为什么我不能用另一个函数替换Python对象的__str__方法？

Here is the code:

这是代码:

class Dummy(object):
    def __init__(self, v):
        self.ticker = v


def main():
        def _assign_custom_str(x):
            def _show_ticker(t):                
                return t.ticker
            x.__str__ = _show_ticker
            x.__repr__ = _show_ticker
            return x


    a = [Dummy(1), Dummy(2)]

    a1 = [_assign_custom_str(t) for t in a]
    print a1[1]
    # print a1[1].__str__ # test to if orig __str__ is replaced

I was hoping to see the output like this

我希望看到这样的输出

However, instead I see the standard representation:

但是,相反,我看到了标准表示:

<__main__.Dummy object at 0x01237730>

Why?

2 个解决方案

#1

Magic methods are only guaranteed to work if they're defined on the type rather than on the object.

只有在类型而不是对象上定义魔术方法时,才能保证魔术方法有效。

For example:

def _assign_custom_str(x):
        def _show_ticker(self):                
            return self.ticker
        x.__class__.__str__ = _show_ticker
        x.__class__.__repr__ = _show_ticker
        return x

But note that will affect all Dummy objects, not just the one you're using to access the class.

但请注意,这将影响所有虚拟对象,而不仅仅是您用于访问该类的对象。

#2

if you want to custmize __str__ for every instance, you can call another method _str in __str__, and custmize _str:

如果你想为每个实例保留__str__,你可以在__str__中调用另一个方法_str,并调用_str:

class Dummy(object):
    def __init__(self, v):
        self.ticker = v

    def __str__(self):
        return self._str()

    def _str(self):
        return super(Dummy, self).__str__()

def main():
    a1 = Dummy(1)
    a2 = Dummy(2)

    a1._str = lambda self=a1:"a1: %d" % self.ticker
    a2._str = lambda self=a2:"a2: %d" % self.ticker

    print a1
    print a2    

    a1.ticker = 100
    print a1

main()

the output is :

输出是:

a1: 1
a2: 2
a1: 100

#1