我有一个在模块中声明如下的python类
class Position:
def __init__(self, x, y):
self.x = int(x)
self.y = int(y)
def __str__(self):
return self.toString()
def toString(self): #deprecated
return "{x:"+str(self.x)+" y:"+str(self.y)+"}"
现在,在主程序的后面,我做了这样的比较:
can_pos = somestreet.endOfStreet(curPos).getPos() #returns a Position object
if(can_pos == atPos): # this returns False
#blafoo
#if(can_pos.x == atPos.x and can_pos.y == atPos.y): #this returns True (and is expected)
我不明白不同行为的原因是什么......
如果有人可以给我一个提示,那将是非常好的:)
提前致谢
答案 0 :(得分:3)
如评论中所述,您需要明确定义至少__eq__
和__ne__
:
class Position:
def __init__(self, x, y):
self.x = int(x)
self.y = int(y)
def __eq__(self, other):
return self.x == other.x and self.y == other.y
def __ne__(self, other):
return not self == other
给出了
>>> a = Position(1,2)
>>> b = Position(1,2)
>>> c = Position(2,3)
>>> a == b
True
>>> a == c
False
>>> b == c
False
>>> a != a
False
>>> a != b
False
>>> a != c
True
但是,请注意,与Python 2相比,您将拥有:
>>> a > c
True
和其他可能不受欢迎的行为,而在Python 3(您正在使用)中,您将获得
TypeError: unorderable types: Position() > Position()