[英]Why can't I replace the __str__ method of a Python object with another function?
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
2
However, instead I see the standard representation:但是,我看到的是标准表示:
<__main__.Dummy object at 0x01237730>
Why?为什么?
Magic methods are only guaranteed to work if they're defined on the type rather than on the object .魔术方法只有在定义在类型上而不是在对象上时才能保证有效。
For example:例如:
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Dummy
对象,而不仅仅是您用来访问该类的对象。
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自定义__str__
,你可以在__str__
调用另一个方法__str__
,并自定义 _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
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