We Keep Coding

Issue using deepcopy with dictionary inside object

Reading the documentation, I understood that copy.deepcopy(obj) copies recursively any other object inside this one, but when I run:
>>> import copy
>>> class SomeObject:
... a=1
... b={1:1,2:2}
...
>>> o1=SomeObject()
>>> o2=copy.deepcopy(o1)
>>> id(o1)
140041523635624
>>> id(o2)
140041523635912
>>> id(o1.b)
30087968
>>> id(o2.b)
30087968
It does not seem to be copying the dictionary inside 'o1'. Can anyone tell me if I am doing something wrong, or how can I get a copy of the dictionary inside the object?
Thanks

Deepcopy only copies instance attributes. Your b attribute is a class attribute, instead.
Even if you did not create a copy but a new instance of SomeObject manually, b would still be shared:
>>> class SomeObject:
... a=1
... b={1:1,2:2}
...
>>> so1 = SomeObject()
>>> so2 = SomeObject()
>>> so1.b is so2.b
True
>>> so1.b is SomeObject.b
True
Make b an instance attribute:
>>> import copy
>>> class SomeObject:
... a = 1
... def __init__(self):
... self.b = {1: 1, 2: 2}
...
>>> so1 = SomeObject()
>>> so2 = copy.deepcopy(so1)
>>> so1.b is so2.b
False

Python subclass does not have __dict__ while the superclass has?

In the below example, the superclass has a __dict__ attribute, while the subclass does not have it.
>>> class Super(object):
... def hello(self):
... self.data1="hello"
...
>>>
>>> class Sub(Super):
... def hola(self):
... self.data2="hola"
...
>>>
>>> Super.__dict__
<dictproxy object at 0x108794868>
>>> Super.__dict__.keys()
['__dict__', '__module__', '__weakref__', 'hello', '__doc__'] # note __dict__
>>> Sub.__dict__.keys()
['__module__', '__doc__', 'hola'] #__dict__ absent here
>>> Sub.__dict__
<dictproxy object at 0x108794868>
Q1: The comments on the above shows where dict is present. why the superclass has it but not the sublcass.
while trying to find out the answer for this, I came across this post. and this confused me further.
>>> class Foo(object):
... __slots__ = ('bar',)
... bar="spam"
...
>>> f = Foo()
>>> f.__dict__
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'Foo' object has no attribute '__dict__'
>>> class A(object):
... pass
...
>>> b = A()
>>> b.__dict__
{}
Q2: why the instance of Foo throws AttributeError but that of A has empty dict.

Class with slots hasn't dict. Here there is conflict between 'bar' and slots. Delete 'bar' and it will work fine.

How to change Method of Object in python

I write a python file like :
class A(object):
def update(self, str):
pass
def say(self, str):
print "I update: " + str
def fun(obj, str):
obj.say(str)
a = A()
import types
setattr(A, "update", types.MethodType(fun, None, A))
a.update("hello")
b = A()
b.update("world?")
It change behave of class, the object b have been changed. but, I want to only change object a.
How to change Method of Object in python?

Here is a way to do it:
a.update = lambda x: fun(a, x)

You are setting the class method, while you want to set only the method bound to some instance.
>>> class MyClass(object):
... def a(self): pass
...
>>> MyClass.a = lambda x: x
>>> MyClass.a
<unbound method MyClass.<lambda>>
>>> a = MyClass()
>>> a.a
<bound method MyClass.<lambda> of <__main__.MyClass object at 0x1d7fed0>>
Changing the a method at class level changes also the a methods of all instances.
>>> class MyClass(object):
... def a(self): pass
...
>>> b = MyClass()
>>> b.a = lambda x: x
>>> MyClass.a
<unbound method MyClass.a>
>>> b.a
<function <lambda> at 0x1d88938>
>>> c = MyClass()
>>> c.a
<bound method MyClass.a of <__main__.MyClass object at 0x1d8d110>>
Changing the a method of an instance does not change the method of the class or other instances.

How do I assign a property to an instance in Python?

Using python, one can set an attribute of a instance via either of the two methods below:
>>> class Foo(object):
pass
>>> a = Foo()
>>> a.x = 1
>>> a.x
1
>>> setattr(a, 'b', 2)
>>> a.b
2
One can also assign properties via the property decorator.
>>> class Bar(object):
#property
def x(self):
return 0
>>> a = Bar()
>>> a.x
0
My question is, how can I assign a property to an instance?
My intuition was to try something like this...
>>> class Doo(object):
pass
>>> a = Doo()
>>> def k():
return 0
>>> a.m = property(k)
>>> a.m
<property object at 0x0380F540>
... but, I get this weird property object. Similar experimentation yielded similar results. My guess is that properties are more closely related to classes than instances in some respect, but I don't know the inner workings well enough to understand what's going on here.

It is possible to dynamically add properties to a class after it's already created:
class Bar(object):
def x(self):
return 0
setattr(Bar, 'x', property(Bar.x))
print Bar.x
# <property object at 0x04D37270>
print Bar().x
# 0
However, you can't set a property on an instance, only on a class. You can use an instance to do it:
class Bar(object):
def x(self):
return 0
bar = Bar()
setattr(bar.__class__, 'x', property(bar.__class__.x))
print Bar.x
# <property object at 0x04D306F0>
print bar.x
# 0
See How to add property to a class dynamically? for more information.

Properties use descriptors which only work on classes and thus
for all instances. But you could use a combination of a descriptor on
a class that would consult a per-instance function.
>>> class Foo(object):
... #property
... def x(self):
... if 'x' in self.__dict__:
... return self.__dict__['x'](self)
...
>>> a = Foo()
>>> def k(self):
... return 0
...
>>> a.__dict__['x'] = k
>>> a.x
0

You can assign the property directly to the class object:
>>> class Foo(object):
pass
>>> a = Foo()
>>> a.__class__
__main__.Foo
>>> setattr(a.__class__, 'm', property(lambda self: 0))
>>> a.m
0
>>> a.m = 24
AttributeError: can't set attribute

Here we have taken #agf's solution and used a lambda function to define the class property.
class A(object):
pass
a = A()
a.__class__.f = property(lambda self: 57)
a.f # 57
The following post provides more context: https://crosscompute.com/n/jAbsB6OIm6oCCJX9PBIbY5FECFKCClyV/_/Assign%20a%20class%20property%20to%20an%20instance

Getting the class name of an instance

How do I find out the name of the class used to create an instance of an object in Python?
I'm not sure if I should use the inspect module or parse the __class__ attribute.

Have you tried the __name__ attribute of the class? ie type(x).__name__ will give you the name of the class, which I think is what you want.
>>> import itertools
>>> x = itertools.count(0)
>>> type(x).__name__
'count'
If you're still using Python 2, note that the above method works with new-style classes only (in Python 3+ all classes are "new-style" classes). Your code might use some old-style classes. The following works for both:
x.__class__.__name__

Do you want the name of the class as a string?
instance.__class__.__name__

type() ?
>>> class A:
... def whoami(self):
... print(type(self).__name__)
...
>>>
>>> class B(A):
... pass
...
>>>
>>>
>>> o = B()
>>> o.whoami()
'B'
>>>

class A:
pass
a = A()
str(a.__class__)
The sample code above (when input in the interactive interpreter) will produce '__main__.A' as opposed to 'A' which is produced if the __name__ attribute is invoked. By simply passing the result of A.__class__ to the str constructor the parsing is handled for you. However, you could also use the following code if you want something more explicit.
"{0}.{1}".format(a.__class__.__module__,a.__class__.__name__)
This behavior can be preferable if you have classes with the same name defined in separate modules.
The sample code provided above was tested in Python 2.7.5.

In Python 2,
type(instance).__name__ != instance.__class__.__name__
# if class A is defined like
class A():
...
type(instance) == instance.__class__
# if class A is defined like
class A(object):
...
Example:
>>> class aclass(object):
... pass
...
>>> a = aclass()
>>> type(a)
<class '__main__.aclass'>
>>> a.__class__
<class '__main__.aclass'>
>>>
>>> type(a).__name__
'aclass'
>>>
>>> a.__class__.__name__
'aclass'
>>>
>>> class bclass():
... pass
...
>>> b = bclass()
>>>
>>> type(b)
<type 'instance'>
>>> b.__class__
<class __main__.bclass at 0xb765047c>
>>> type(b).__name__
'instance'
>>>
>>> b.__class__.__name__
'bclass'
>>>

Alternatively you can use the classmethod decorator:
class A:
#classmethod
def get_classname(cls):
return cls.__name__
def use_classname(self):
return self.get_classname()
Usage:
>>> A.get_classname()
'A'
>>> a = A()
>>> a.get_classname()
'A'
>>> a.use_classname()
'A'

Good question.
Here's a simple example based on GHZ's which might help someone:
>>> class person(object):
def init(self,name):
self.name=name
def info(self)
print "My name is {0}, I am a {1}".format(self.name,self.__class__.__name__)
>>> bob = person(name='Robert')
>>> bob.info()
My name is Robert, I am a person

Apart from grabbing the special __name__ attribute, you might find yourself in need of the qualified name for a given class/function. This is done by grabbing the types __qualname__.
In most cases, these will be exactly the same, but, when dealing with nested classes/methods these differ in the output you get. For example:
class Spam:
def meth(self):
pass
class Bar:
pass
>>> s = Spam()
>>> type(s).__name__
'Spam'
>>> type(s).__qualname__
'Spam'
>>> type(s).Bar.__name__ # type not needed here
'Bar'
>>> type(s).Bar.__qualname__ # type not needed here
'Spam.Bar'
>>> type(s).meth.__name__
'meth'
>>> type(s).meth.__qualname__
'Spam.meth'
Since introspection is what you're after, this is always you might want to consider.

You can simply use __qualname__ which stands for qualified name of a function or class
Example:
>>> class C:
... class D:
... def meth(self):
... pass
...
>>> C.__qualname__
'C'
>>> C.D.__qualname__
'C.D'
>>> C.D.meth.__qualname__
'C.D.meth'
documentation link qualname

To get instance classname:
type(instance).__name__
or
instance.__class__.__name__
both are the same

You can first use type and then str to extract class name from it.
class foo:pass;
bar:foo=foo();
print(str(type(bar))[8:-2][len(str(type(bar).__module__))+1:]);
Result
foo

If you're looking to solve this for a list (or iterable collection) of objects, here's how I would solve:
from operator import attrgetter
# Will use a few data types to show a point
my_list = [1, "2", 3.0, [4], object(), type, None]
# I specifically want to create a generator
my_class_names = list(map(attrgetter("__name__"), map(type, my_list))))
# Result:
['int', 'str', 'float', 'list', 'object', 'type', 'NoneType']
# Alternatively, use a lambda
my_class_names = list(map(lambda x: type(x).__name__, my_list))