I want to extract the python class name while using abstract classes with abc library. I unfortunately instead receive the class name ABCMeta.
import abc
class A(abc.ABC)
pass
class B(A)
pass
print(A.__class__.__name__) # output: 'ABCMeta'
print(B.__class__.__name__) # output: 'ABCMeta'
print(str(A)) # output: "<class '__main__.A'>"
print(str(B)) # output: "<class '__main__.B'>"
I expect that I should receive the output as below
print(A.__class__.__name__) # output: 'A'
print(B.__class__.__name__) # output: 'B'
The str(A) and str(B) seems to print the class name so I assume the class name can be extracted from somewhere. But nonetheless, I am not interested to use str to parse and get the class name.
Recall that a metaclass is the type of a class, while a class is the type of its instances.
If we have a = A(), a is of type A, and A is of type abc.ABCMeta. Therefore, you should naturally expect that A.__class__ and B.__class__ both return abc.ABCMeta, since they are instances of it!
What you want is the names of A and B themselves, which you can get with A.__name__ and B.__name__ respectively.
Use just the __name__ property
print(A.__name__)
#A
A in itself is a class, if you use A.__class__ you are getting it’s metaclass therefore it’s metaclass’ name.
Related
Let's say I have 2 classes A and B, where B inherits from A. B overrides some methods of A and B have a couple more attributes. Once I created an object b of type B, is it possible to convert it into the type A and only A ? This is to get the primitive behavior of the methods
I don't know how safe it is, but you can reassign the __class__ attribute of the object.
class A:
def f(self):
print("A")
class B(A):
def f(self):
print("B")
b = B()
b.f() # prints B
b.__class__ = A
b.f() # prints A
This only changes the class of the object, it doesn't update any of the attributes. In Python, attributes are added dynamically to objects, and nothing intrinsically links them to specific classes, so there's no way to automatically update the attributes if you change the class.
In Python, we can define a class and print it as such:
class a:
num = 1
b = a()
print(b)
And we would get the output:
<class '__main__.a'>
I'm trying to identify unique classes, and I have some classes with longer paths. I would like to extract the "class path", or __main__.a in the case above. For example, if I print some longer class I would get:
<class 'django_seed.tests.Game'>
<class 'django_seed.tests.Player'>
<class 'django_seed.tests.Action'>
And I would like to extract:
'django_seed.tests.Game'
'django_seed.tests.Player'
'django_seed.tests.Action'
Since I can cast <class 'django_seed.tests.Game'> to a string, I can substring it quite easily with '<class 'django_seed.tests.Game'>'[8:-2], but I'm sure there must be a cleaner way. Thanks!
The __module__ attribute can be used to access the "path" of a class (where it was defined) and the __name__ attribute returns the name of the class as a string
print(f'{YourClass.__module__}.{YourClass.__name__}')
You can use inspect:
import inspect
print((inspect.getmodule(a).__file__))
I have a class A
class A(object):
a = 1
def __init__(self):
self.b = 10
def foo(self):
print type(self).a
print self.b
Then I want to create a class B, which equivalent as A but with different name and value of class member a:
This is what I have tried:
class A(object):
a = 1
def __init__(self):
self.b = 10
def foo(self):
print type(self).a
print self.b
A_dummy = type('A_dummy',(object,),{})
A_attrs = {attr:getattr(A,attr) for attr in dir(A) if (not attr in dir(A_dummy))}
B = type('B',(object,),A_attrs)
B.a = 2
a = A()
a.foo()
b = B()
b.foo()
However I got an Error:
File "test.py", line 31, in main
b.foo()
TypeError: unbound method foo() must be called with A instance as first argument (got nothing instead)
So How I can cope with this sort of jobs (create a copy of an exists class)? Maybe a meta class is needed? But What I prefer is just a function FooCopyClass, such that:
B = FooCopyClass('B',A)
A.a = 10
B.a = 100
print A.a # get 10 as output
print B.a # get 100 as output
In this case, modifying the class member of B won't influence the A, vice versa.
The problem you're encountering is that looking up a method attribute on a Python 2 class creates an unbound method, it doesn't return the underlying raw function (on Python 3, unbound methods are abolished, and what you're attempting would work just fine). You need to bypass the descriptor protocol machinery that converts from function to unbound method. The easiest way is to use vars to grab the class's attribute dictionary directly:
# Make copy of A's attributes
Bvars = vars(A).copy()
# Modify the desired attribute
Bvars['a'] = 2
# Construct the new class from it
B = type('B', (object,), Bvars)
Equivalently, you could copy and initialize B in one step, then reassign B.a after:
# Still need to copy; can't initialize from the proxy type vars(SOMECLASS)
# returns to protect the class internals
B = type('B', (object,), vars(A).copy())
B.a = 2
Or for slightly non-idiomatic one-liner fun:
B = type('B', (object,), dict(vars(A), a=2))
Either way, when you're done:
B().foo()
will output:
2
10
as expected.
You may be trying to (1) create copies of classes for some reason for some real app:
in that case, try using copy.deepcopy - it includes the mechanisms to copy classes. Just change the copy __name__ attribute afterwards if needed. Works both in Python 2 or Python 3.
(2) Trying to learn and understand about Python internal class organization: in that case, there is no reason to fight with Python 2, as some wrinkles there were fixed for Python 3.
In any case, if you try using dir for fetching a class attributes, you will end up with more than you want - as dir also retrieves the methods and attributes of all superclasses. So, even if your method is made to work (in Python 2 that means getting the .im_func attribute of retrieved unbound methods, to use as raw functions on creating a new class), your class would have more methods than the original one.
Actually, both in Python 2 and Python 3, copying a class __dict__ will suffice. If you want mutable objects that are class attributes not to be shared, you should resort again to deepcopy. In Python 3:
class A(object):
b = []
def foo(self):
print(self.b)
from copy import deepcopy
def copy_class(cls, new_name):
new_cls = type(new_name, cls.__bases__, deepcopy(A.__dict__))
new_cls.__name__ = new_name
return new_cls
In Python 2, it would work almost the same, but there is no convenient way to get the explicit bases of an existing class (i.e. __bases__ is not set). You can use __mro__ for the same effect. The only thing is that all ancestor classes are passed in a hardcoded order as bases of the new class, and in a complex hierarchy you could have differences between the behaviors of B descendants and A descendants if multiple-inheritance is used.
I saw the following Python documentation which says that "define variables in a Class" will be class variables:
"Programmer's note: Variables defined in the class definition are
class variables; they are shared by all instances. "
but as I wrote sample code like this:
class CustomizedMethods(object):
class_var1 = 'foo'
class_var2 = 'bar'
cm1 = CustomizedMethods()
cm2 = CustomizedMethods()
print cm1.class_var1, cm1.class_var2 #'foo bar'
print cm2.class_var1, cm2.class_var2 #'foo bar'
cm2.class_var1, cm2.class_var2 = 'bar','for'
print cm1.class_var1, cm1.class_var2 #'foo bar' #here not changed as my expectation
print cm2.class_var1, cm2.class_var2 #'bar foo' #here has changed but they seemed to become instance variables.
I'm confused since what I tried is different from Python's official documentation.
When you assign an attribute on the instance, it is assigned on the instance, even if it previously existed on the class. At first, class_var1 and class_var2 are indeed class attributes. But when you do cm1.class_var1 = "bar", you are not changing this class attribute. Rather, you are creating a new attribute, also called class_var1, but this one is an instance attribute on the instance cm1.
Here is another example showing the difference, although it still may be a bit tough to grasp:
>>> class A(object):
... var = []
>>> a = A()
>>> a.var is A.var
True
>>> a.var = []
>>> a.var is A.var
False
At first, a.var is A.var is true (i.e., they are the same object): since a doesn't have it's own attribute called var, trying to access that goes through to the class. After you give a its own instance attribute, it is no longer the same as the one on the class.
You're assigning attributes on the instances, so yes, they become instance variables at that point. Python looks for attributes on whatever object you specify, then if it can't find them there, looks up the inheritance chain (to the class, the class's parents, etc.). So the attribute you assign on the instance "shadows" or "hides" the class's attribute of the same name.
Strings are immutable, so the difference between a class and instance variable isn't as noticable. For immutable variables in a class definition, the main thing to notice is less use of memory (i.e., if you have 1,000 instances of CustomizedMethods, there's still only one instance of the string "foo" stored in memory.)
However, using mutable variables in a class can introduce subtle bugs if you don't know what you're doing.
Consider:
class CustomizedMethods(object):
class_var = {}
cm1 = CustomizedMethods()
cm2 = CustomizedMethods()
cm1.class_var['test'] = 'foo'
print cm2.class_var
'foo'
cm2.class_var['test'] = 'bar'
print cm1.class_var
'bar'
When you reassigned the cm2 variables, you created new instance variables that "hid" the class variables.
>>> CustomizedMethods.class_var1 = 'one'
>>> CustomizedMethods.class_var2 = 'two'
>>> print cm1.class_var1, cm1.class_var2
one two
>>> print cm2.class_var1, cm2.class_var2
bar for
Try to
print cm1.__dict__
print cm2.__dict__
it will be enlightning...
When you ask cm2 for an attribute it first looks among the attributes of the instance (if one matches the name) and then if there is no matching attribute among the class attributes.
So class_var1 and class_var2 are the names of the class attributes.
Try also the following:
cm2.__class__.class_var1 = "bar_foo"
print cm1.class_var1
what do you expect?
I have a class:
class A:
s = 'some string'
b = <SOME OTHER INSTANCE>
now I want this class to have the functionality of a string whenever it can. That is:
a = A()
print a.b
will print b's value. But I want functions that expect a string (for example replace) to work. For example:
'aaaa'.replace('a', a)
to actually do:
'aaa'.replace('a', a.s)
I tried overidding __get__ but this isn't correct.
I see that you can do this by subclassing str, but is there a way without it?
If you want your class to have the functionality of a string, just extend the built in string class.
>>> class A(str):
... b = 'some other value'
...
>>> a = A('x')
>>> a
'x'
>>> a.b
'some other value'
>>> 'aaa'.replace('a',a)
'xxx'
I found an answer in Subclassing Python tuple with multiple __init__ arguments .
I used Dave's solution and extended str, and then added a new function:
def __new__(self,a,b):
s=a
return str.__new__(A,s)
Override __str__ or __unicode__ to set the string representation of an object (Python documentation).