This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is there a function in Python to print all the current properties and values of an object?
In interactive python sessions I make a lot of use of the dir function, to get a sense of an object's structure. Unfortunately, dir only shows the names of attributes, not their values, so it is nowhere near as informative as it could be. Furthermore, dir's print out makes no attempt to format the output for ease of reading (IOW: dir don't do no stinkin' pretty-print).
Where can I find an "off-the-shelf" data-inspection utility that is more informative, and better formatted, dir?
For example, a more useful alternative to dir would print the values (suitably formatted, as needed) associated with each attribute, and would format this output for easy reading. For attributes whose values are callable, it would print their signatures and/or the first lines of their docstrings.
Thanks!
Try the inspect module, which can give you all sorts of things, including the original source code for a function, stack frames and more:
>>> class Foo:
... def __init__(self, a, b, c):
... self.a = a
... self.b = b
... self.c = c
... def foobar(self):
... return 100
...
>>> f = Foo(50, 'abc', 2.5)
>>> import inspect
>>> inspect.getmembers(f)
[('__doc__', None), ('__init__', <bound method Foo.__init__ of <__main__.Foo instance at 0x7f0d76c440e0>>), ('__module__', '__main__'), ('a', 50), ('b', 'abc'), ('c', 2.5), ('foobar', <bound method Foo.foobar of <__main__.Foo instance at 0x7f0d76c440e0>>)]
>>>
>>> def add5(x):
... """adds 5 to arg"""
... return 5 + x
...
>>> inspect.getdoc(add5)
'adds 5 to arg'
>>> # What arguments does add5 take?
>>> inspect.getargspec(add5)
ArgSpec(args=['x'], varargs=None, keywords=None, defaults=None)
>>> # Format that nicely...
>>> inspect.formatargspec(inspect.getargspec(add5))
'((x,), None, None, None)'
>>>
If , by "object's structure" , you mean the instance attributes of the object, note that dir(something) calls something.__dir()__ if a user-defined method with name __dir__() exists, otherwise it inspects something.__dict__ and the type of something (I suppose it then calls the type's __dir__() or inspects the type's __dict__ )
Hence I believe that the instance attributes of an objects are given by its __dict__ attribute.
But some objects have no instance attributes, for example integers.
So, I think that if you test the existence of the __dict__ attribute of an object and, if it exists, obtaining its value will give you what you need to obtain : not only the names of the attributes but the objects so named.
Since you ask for formating of the output, I give this example with cPickle (a module is an object, as everything in Python)
import cPickle
for name,obj in cPickle.__dict__.iteritems() :
print '%-25s --> %r' % (name, obj)
result
load --> <built-in function load>
PicklingError --> <class 'cPickle.PicklingError'>
__version__ --> '1.71'
UnpickleableError --> <class 'cPickle.UnpickleableError'>
dump --> <built-in function dump>
__builtins__ --> <module '__builtin__' (built-in)>
Unpickler --> <built-in function Unpickler>
compatible_formats --> ['1.0', '1.1', '1.2', '1.3', '2.0']
BadPickleGet --> <class 'cPickle.BadPickleGet'>
__package__ --> None
dumps --> <built-in function dumps>
UnpicklingError --> <class 'cPickle.UnpicklingError'>
PickleError --> <class 'cPickle.PickleError'>
HIGHEST_PROTOCOL --> 2
__name__ --> 'cPickle'
loads --> <built-in function loads>
Pickler --> <built-in function Pickler>
__doc__ --> 'C implementation and optimization of the Python pickle module.'
format_version --> '2.0'
.
There's also the function help() that brings documentation you are searching for.
Check out pprint
from pprint import pprint
pprint(object_you_want_to_see)
Related
Suppose I have a class with __slots__
class A:
__slots__ = ['x']
a = A()
a.x = 1 # works fine
a.y = 1 # AttributeError (as expected)
Now I am going to change __slots__ of A.
A.__slots__.append('y')
print(A.__slots__) # ['x', 'y']
b = A()
b.x = 1 # OK
b.y = 1 # AttributeError (why?)
b was created after __slots__ of A had changed, so Python, in principle, could allocate memory for b.y. Why it didn't?
How to properly modify __slots__ of a class, so that new instances have the modified attributes?
You cannot dynamically alter the __slots__ attribute after creating the class, no. That's because the value is used to create special descriptors for each slot. From the __slots__ documentation:
__slots__ are implemented at the class level by creating descriptors (Implementing Descriptors) for each variable name. As a result, class attributes cannot be used to set default values for instance variables defined by __slots__; otherwise, the class attribute would overwrite the descriptor assignment.
You can see the descriptors in the class __dict__:
>>> class A:
... __slots__ = ['x']
...
>>> A.__dict__
mappingproxy({'__module__': '__main__', '__doc__': None, 'x': <member 'x' of 'A' objects>, '__slots__': ['x']})
>>> A.__dict__['x']
<member 'x' of 'A' objects>
>>> a = A()
>>> A.__dict__['x'].__get__(a, A)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: x
>>> A.__dict__['x'].__set__(a, 'foobar')
>>> A.__dict__['x'].__get__(a, A)
'foobar'
>>> a.x
'foobar'
You cannot yourself create these additional descriptors. Even if you could, you cannot allocate more memory space for the extra slot references on the instances produced for this class, as that's information stored in the C struct for the class, and not in a manner accessible to Python code.
That's all because __slots__ is only an extension of the low-level handling of the elements that make up Python instances to Python code; the __dict__ and __weakref__ attributes on regular Python instances were always implemented as slots:
>>> class Regular: pass
...
>>> Regular.__dict__['__dict__']
<attribute '__dict__' of 'Regular' objects>
>>> Regular.__dict__['__weakref__']
<attribute '__weakref__' of 'Regular' objects>
>>> r = Regular()
>>> Regular.__dict__['__dict__'].__get__(r, Regular) is r.__dict__
True
All the Python developers did here was extend the system to add a few more of such slots using arbitrary names, with those names taken from the __slots__ attribute on the class being created, so that you can save memory; dictionaries take more memory than simple references to values in slots do. By specifying __slots__ you disable the __dict__ and __weakref__ slots, unless you explicitly include those in the __slots__ sequence.
The only way to extend slots then is to subclass; you can dynamically create a subclass with the type() function or by using a factory function:
def extra_slots_subclass(base, *slots):
class ExtraSlots(base):
__slots__ = slots
ExtraSlots.__name__ = base.__name__
return ExtraSlots
It appears to me a type turns __slots__ into a tuple as one of it's first orders of action. It then stores the tuple on the extended type object. Since beneath it all, the python is looking at a tuple, there is no way to mutate it. Indeed, I'm not even sure you can access it unless you pass a tuple in to the instance in the first place.
The fact that the original object that you set still remains as an attribute on the type is (perhaps) just a convenience for introspection.
You can't modify __slots__ and expect to have that show up somewhere (and really -- from a readability perspective, You probably don't really want to do that anyway, right?)...
Of course, you can always subclass to extend the slots:
>>> class C(A):
... __slots__ = ['z']
...
>>> c = C()
>>> c.x = 1
>>> c.z = 1
You cannot modify the __slots__ attribute after class creation. This is because it would leade to strange behaviour.
Imagine the following.
class A:
__slots__ = ["x"]
a = A()
A.__slots__.append("y")
a.y = None
What should happen in this scenario? No space was originally allocated for a second slot, but according to the slots attribute, a should be able have space for y.
__slots__ is not about protecting what names can and cannot be accessed. Rather __slots__ is about reducing the memory footprint of an object. By attempting to modify __slots__ you would defeat the optimisations that __slots__ is meant to achieve.
How __slots__ reduces memory footprint
Normally, an object's attributes are stored in a dict, which requires a fair bit of memory itself. If you are creating millions of objects then the space required by these dicts becomes prohibitive. __slots__ informs the python machinery that makes the class object that there will only be so many attributes refered to by instances of this class and what the names of the attributes will be. Therefore, the class can make an optimisation by storing the attributes directly on the instance rather than in a dict. It places the memory for the (pointers to the) attributes directly on the object, rather than creating a new dict for the object.
Putting answers to this and related question together, I want to make an accent on a solution to this problem:
You can kind of modify __slots__ by creating a subclass with the same name and then replacing parent class with its child. Note that you can do this for classes declared and used in any module, not just yours!
Consider the following module which declares some classes:
module.py:
class A(object):
# some class a user should import
__slots__ = ('x', 'b')
def __init__(self):
self.b = B()
class B(object):
# let's suppose we can't use it directly,
# it's returned as a part of another class
__slots__ = ('z',)
Here's how you can add attributes to these classes:
>>> import module
>>> from module import A
>>>
>>> # for classes imported into your module:
>>> A = type('A', (A,), {'__slots__': ('foo',)})
>>> # for classes which will be instantiated by the `module` itself:
>>> module.B = type('B', (module.B,), {'__slots__': ('bar',)})
>>>
>>> a = A()
>>> a.x = 1
>>> a.foo = 2
>>>
>>> b = a.b
>>> b.z = 3
>>> b.bar = 4
>>>
But what if you receive class instances from some third-party module using the module?
module_3rd_party.py:
from module import A
def get_instance():
return A()
No problem, it will also work! The only difference is that you may need to patch them before you import third-party module (in case it imports classes from the module):
>>> import module
>>>
>>> module.A = type('A', (module.A,), {'__slots__': ('foo',)})
>>> module.B = type('B', (module.B,), {'__slots__': ('bar',)})
>>>
>>> # note that we import `module_3rd_party` AFTER we patch the `module`
>>> from module_3rd_party import get_instance
>>>
>>> a = get_instance()
>>> a.x = 1
>>> a.foo = 2
>>>
>>> b = a.b
>>> b.z = 3
>>> b.bar = 4
>>>
It works because Python imports modules only once and then shares them between all other modules, so the changes you make to modules affect all code running along yours.
Suppose I have a class with __slots__
class A:
__slots__ = ['x']
a = A()
a.x = 1 # works fine
a.y = 1 # AttributeError (as expected)
Now I am going to change __slots__ of A.
A.__slots__.append('y')
print(A.__slots__) # ['x', 'y']
b = A()
b.x = 1 # OK
b.y = 1 # AttributeError (why?)
b was created after __slots__ of A had changed, so Python, in principle, could allocate memory for b.y. Why it didn't?
How to properly modify __slots__ of a class, so that new instances have the modified attributes?
You cannot dynamically alter the __slots__ attribute after creating the class, no. That's because the value is used to create special descriptors for each slot. From the __slots__ documentation:
__slots__ are implemented at the class level by creating descriptors (Implementing Descriptors) for each variable name. As a result, class attributes cannot be used to set default values for instance variables defined by __slots__; otherwise, the class attribute would overwrite the descriptor assignment.
You can see the descriptors in the class __dict__:
>>> class A:
... __slots__ = ['x']
...
>>> A.__dict__
mappingproxy({'__module__': '__main__', '__doc__': None, 'x': <member 'x' of 'A' objects>, '__slots__': ['x']})
>>> A.__dict__['x']
<member 'x' of 'A' objects>
>>> a = A()
>>> A.__dict__['x'].__get__(a, A)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: x
>>> A.__dict__['x'].__set__(a, 'foobar')
>>> A.__dict__['x'].__get__(a, A)
'foobar'
>>> a.x
'foobar'
You cannot yourself create these additional descriptors. Even if you could, you cannot allocate more memory space for the extra slot references on the instances produced for this class, as that's information stored in the C struct for the class, and not in a manner accessible to Python code.
That's all because __slots__ is only an extension of the low-level handling of the elements that make up Python instances to Python code; the __dict__ and __weakref__ attributes on regular Python instances were always implemented as slots:
>>> class Regular: pass
...
>>> Regular.__dict__['__dict__']
<attribute '__dict__' of 'Regular' objects>
>>> Regular.__dict__['__weakref__']
<attribute '__weakref__' of 'Regular' objects>
>>> r = Regular()
>>> Regular.__dict__['__dict__'].__get__(r, Regular) is r.__dict__
True
All the Python developers did here was extend the system to add a few more of such slots using arbitrary names, with those names taken from the __slots__ attribute on the class being created, so that you can save memory; dictionaries take more memory than simple references to values in slots do. By specifying __slots__ you disable the __dict__ and __weakref__ slots, unless you explicitly include those in the __slots__ sequence.
The only way to extend slots then is to subclass; you can dynamically create a subclass with the type() function or by using a factory function:
def extra_slots_subclass(base, *slots):
class ExtraSlots(base):
__slots__ = slots
ExtraSlots.__name__ = base.__name__
return ExtraSlots
It appears to me a type turns __slots__ into a tuple as one of it's first orders of action. It then stores the tuple on the extended type object. Since beneath it all, the python is looking at a tuple, there is no way to mutate it. Indeed, I'm not even sure you can access it unless you pass a tuple in to the instance in the first place.
The fact that the original object that you set still remains as an attribute on the type is (perhaps) just a convenience for introspection.
You can't modify __slots__ and expect to have that show up somewhere (and really -- from a readability perspective, You probably don't really want to do that anyway, right?)...
Of course, you can always subclass to extend the slots:
>>> class C(A):
... __slots__ = ['z']
...
>>> c = C()
>>> c.x = 1
>>> c.z = 1
You cannot modify the __slots__ attribute after class creation. This is because it would leade to strange behaviour.
Imagine the following.
class A:
__slots__ = ["x"]
a = A()
A.__slots__.append("y")
a.y = None
What should happen in this scenario? No space was originally allocated for a second slot, but according to the slots attribute, a should be able have space for y.
__slots__ is not about protecting what names can and cannot be accessed. Rather __slots__ is about reducing the memory footprint of an object. By attempting to modify __slots__ you would defeat the optimisations that __slots__ is meant to achieve.
How __slots__ reduces memory footprint
Normally, an object's attributes are stored in a dict, which requires a fair bit of memory itself. If you are creating millions of objects then the space required by these dicts becomes prohibitive. __slots__ informs the python machinery that makes the class object that there will only be so many attributes refered to by instances of this class and what the names of the attributes will be. Therefore, the class can make an optimisation by storing the attributes directly on the instance rather than in a dict. It places the memory for the (pointers to the) attributes directly on the object, rather than creating a new dict for the object.
Putting answers to this and related question together, I want to make an accent on a solution to this problem:
You can kind of modify __slots__ by creating a subclass with the same name and then replacing parent class with its child. Note that you can do this for classes declared and used in any module, not just yours!
Consider the following module which declares some classes:
module.py:
class A(object):
# some class a user should import
__slots__ = ('x', 'b')
def __init__(self):
self.b = B()
class B(object):
# let's suppose we can't use it directly,
# it's returned as a part of another class
__slots__ = ('z',)
Here's how you can add attributes to these classes:
>>> import module
>>> from module import A
>>>
>>> # for classes imported into your module:
>>> A = type('A', (A,), {'__slots__': ('foo',)})
>>> # for classes which will be instantiated by the `module` itself:
>>> module.B = type('B', (module.B,), {'__slots__': ('bar',)})
>>>
>>> a = A()
>>> a.x = 1
>>> a.foo = 2
>>>
>>> b = a.b
>>> b.z = 3
>>> b.bar = 4
>>>
But what if you receive class instances from some third-party module using the module?
module_3rd_party.py:
from module import A
def get_instance():
return A()
No problem, it will also work! The only difference is that you may need to patch them before you import third-party module (in case it imports classes from the module):
>>> import module
>>>
>>> module.A = type('A', (module.A,), {'__slots__': ('foo',)})
>>> module.B = type('B', (module.B,), {'__slots__': ('bar',)})
>>>
>>> # note that we import `module_3rd_party` AFTER we patch the `module`
>>> from module_3rd_party import get_instance
>>>
>>> a = get_instance()
>>> a.x = 1
>>> a.foo = 2
>>>
>>> b = a.b
>>> b.z = 3
>>> b.bar = 4
>>>
It works because Python imports modules only once and then shares them between all other modules, so the changes you make to modules affect all code running along yours.
I am learning a tutorial on python.It is explaining how functions are first class objects in Python.
def foo():
pass
print(foo.__class__)
print(issubclass(foo.__class__,object))
The output that I get for the above code is
<type 'function'>
True
This program is supposed to demonstrate that functions are first class objects in python? My questions are as follows.
How does the above code prove that functions are fist class objects?
What are the attributes of a first class object?
what does function.__class__ signify? It returns a tuple <type,function> which doesn't mean much?
Here's what Guido says about first class objects in his blog:
One of my goals for Python was to make it so that all objects were "first class." By this, I meant that I wanted all objects that could be named in the language (e.g., integers, strings, functions, classes, modules, methods, etc.) to have equal status. That is, they can be assigned to variables, placed in lists, stored in dictionaries, passed as arguments, and so forth.
The whole blog post is worth reading.
In the example you posted, the tutorial may be making the point that first class objects are generally descendents of the "object" class.
First-class simply means that functions can be treated as a value -- that is you can assign them to variables, return them from functions, as well as pass them in as a parameter. That is you can do code like:
>>> def say_hi():
print "hi"
>>> def say_bye():
print "bye"
>>> f = say_hi
>>> f()
hi
>>> f = say_bye
>>> f()
bye
This is useful as you can now assign functions to variables like any ordinary variable:
>>> for f in (say_hi, say_bye):
f()
hi
bye
Or write higher order functions (that take functions as parameters):
>>> def call_func_n_times(f, n):
for i in range(n):
f()
>>> call_func_n_times(say_hi, 3)
hi
hi
hi
>>> call_func_n_times(say_bye, 2)
bye
bye
About __class__ in python tells what type of object you have. E.g., if you define an list object in python: a = [1,2,3], then a.__class__ will be <type 'list'>. If you have a datetime (from datetime import datetime and then d = datetime.now(), then the type of d instance will be <type 'datetime.datetime'>. They were just showing that in python a function is not a brand new concept. It's just an ordinary object of <type 'function'>.
You proved that functions are first class objects because you were allowed to pass foo as an argument to a method.
The attributes of first class objects was nicely summarised in this post: https://stackoverflow.com/a/245208/3248346
Depending on the language, this can
imply:
being expressible as an anonymous literal value
being storable in variables
being storable in data structures
having an intrinsic identity (independent of any given name)
being comparable for equality with other entities
being passable as a parameter to a procedure/function
being returnable as the result of a procedure/function
being constructible at runtime
being printable
being readable
being transmissible among distributed processes
being storable outside running processes
Regarding your third question, <type 'function'> isn't a tuple. Python's tuple notation is (a,b), not angle brackets.
foo.__class__ returns a class object, that is, an object which represents the class to which foo belongs; class objects happen to produce descriptive strings in the interpreter, in this case telling you that the class of foo is the type called 'function'. (Classes and types are basically the same in modern Python.)
It doesn't mean a whole lot other than that, like any other object, functions have a type:
>>> x = 1
>>> x.__class__
<type 'int'>
>>> y = "bar"
>>> y.__class__
<type 'str'>
>>> def foo(): pass
...
>>> foo.__class__
<type 'function'>
Regarding your comment to #I.K.s answer, f_at_2() in the following would be the method.
def f_at_2(f):
return f(2)
def foo(n):
return n ** n
def bar(n):
return n * n
def baz(n):
return n / 2
funcs = [foo, bar, baz]
for f in funcs:
print f.func_name, f_at_2(f)
...
>>>
foo 4
bar 4
baz 1
>>>
A method is a function of/in a class, but the concept also applies to a function (outside of a class). The functions (as objects) are contained in a data structure and passed to another object.
I am really confused about the __dict__ attribute. I have searched a lot but still I am not sure about the output.
Could someone explain the use of this attribute from zero, in cases when it is used in a object, a class, or a function?
Basically it contains all the attributes which describe the object in question. It can be used to alter or read the attributes.
Quoting from the documentation for __dict__
A dictionary or other mapping object used to store an object's (writable) attributes.
Remember, everything is an object in Python. When I say everything, I mean everything like functions, classes, objects etc (Ya you read it right, classes. Classes are also objects). For example:
def func():
pass
func.temp = 1
print(func.__dict__)
class TempClass:
a = 1
def temp_function(self):
pass
print(TempClass.__dict__)
will output
{'temp': 1}
{'__module__': '__main__',
'a': 1,
'temp_function': <function TempClass.temp_function at 0x10a3a2950>,
'__dict__': <attribute '__dict__' of 'TempClass' objects>,
'__weakref__': <attribute '__weakref__' of 'TempClass' objects>,
'__doc__': None}
Python documentation defines __dict__ as:
A dictionary or other mapping object used to store an object’s (writable) attributes.
This definition is however a bit fuzzy, which leads to a lot of wrong usage of __dict__.
Indeed, when you read this definition, can you tell what is a "writable" attribute and what it isn't?
Examples
Let's run a few examples showing how confusing and inaccurate it can be.
class A:
foo = 1
def __init__(self):
self.bar = 2
#property
def baz(self):
return self._baz
#bar.setter
def baz(self, value):
self._baz = value
>>> a = A()
>>> a.foo
1
>>> a.bar
2
Given the above class A and knowing __dict__'s definition, can you guess what would be the value of a.__dict__?
Is foo a writable attribute of a?
Is bar a writable attribute of a?
Is baz a writable attribute of a?
Is _baz a writable attribute of a?
Here is the answer:
>>> a.__dict__
{'bar': 2}
Surprisingly, foo doesn't show up. Indeed, although accessible with a.foo, it is an attribute of the class A, not of the instance a.
Now what happens if we define it explicitly as an attribute of a?
>>> a.foo = 1
>>> a.__dict__
{'bar': 2, 'foo': 1}
From our point of view, nothing really changed, a.foo is still equal to 1, but now it shows up in __dict__. Note that we can keep playing with it by deleting a.foo for instance:
>>> del a.foo
>>> a.__dict__
{'bar': 2}
>>> a.foo
1
What happened here is that we deleted the instance attribute, and calling a.foo falls back again to A.foo.
Let's have a look at baz now. We can assume that we can't find it in a.__dict__ because we didn't give it a value yet.
>>> a.baz = 3
Alright, now we defined a.baz. So, is baz a writable attribute? What about _baz?
>>> a.__dict__
{'bar': 2, '_baz': 3}
From __dict__'s perspective, _baz is a writable attribute, but baz isn't. The explanation, again, is that baz is an attribute of the class A, not the instance a.
>>> A.__dict__['baz']
<property at 0x7fb1e30c9590>
a.baz is only an abstraction layer calling A.baz.fget(a) behind the scenes.
Let's be even more sneaky with our dear friend and challenge its definition of "writable".
class B:
def __init__(self):
self.foobar = 'baz'
def __setattr__(self, name, value):
if name == 'foobar' and 'foobar' in self.__dict__:
# Allow the first definition of foobar
# but prevent any subsequent redefinition
raise TypeError("'foobar' is a read-only attribute")
super().__setattr__(name, value)
>>> b = B()
>>> b.foobar
'baz'
>>> b.foobar = 'bazbar'
TypeError: 'foobar' is a read-only attribute
>>> # From our developer's perspective, 'foobar' is not a writable attribute
>>> # But __dict__ doesn't share this point of view
>>> b.__dict__
{'foobar': 'baz'}
Then what is __dict__ exactly?
Thanks to the behavior noticed in the above examples, we can now have a better understanding of what __dict__ actually does. But we need to switch from the developer's to the computer's perspective:
__dict__ contains the data stored in the program's memory for this specific object.
That's it, __dict__ exposes what's actually stored in memory at our object's address.
Python documentation on data model also defines it as the object's namespace:
A class instance has a namespace implemented as a dictionary which is the first place in which attribute references are searched. When an attribute is not found there, and the instance’s class has an attribute by that name, the search continues with the class attributes.
However, I believe thinking of __dict__ as the object's memory table gives a much better visualization of what's included in this namespace, and what's not.
But! There is a catch...
You thought we were done with __dict__?
__dict__ is not the way to deal with the object's memory footprint, but a way.
There is indeed another way: __slots__. I won't detail here how it works, there is already a very complete answer about __slots__ if you want to learn more about it. But the important thing for us is that, if slots are defined:
class C:
__slots__ = ('foo', 'bar')
def __init__(self):
self.foo = 1
self.bar = 2
>>> c = C()
>>> c.foo
1
>>> c.bar
2
>>> c.__dict__
AttributeError: 'C' object has no attribute '__dict__'
We can say "goodbye" to __dict__.
So when should I use __dict__?
As we saw, __dict__ must be seen from the computer's perspective, not from the seat of the developer. More often than not, what we consider "attributes" of our object is not directly connected to what's actually stored in memory. Especially with the use of properties or __getattr__ for instance, that add a level of abstraction for our comfort.
Although the use of __dict__ to inspect the attributes of an object will work in most trivial cases, we can't rely 100% rely on it. Which is a shame for something used to write generic logic.
The use case of __dict__ should probably be limited to inspecting an object's memory contents. Which is not so common. And keep in mind that __dict__ might not be defined at all (or lack some attributes actually stored in memory) when slots are defined.
It can also be very useful in Python's console to quickly check a class' attributes and methods. Or an object's attributes (I know I just said we can't rely on it, but in the console who cares if it fails sometimes or if it's not accurate).
Thanks but... how do I browse my object's attribute then?
We saw that __dict__ is often misused and that we can't really rely on it to inspect an object's attributes. But then, what is the correct way to do it? Is there any way to browse an objects attributes from the developer's abstracted point of view?
Yes. There are several ways to do introspection, and the correct way will depend on what you actually want to get. Instance attributes, class attributes, properties, private attributes, even methods, ... Technically, all these are attributes, and according to your situation, you might want to include some but exclude others. The context is also important. Maybe you are using a library that already exposes the attributes you want through their API.
But in general, you can rely on the inspect module.
class D:
foo = 1
__slots = ('bar', '_baz')
#property
def baz(self):
return self._baz
#baz.setter
def baz(self, value):
self._baz = value
def __init__(self):
self.bar = 2
self.baz = 3
def pointless_method(self):
pass
>>> import inspect
>>> dict((k, v) for k, v in inspect.getmembers(d) if k[0] != '_')
{'bar': 2, 'baz': 3, 'foo': 1}
>>> dict((k, getattr(d, k)) for k, v in inspect.getmembers(D) if inspect.isdatadescriptor(v) or inspect.isfunction(v))
{
'__init__': <bound method D.__init__ of <__main__.D object at 0x7fb1e26a5b40>>,
'_baz': 3,
'bar': 2,
'baz': 3,
'pointless_method': <bound method D.pointless_method of <__main__.D object at 0x7fb1e26a5b40>>
}
__dict__ can get the instance variables (data attributes) in an object as a dictionary.
So, if there is Person class below:
class Person:
x1 = "Hello"
x2 = "World"
def __init__(self, name, age):
self.name = name
self.age = age
def test1(self):
pass
#classmethod
def test2(cls):
pass
#staticmethod
def test3():
pass
obj = Person("John", 27)
print(obj.__dict__) # Here
__dict__ gets name and age with their values in a dictionary as shown below:
{'name': 'John', 'age': 27}
And, if the new instance variable gender is added after instanciation as shown below:
# ...
obj = Person("John", 27)
obj.gender = "Male" # Here
print(obj.__dict__)
__dict__ gets name, age and gender with their values in a dictionary as shown below:
{'name': 'John', 'age': 27, 'gender': 'Male'}
In addition, if using dir() as shown below:
# ...
obj = Person("John", 27)
obj.gender = "Male"
print(dir(obj)) # Here
We can get all in an object as a list as shown below:
['__class__', '__delattr__', '__dict__', '__dir__', '__doc__', '__eq__',
'__format__', '__ge__', '__getattribute__', '__gt__', '__hash__',
'__init__', '__init_subclass__', '__le__', '__lt__', '__module__',
'__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__',
'__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__',
'age', 'gender', 'name', 'test1', 'test2', 'test3', 'x1', 'x2']
And, as far as I researched and as I asked this question, there are no functions to get only the static or special variables or normal, class, static or special methods in an object in Python.
import pickle
class A:
pass
pickle.dumps(B().__reduce__())
yields
(<function _reconstructor at 0x1010143b0>, (<class '__main__.B'>, <class 'object'>, None))
What is this function "_reconstructor". It's neither B, B.__init__, nor B.__new__ as I expected.
I have to make 2 changes to get that result:
Change the name of your class from A to B.
Remove the outer pickle.dumps() call.
In any case, pickle is free to do anything it likes to reconstruct the object ;-) In this case, you can find the _reconstructor() function in Lib/copyreg.py.