python - 比较python中不同键的两个字典

Question

如何比较两个不同键的字典，例如下面的字典？

dictionary={"name":"abc","age":23,"male":True}
new_dictionary={"my_name":"abc","my_age":23,"male":1}

当比较示例中的两个字典时，比较应该返回true。

Answer 1

>>> 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，如果您只是对确保值相同而不进行任何密钥检查感兴趣：

>>> set(dictionary.values()) == set(new_dictionary.values())
True

编辑：正如 Tadeck 在评论中指出的那样，在sorted()这里使用比set().

（是的，即使一本字典有1而另一本有，这也有效True）

Answer 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，无论它们分配给什么键，都可以使用：

按值比较（忽略键）

根据您的更新要求，其中一种解决方案是：

>>> 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

Answer 3

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