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
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
其他回答
总结如下:
It's advised to implement __eq__ rather than __cmp__, except if you run python <= 2.0 (__eq__ has been added in 2.1) Don't forget to also implement __ne__ (should be something like return not self.__eq__(other) or return not self == other except very special case) Don`t forget that the operator must be implemented in each custom class you want to compare (see example below). If you want to compare with object that can be None, you must implement it. The interpreter cannot guess it ... (see example below) class B(object): def __init__(self): self.name = "toto" def __eq__(self, other): if other is None: return False return self.name == other.name class A(object): def __init__(self): self.toto = "titi" self.b_inst = B() def __eq__(self, other): if other is None: return False return (self.toto, self.b_inst) == (other.toto, other.b_inst)
类的实例与==比较是非相等的。最好的方法是将cmp函数赋给你的类，它会做这些事情。
如果你想通过内容进行比较你可以使用cmp(obj1,obj2)
在你的例子中，cmp(doc1,doc2)如果它们的内容相同，它将返回-1。
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。
重写对象中的富比较运算符。
class MyClass:
def __lt__(self, other):
# return comparison
def __le__(self, other):
# return comparison
def __eq__(self, other):
# return comparison
def __ne__(self, other):
# return comparison
def __gt__(self, other):
# return comparison
def __ge__(self, other):
# return comparison
是这样的:
def __eq__(self, other):
return self._id == other._id
我写了这个，并把它放在我的项目中的test/utils模块中。对于情况下，当它不是一个类，只是计划ol' dict，这将遍历两个对象，并确保
每个属性都与其对应的属性相等 不存在悬空属性(只存在于一个对象上的attrs)
它的大…这一点都不性感……但是，哦，boi，它工作!
def assertObjectsEqual(obj_a, obj_b):
def _assert(a, b):
if a == b:
return
raise AssertionError(f'{a} !== {b} inside assertObjectsEqual')
def _check(a, b):
if a is None or b is None:
_assert(a, b)
for k,v in a.items():
if isinstance(v, dict):
assertObjectsEqual(v, b[k])
else:
_assert(v, b[k])
# Asserting both directions is more work
# but it ensures no dangling values on
# on either object
_check(obj_a, obj_b)
_check(obj_b, obj_a)
您可以通过删除_assert并使用普通的ol' assert来稍微清理它，但当它失败时，您得到的消息是非常没有帮助的。
