":"abc","age":23,"male":True}
>>> new_dictionary={"my_name":"abc","my_age":23,"male":1}
>>> key_map = {"name": "my_name", "age": "my_age"}
>>> all(new_dictionary[key_map.get(k, k)] == v for k, v in dictionary.items())
True
或者,如果您只是想确保在没有任何密钥检查的情况下值相同的话:
>>> set(dictionary.values()) == set(new_dictionary.values())
True
编辑:正如Tadeck在评论中指出的那样,sorted()
比set()
更安全。
(是的,即使一个词典有1
而另一个词典有True
)
答案 1 :(得分:2)
您需要定义比较发生的确切方式。如果您只想比较这些值,无论它们被分配给哪些键,您都可以使用:
根据您的更新要求,其中一个解决方案是:
>>> def compare(dict1, dict2):
return sorted(dict1.values()) == sorted(dict2.values())
>>> compare({"name":"abc","age":23,"male":True},
{"my_name":"abc","my_age":23,"male":1})
True
>>> compare({"name":"abc","age":23,"male":True},
{"my_name":"abc","my_age":24,"male":1})
False
答案 2 :(得分:1)
dictionary={"name":"abc","age":23,"male":True}
new_dictionary={"my_name":"abc","my_age":23,"male":1}
dict_alias = {"name":"my_name","age":"my_age","male":"male"}
def compare(dictionary,new_dictionary,dict_alias):
same = True
for key in dictionary.keys():
if dictionary[key] == new_dictionary[dict_alias[key]]:
continue
else:
same = False
break
return same