This returns what you'd expect:
class My(object):
def __len__(self):
return 6
a = My()
print(len(a))
But this throws an error:
class My2(object):
pass
b = My2()
b.__len__ = lambda x: 6
print(len(b))
TypeError: object of type 'My2' has no len()
Why?
It is because you are trying to define the method on the instance not on the underlying class. For example, this would work:
class My2(object):
pass
b = My2()
b.__class__.__len__ = lambda x: 6
print(len(b)) # Prints 6
Furthermore to clarify the comments, it is not a matter of it being a lambda or a function as proven by this not working:
class My2(object):
pass
def l(x):
return 6
b = My2()
b.__len__ = l
print(len(b)) # TypeError: object of type 'My2' has no len()
b.__class__.__len__ = lambda x: 6
In Python len(a) as other "magic" functions, is a shorthand for type(a).__len__(a).
Therefore the function (as opposed to method) __len__ you define on the instance b cannot be accessed through len since the call would be resolved as:
type(b).__len__(b)
But type(b) doesn't have a method __len__.
I asked a similar question and got a very good answer here
The documentation is worth reading. Your example is almost identical.
callable(lambda x: x)returningTrue. And next time maybe you'll want to avoid ending your statements with question marks.