Python set([]) 如何检查两个对象是否相等?一个对象需要定义哪些方法来自定义它?

2022-01-17 00:00:00 python set methods comparison

问题描述

我需要在 Python 中创建一个容器"对象或类，它会记录我还定义的其他对象.此容器的一个要求是，如果两个对象被认为是相同的，则删除一个(其中一个).我的第一个想法是使用 set([]) 作为包含对象，来完成这个要求.

I need to create a 'container' object or class in Python, which keeps a record of other objects which I also define. One requirement of this container is that if two objects are deemed to be identical, one (either one) is removed. My first thought was to use a set([]) as the containing object, to complete this requirement.

但是，该集合不会删除两个相同的对象实例之一.我必须定义什么来创建一个?

However, the set does not remove one of the two identical object instances. What must I define to create one?

这是 Python 代码.

Here is the Python code.

class Item(object):
  def __init__(self, foo, bar):
    self.foo = foo
    self.bar = bar
  def __repr__(self):
    return "Item(%s, %s)" % (self.foo, self.bar)
  def __eq__(self, other):
    if isinstance(other, Item):
      return ((self.foo == other.foo) and (self.bar == other.bar))
    else:
      return False
  def __ne__(self, other):
    return (not self.__eq__(other))

口译员

>>> set([Item(1,2), Item(1,2)])
set([Item(1, 2), Item(1, 2)])

很明显，x == y调用的__eq__()并不是集合调用的方法.什么叫做?我还必须定义什么其他方法?

It is clear that __eq__(), which is called by x == y, is not the method called by the set. What is called? What other method must I define?

注意:Items 必须保持可变，并且可以更改，因此我无法提供 __hash__()代码> 方法.如果这是唯一的方法，那么我将重写以使用不可变的 Items.

解决方案

恐怕你得提供一个 __hash__() 方法.但是你可以这样编码，它不依赖于你的 Item 的可变属性.

I am afraid you will have to provide a __hash__() method. But you can code it the way, that it does not depend on the mutable attributes of your Item.

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