Question

我有一个看起来像这样的课程：

class Foo(object):
   def __init__(self, a, b, c=None):
       self.a = a
       self.b = b
       self.c = c  # c is presumed to be a list
   def __eq__(self, other):
       return self.a == other.a and self.b == other.b

然而，在这种情况下＆＃34; c＆＃34;可能是一个Foos列表，其中＆＃34; c＆＃34; s包含Foos列表，例如：

[Foo(1,2), Foo(3,4,[Foo(5,6)])]

在给定列表结构/对象结构的情况下，处理这种类型的对象比较有什么好方法？我假设只是做一个self.c == other.c就不够了。

Answer 1

修复__eq__方法

class Foo(object):
   def __init__(self, a, b, c=None):
       self.a = a
       self.b = b
       self.c = c  # c is presumed to be a list
   def __eq__(self, other):
       return self.a == other.a \
               and self.b == other.b and self.c == other.c

a,b = Foo(2,3), Foo(5,6)
c = Foo(1,2, [a,b])
d = Foo(1,2)
e,f = Foo(2,3), Foo(5,6)
g = Foo(1,2, [e,f])

print c == d #False
print c == g #True

Answer 2

Foo中n属性的通用解决方案：

class Foo(object):
    def __init__(self, a, b, c=None):
        self.a = a
        self.b = b
        self.c = c  # c is presumed to be a list

    def __eq__(self, other):
        for attr, value in self.__dict__.iteritems():
            if not value == getattr(other, attr):
                return False
        return True


item1 = Foo(1, 2)
item2 = Foo(3, 4, [Foo(5, 6)])
item3 = Foo(3, 4, [Foo(5, 6)])

print(item1 == item2)  # False
print(item3 == item2)  # True

将两个对象与嵌套列表作为属性进行比较

2 个答案: