我有一个类MyClass，它包含两个成员变量foo和bar:
class MyClass:
def __init__(self, foo, bar):
self.foo = foo
self.bar = bar
我有这个类的两个实例，每个实例都有相同的foo和bar值:
x = MyClass('foo', 'bar')
y = MyClass('foo', 'bar')
然而，当我比较它们是否相等时，Python返回False:
>>> x == y
False
我如何使python认为这两个对象相等?
当前回答
class Node:
def __init__(self, value):
self.value = value
self.next = None
def __repr__(self):
return str(self.value)
def __eq__(self,other):
return self.value == other.value
node1 = Node(1)
node2 = Node(1)
print(f'node1 id:{id(node1)}')
print(f'node2 id:{id(node2)}')
print(node1 == node2)
>>> node1 id:4396696848
>>> node2 id:4396698000
>>> True
其他回答
在你的类中实现__eq__方法;就像这样:
def __eq__(self, other):
return self.path == other.path and self.title == other.title
编辑:如果你想让你的对象比较相等当且仅当它们有相等的实例字典:
def __eq__(self, other):
return self.__dict__ == other.__dict__
在Python 3.7(及以上版本)中的数据类中，对象实例的相等性比较是一个内置特性。
Python 3.6提供了数据类的后端端口。
(Py37) nsc@nsc-vbox:~$ python
Python 3.7.5 (default, Nov 7 2019, 10:50:52)
[GCC 8.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from dataclasses import dataclass
>>> @dataclass
... class MyClass():
... foo: str
... bar: str
...
>>> x = MyClass(foo="foo", bar="bar")
>>> y = MyClass(foo="foo", bar="bar")
>>> x == y
True
你应该实现方法__eq__:
class MyClass:
def __init__(self, foo, bar, name):
self.foo = foo
self.bar = bar
self.name = name
def __eq__(self,other):
if not isinstance(other,MyClass):
return NotImplemented
else:
#string lists of all method names and properties of each of these objects
prop_names1 = list(self.__dict__)
prop_names2 = list(other.__dict__)
n = len(prop_names1) #number of properties
for i in range(n):
if getattr(self,prop_names1[i]) != getattr(other,prop_names2[i]):
return False
return True
下面通过在两个对象层次结构之间进行深度比较来工作(在我有限的测试中)。In处理各种情况，包括对象本身或其属性是字典的情况。
def deep_comp(o1:Any, o2:Any)->bool:
# NOTE: dict don't have __dict__
o1d = getattr(o1, '__dict__', None)
o2d = getattr(o2, '__dict__', None)
# if both are objects
if o1d is not None and o2d is not None:
# we will compare their dictionaries
o1, o2 = o1.__dict__, o2.__dict__
if o1 is not None and o2 is not None:
# if both are dictionaries, we will compare each key
if isinstance(o1, dict) and isinstance(o2, dict):
for k in set().union(o1.keys() ,o2.keys()):
if k in o1 and k in o2:
if not deep_comp(o1[k], o2[k]):
return False
else:
return False # some key missing
return True
# mismatched object types or both are scalers, or one or both None
return o1 == o2
这是一个非常棘手的代码，所以请在注释中添加任何可能不适合你的情况。
如果你想逐个属性进行比较，并查看它是否以及在哪里失败，你可以使用下面的列表推导式:
[i for i,j in
zip([getattr(obj_1, attr) for attr in dir(obj_1)],
[getattr(obj_2, attr) for attr in dir(obj_2)])
if not i==j]
这里的额外优势是，当在PyCharm中调试时，您可以压缩一行并在“Evaluate Expression”窗口中输入。
