问题描述
我有一个像这样的 Person
类:
I have a Person
class like this:
class Person:
def __init__(self, name, age):
self.name = name
self.age = age
def __repr__(self):
return '<Person {}>'.format(self.name)
我想将此类的一些实例添加到集合中,如下所示:
I want to add some instances of this class to a set, like this:
tom = Person('tom', 18)
mary = Person('mary', 22)
mary2 = Person('mary2', 22)
person_set = {tom, mary, mary2}
print(person_set)
# output: {<Person tom>, <Person mary>, <Person mary2>}
如您所见,集合中有 2 个玛丽.如何使具有相同年龄的 Person
实例被视为同一个人,并且只添加到集合中一次?
As you can see, there are 2 Marys in the set. How can I make it so that Person
instances with the same age are considered the same person, and only added to the set once?
换句话说,我怎样才能得到 {<Person tom>, <Person mary>}
的结果?
In other words, how can I get a result of {<Person tom>, <Person mary>}
?
推荐答案
当一个新对象被添加到一个python集合中时,首先计算对象的哈希码,然后,如果一个或多个对象具有相同的哈希值代码已经/已经在集合中,这些对象将被测试是否与新对象相等.
When a new object is being added to a python set, the hash code of the object is first computed and then, if one or more objects with the same hash code is/are already in the set, these objects are tested for equality with the new object.
这样做的结果是你需要实现 __hash__(...)
和 __eq__(...)
类的方法.例如:
The upshot of this is that you need to implement the __hash__(...)
and __eq__(...)
methods on your class. For example:
class Person:
def __init__(self, name, age):
self.name = name
self.age = age
def __eq__(self, other):
return self.age == other.age
def __hash__(self):
return hash(self.age)
def __repr__(self):
return '<Person {}>'.format(self.name)
tom = Person('tom', 18)
mary = Person('mary', 22)
mary2 = Person('mary2', 22)
person_set = {tom, mary, mary2}
print(person_set)
# output: {<Person tom>, <Person mary>}
但是,您应该非常仔细地考虑 __hash__
和 __eq__
的正确实现对于您的类应该是什么.上面的示例有效,但没有意义(例如,__hash__
和 __eq__
都仅根据年龄定义).
However, you should think very carefully about what the correct implementation of __hash__
and __eq__
should be for your class. The above example works, but is non-sensical (e.g. in that both __hash__
and __eq__
are defined only in terms of age).
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