How to define a __str__ method for a class?
Question:
In Python, the object
class serves as the root superclass for all the (new-style) classes. By default at least, applying str
and repr
to the “class instance” of any subclass of object
produces the same result:
>>> class spam(object): pass
...
>>> str(spam)
"<class '__main__.spam'>"
>>> str(spam) == repr(spam)
I would like to define a subclass of object
, say fancyobject
, that is identical to object
in every way, except that applying str
and repr
to fancyobject
itself produces different outputs:
>>> class ham(fancyobject): pass
...
>>> str(ham)
'ham'
>>> repr(ham)
"<class '__main__.ham'>"
Is there a way to do this in Python?
PS: I’m aware of the __str__
special method, but it is my understanding that if class A
overrides __str__
, then the overriding method is called only when str
is called on instances of A
, not when it is called on A
itself. I.e.:
>>> class A(object):
... def __str__(self):
... return 'from new __str__: ' + object.__str__(self)
...
>>> str(A())
'from new __str__: <__main__.A object at 0x7f79c62a5310>'
>>> str(A)
"<class '__main__.A'>"
Answers:
Actually the same mechanism as for object instances applies for types. Types are just objects themselves, so they are converted to strings by calling the __str__()
method on their type, which is called the “metaclass”. So you have to overwrite the __str__()
method on the metaclass:
class fancytype(type):
def __str__(self):
return self.__name__
class ham(object):
__metaclass__ = fancytype
print ham
prints
ham
You can also set the default metaclass for a whole module like this
class fancytype(type):
def __str__(self):
return self.__name__
__metaclass__ = fancytype
class ham:
pass
print ham
Here’s the new answer for Python 3. Basically, you pass in a metaclass as a keyword parameter to the class definition.
class fancytype(type):
def __str__(self):
return self.__name__
class ham(metaclass=fancytype):
pass
print(ham)
prints
ham
In Python, the object
class serves as the root superclass for all the (new-style) classes. By default at least, applying str
and repr
to the “class instance” of any subclass of object
produces the same result:
>>> class spam(object): pass
...
>>> str(spam)
"<class '__main__.spam'>"
>>> str(spam) == repr(spam)
I would like to define a subclass of object
, say fancyobject
, that is identical to object
in every way, except that applying str
and repr
to fancyobject
itself produces different outputs:
>>> class ham(fancyobject): pass
...
>>> str(ham)
'ham'
>>> repr(ham)
"<class '__main__.ham'>"
Is there a way to do this in Python?
PS: I’m aware of the __str__
special method, but it is my understanding that if class A
overrides __str__
, then the overriding method is called only when str
is called on instances of A
, not when it is called on A
itself. I.e.:
>>> class A(object):
... def __str__(self):
... return 'from new __str__: ' + object.__str__(self)
...
>>> str(A())
'from new __str__: <__main__.A object at 0x7f79c62a5310>'
>>> str(A)
"<class '__main__.A'>"
Actually the same mechanism as for object instances applies for types. Types are just objects themselves, so they are converted to strings by calling the __str__()
method on their type, which is called the “metaclass”. So you have to overwrite the __str__()
method on the metaclass:
class fancytype(type):
def __str__(self):
return self.__name__
class ham(object):
__metaclass__ = fancytype
print ham
prints
ham
You can also set the default metaclass for a whole module like this
class fancytype(type):
def __str__(self):
return self.__name__
__metaclass__ = fancytype
class ham:
pass
print ham
Here’s the new answer for Python 3. Basically, you pass in a metaclass as a keyword parameter to the class definition.
class fancytype(type):
def __str__(self):
return self.__name__
class ham(metaclass=fancytype):
pass
print(ham)
prints
ham