Develop Reference

Functions, methods, and how many arguments do I have to give them?

Why do the following lines give me the same result?
str.upper('hello')
and
'hello'.upper()
I tried to do the same with list.append but got a TypeError.
list.append([1])
Is the str type in Python overloaded? How can this be achieved by writing a class/function? I would appreciate an example.

list.append takes two arguments - the list to modify and the element to append. So you need to do it like this:
ls = [1]
list.append(ls, 2)
which is equivalent to the much more popular:
ls.append(2)

str.upper and list.append are both functions.
str.upper takes one argument.
>>> str.upper('test')
'TEST'
list.append takes two arguments.
>>> my_list = []
>>> list.append(my_list, 1)
>>> my_list
[1]
str.upper and list.append (like other functions) are also non-data-descriptors with a __get__ method which in this context has two implications:
When you access the function through the class via the dot notation (str.upper, list.append) the function's __get__ method (i.e. string.upper.__get__ and list.append.__get__) is called but it returns just the function itself.
When you access the function through an instance (my_string.upper, my_list.append) the function's __get__ method is called and it will return a new callable acting like the original function, but with whatever was "in front of the dot" automatically passed as the first argument. .
That's why you need to pass 1 - 1 = 0 arguments when calling my_string.upper() and 2 - 1 = 1 argument when calling my_list.append(1).
>>> 'my_string'.upper()
'MY_STRING'
>>>
>>> my_list = []
>>> my_list.append(1)
>>> my_list
[1]
You could even get these modified callables (methods) by explicitly calling __get__ and passing the argument to be bound (what has been before the dot) as its argument.
>>> my_string = 'my_string'
>>> upper_maker = str.upper.__get__(my_string)
>>> upper_maker()
'MY_STRING'
>>>
>>> my_list = []
>>> appender = list.append.__get__(my_list)
>>> appender(1)
>>> my_list
[1]
Finally, here's a short example demonstrating how descriptor instances can detect whether they are being accessed via their owner-class or via an instance.
class Descriptor:
def __get__(self, instance, owner_class):
if instance is None:
print('accessed through class')
# list.append.__get__ would return list.append here
else:
print('accessed through instance')
# list.append.__get__ would build a new callable here
# that takes one argument x and that internally calls
# list.append(instance, x)
class Class:
attribute = Descriptor()
Class.attribute # prints 'accessed through class'
instance = Class()
instance.attribute # prints 'accessed through instance'

Quoting Dave Kirbys answer from Relationship between string module and str:
There is some overlap between the string module and the str type,
mainly for historical reasons. In early versions of Python str objects
did not have methods, so all string manipulation was done with
functions from the string module. When methods were added to the str
type (in Python 1.5?) the functions were left in the string module for
compatibility, but now just forward to the equivalent str method.
However the string module also contains constants and functions that
are not methods on str, such as formatting, character translation etc.

There is nothing at all magical going on with str (except that we have a nice syntactic shortcut to creating one using ""). You can write a class that behaves like str and list to see more clearly what is happening here.
class MyClass():
def __init__(self, arg):
self.val=str(arg)
def do_thing(self):
self.val = "asdf"
def do_thing_with_arg(self, arg):
self.val = "asdf " + str(arg)
def __repr__(self):
return self.val
my_thing = MyClass("qwerty")
# this is like 'hello'.upper()
my_thing.do_thing()
print(my_thing)
# it prints 'asdf'
my_thing = MyClass("qwerty")
# this is like str.upper('hello')
MyClass.do_thing(my_thing)
print(my_thing)
# it prints 'asdf'
my_thing = MyClass("qwerty")
# this is like my_list.append('qwerty')
my_thing.do_thing_with_arg('zxcv')
print(my_thing)
# it prints 'asdf zxcv'
my_thing = MyClass("qwerty")
# this is like list.append(my_list, 'qwerty')
MyClass.do_thing_with_arg(my_thing, 'zxcv')
print(my_thing)
# it prints 'asdf zxcv'
The short version is, you're invoking what looks like an "instance method" on a class, but you are supplying the instance ('self') yourself as the first argument to the function call.

Creating classes and variable assignment

I'm trying to understand how classes work a bit better "under the hood" of python.
If I create a class Foo like so
class Foo:
bar = True
Foo is then directly accessible, such as print(Foo) or print(Foo.bar)
However, if I dynamically create create a class and don't set it to a variable like so
type('Foo',(),{'bar':True})
If done in the interpreter it shows <class '__main__.Foo'>. However, when I try to print Foo it's undefined...NameError: name 'Foo' is not defined
Does this mean that when a class is created the "traditional" way (the first Foo class above), that python automatically sets a variable for the class of the same name? Sort of like this
# I realize this is not valid, just to convey the idea
Foo = class Foo:
bar = True
If so, then why doesn't python also create a variable named Foo set to class Foo when using type() to create it?

let's compare your problem with function statements and lambdas (because they play the same role here), consider this function f :
def f ():
return 1
the above snippet of code is not an expression at all, it is a python statement that creates a function named f returning 1 upon calling it.
let's now do the same thing, but in a different way :
f = lambda : 1
the above snippet of code is a python expression (an assignment) that assigns the symbol f to the lambda expression (which is our function) lambda : 1. if we didn't do the assignment, the lambda expression would be lost, it is the same as writing >>> 1 in the python REPL and then trying after that to reference it.

Using type with 3 argument is analogous to using the lambda to create a function. Without assignment the evaluated expression is garbage collected.
However, just you can still create an instance of the class, just like you can immediately call a lambda function.
>>> lambda x: True
<function <lambda> at 0x0000022FF95AB598>
>>> type('Test', (), {'x': True})
<class '__main__.Test'>
You can also create an instance of the class, just like you can immediately call a function
>>> t = type('Test', (), {'x': True})()
>>> t.x
True
>>> type('Test2', (), {'y': 123})().y
123
>>> (lambda x: True)(1000) # any input returns True
True

From documentation
class type(name, bases, dict)
With three arguments, return a new type object. This is essentially a dynamic form of the class statement. The name string is the class name and becomes the name attribute; the bases tuple itemizes the base classes and becomes the bases attribute; and the dict dictionary is the namespace containing definitions for class body and becomes the dict attribute. For example, the following two statements create identical type objects:
class X(object):
a = 1
X = type('X', (object,), dict(a=1))
So yes, I think you have the right idea. type() does create a class but a dynamic form.

I think you're making this too complicated. If you don't assign a value / object to a symbol, it is always "lost". Doesn't matter if the value / object is a class or something else. Example:
x = 2 + 2
That assigns the value 4 to the symbol x. Compare to:
2 + 2
The operation is carried out but the result 4 isn't assigned to a symbol.
Exact situation you have with classes.

What datatype does this variable hold? [duplicate]

This question already has answers here:
What's the canonical way to check for type in Python?
(15 answers)
Closed 6 months ago.
Is there a simple way to determine if a variable is a list, dictionary, or something else?

There are two built-in functions that help you identify the type of an object. You can use type() if you need the exact type of an object, and isinstance() to check an object’s type against something. Usually, you want to use isinstance() most of the times since it is very robust and also supports type inheritance.
To get the actual type of an object, you use the built-in type() function. Passing an object as the only parameter will return the type object of that object:
>>> type([]) is list
True
>>> type({}) is dict
True
>>> type('') is str
True
>>> type(0) is int
True
This of course also works for custom types:
>>> class Test1 (object):
pass
>>> class Test2 (Test1):
pass
>>> a = Test1()
>>> b = Test2()
>>> type(a) is Test1
True
>>> type(b) is Test2
True
Note that type() will only return the immediate type of the object, but won’t be able to tell you about type inheritance.
>>> type(b) is Test1
False
To cover that, you should use the isinstance function. This of course also works for built-in types:
>>> isinstance(b, Test1)
True
>>> isinstance(b, Test2)
True
>>> isinstance(a, Test1)
True
>>> isinstance(a, Test2)
False
>>> isinstance([], list)
True
>>> isinstance({}, dict)
True
isinstance() is usually the preferred way to ensure the type of an object because it will also accept derived types. So unless you actually need the type object (for whatever reason), using isinstance() is preferred over type().
The second parameter of isinstance() also accepts a tuple of types, so it’s possible to check for multiple types at once. isinstance will then return true, if the object is of any of those types:
>>> isinstance([], (tuple, list, set))
True

Use type():
>>> a = []
>>> type(a)
<type 'list'>
>>> f = ()
>>> type(f)
<type 'tuple'>

It might be more Pythonic to use a try...except block. That way, if you have a class which quacks like a list, or quacks like a dict, it will behave properly regardless of what its type really is.
To clarify, the preferred method of "telling the difference" between variable types is with something called duck typing: as long as the methods (and return types) that a variable responds to are what your subroutine expects, treat it like what you expect it to be. For example, if you have a class that overloads the bracket operators with getattr and setattr, but uses some funny internal scheme, it would be appropriate for it to behave as a dictionary if that's what it's trying to emulate.
The other problem with the type(A) is type(B) checking is that if A is a subclass of B, it evaluates to false when, programmatically, you would hope it would be true. If an object is a subclass of a list, it should work like a list: checking the type as presented in the other answer will prevent this. (isinstance will work, however).

On instances of object you also have the:
__class__
attribute. Here is a sample taken from Python 3.3 console
>>> str = "str"
>>> str.__class__
<class 'str'>
>>> i = 2
>>> i.__class__
<class 'int'>
>>> class Test():
... pass
...
>>> a = Test()
>>> a.__class__
<class '__main__.Test'>
Beware that in python 3.x and in New-Style classes (aviable optionally from Python 2.6) class and type have been merged and this can sometime lead to unexpected results. Mainly for this reason my favorite way of testing types/classes is to the isinstance built in function.

Determine the type of a Python object
Determine the type of an object with type
>>> obj = object()
>>> type(obj)
<class 'object'>
Although it works, avoid double underscore attributes like __class__ - they're not semantically public, and, while perhaps not in this case, the builtin functions usually have better behavior.
>>> obj.__class__ # avoid this!
<class 'object'>
type checking
Is there a simple way to determine if a variable is a list, dictionary, or something else? I am getting an object back that may be either type and I need to be able to tell the difference.
Well that's a different question, don't use type - use isinstance:
def foo(obj):
"""given a string with items separated by spaces,
or a list or tuple,
do something sensible
"""
if isinstance(obj, str):
obj = str.split()
return _foo_handles_only_lists_or_tuples(obj)
This covers the case where your user might be doing something clever or sensible by subclassing str - according to the principle of Liskov Substitution, you want to be able to use subclass instances without breaking your code - and isinstance supports this.
Use Abstractions
Even better, you might look for a specific Abstract Base Class from collections or numbers:
from collections import Iterable
from numbers import Number
def bar(obj):
"""does something sensible with an iterable of numbers,
or just one number
"""
if isinstance(obj, Number): # make it a 1-tuple
obj = (obj,)
if not isinstance(obj, Iterable):
raise TypeError('obj must be either a number or iterable of numbers')
return _bar_sensible_with_iterable(obj)
Or Just Don't explicitly Type-check
Or, perhaps best of all, use duck-typing, and don't explicitly type-check your code. Duck-typing supports Liskov Substitution with more elegance and less verbosity.
def baz(obj):
"""given an obj, a dict (or anything with an .items method)
do something sensible with each key-value pair
"""
for key, value in obj.items():
_baz_something_sensible(key, value)
Conclusion
Use type to actually get an instance's class.
Use isinstance to explicitly check for actual subclasses or registered abstractions.
And just avoid type-checking where it makes sense.

You can use type() or isinstance().
>>> type([]) is list
True
Be warned that you can clobber list or any other type by assigning a variable in the current scope of the same name.
>>> the_d = {}
>>> t = lambda x: "aight" if type(x) is dict else "NOPE"
>>> t(the_d) 'aight'
>>> dict = "dude."
>>> t(the_d) 'NOPE'
Above we see that dict gets reassigned to a string, therefore the test:
type({}) is dict
...fails.
To get around this and use type() more cautiously:
>>> import __builtin__
>>> the_d = {}
>>> type({}) is dict
True
>>> dict =""
>>> type({}) is dict
False
>>> type({}) is __builtin__.dict
True

be careful using isinstance
isinstance(True, bool)
True
>>> isinstance(True, int)
True
but type
type(True) == bool
True
>>> type(True) == int
False

While the questions is pretty old, I stumbled across this while finding out a proper way myself, and I think it still needs clarifying, at least for Python 2.x (did not check on Python 3, but since the issue arises with classic classes which are gone on such version, it probably doesn't matter).
Here I'm trying to answer the title's question: how can I determine the type of an arbitrary object? Other suggestions about using or not using isinstance are fine in many comments and answers, but I'm not addressing those concerns.
The main issue with the type() approach is that it doesn't work properly with old-style instances:
class One:
pass
class Two:
pass
o = One()
t = Two()
o_type = type(o)
t_type = type(t)
print "Are o and t instances of the same class?", o_type is t_type
Executing this snippet would yield:
Are o and t instances of the same class? True
Which, I argue, is not what most people would expect.
The __class__ approach is the most close to correctness, but it won't work in one crucial case: when the passed-in object is an old-style class (not an instance!), since those objects lack such attribute.
This is the smallest snippet of code I could think of that satisfies such legitimate question in a consistent fashion:
#!/usr/bin/env python
from types import ClassType
#we adopt the null object pattern in the (unlikely) case
#that __class__ is None for some strange reason
_NO_CLASS=object()
def get_object_type(obj):
obj_type = getattr(obj, "__class__", _NO_CLASS)
if obj_type is not _NO_CLASS:
return obj_type
# AFAIK the only situation where this happens is an old-style class
obj_type = type(obj)
if obj_type is not ClassType:
raise ValueError("Could not determine object '{}' type.".format(obj_type))
return obj_type

using type()
x='hello this is a string'
print(type(x))
output
<class 'str'>
to extract only the str use this
x='this is a string'
print(type(x).__name__)#you can use__name__to find class
output
str
if you use type(variable).__name__ it can be read by us

In many practical cases instead of using type or isinstance you can also use #functools.singledispatch, which is used to define generic functions (function composed of multiple functions implementing the same operation for different types).
In other words, you would want to use it when you have a code like the following:
def do_something(arg):
if isinstance(arg, int):
... # some code specific to processing integers
if isinstance(arg, str):
... # some code specific to processing strings
if isinstance(arg, list):
... # some code specific to processing lists
... # etc
Here is a small example of how it works:
from functools import singledispatch
#singledispatch
def say_type(arg):
raise NotImplementedError(f"I don't work with {type(arg)}")
#say_type.register
def _(arg: int):
print(f"{arg} is an integer")
#say_type.register
def _(arg: bool):
print(f"{arg} is a boolean")
>>> say_type(0)
0 is an integer
>>> say_type(False)
False is a boolean
>>> say_type(dict())
# long error traceback ending with:
NotImplementedError: I don't work with <class 'dict'>
Additionaly we can use abstract classes to cover several types at once:
from collections.abc import Sequence
#say_type.register
def _(arg: Sequence):
print(f"{arg} is a sequence!")
>>> say_type([0, 1, 2])
[0, 1, 2] is a sequence!
>>> say_type((1, 2, 3))
(1, 2, 3) is a sequence!

As an aside to the previous answers, it's worth mentioning the existence of collections.abc which contains several abstract base classes (ABCs) that complement duck-typing.
For example, instead of explicitly checking if something is a list with:
isinstance(my_obj, list)
you could, if you're only interested in seeing if the object you have allows getting items, use collections.abc.Sequence:
from collections.abc import Sequence
isinstance(my_obj, Sequence)
if you're strictly interested in objects that allow getting, setting and deleting items (i.e mutable sequences), you'd opt for collections.abc.MutableSequence.
Many other ABCs are defined there, Mapping for objects that can be used as maps, Iterable, Callable, et cetera. A full list of all these can be seen in the documentation for collections.abc.

value = 12
print(type(value)) # will return <class 'int'> (means integer)
or you can do something like this
value = 12
print(type(value) == int) # will return true

type() is a better solution than isinstance(), particularly for booleans:
True and False are just keywords that mean 1 and 0 in python. Thus,
isinstance(True, int)
and
isinstance(False, int)
both return True. Both booleans are an instance of an integer. type(), however, is more clever:
type(True) == int
returns False.

In general you can extract a string from object with the class name,
str_class = object.__class__.__name__
and using it for comparison,
if str_class == 'dict':
# blablabla..
elif str_class == 'customclass':
# blebleble..

For the sake of completeness, isinstance will not work for type checking of a subtype that is not an instance. While that makes perfect sense, none of the answers (including the accepted one) covers it. Use issubclass for that.
>>> class a(list):
... pass
...
>>> isinstance(a, list)
False
>>> issubclass(a, list)
True

what is the significance of `__repr__` function over normal function [duplicate]

This question already has answers here:
Purpose of __repr__ method?
(6 answers)
Closed 5 years ago.
I am trying to learn python with my own and i stucked at __repr__ function. Though i have read lots of post on __repr__ along with the python document. so i have decided to ask this Question here. The code bellow explains my confusion.
class Point:
def __init__(self,x,y):
self.x, self.y = x,y
def __repr__(self):
return 'Point(x=%s, y=%s)'%(self.x, self.y)
def print_class(self):
return 'Point(x=%s, y=%s)'%(self.x, self.y)
p = Point(1,2)
print p
print p.print_class()
Point(x=1, y=2)
Point(x=1, y=2)
If a normal function can also perform similar task then what is the extra advantage of __repr__ over print_class() (in my case a normal function) function.

The __repr__ function is called by repr() internally. repr() is called when you are printing the object directly , and the class does not define a __str__() . From documentation -
object.__repr__(self)
Called by the repr() built-in function and by string conversions (reverse quotes) to compute the “official” string representation of an object. If at all possible, this should look like a valid Python expression that could be used to recreate an object with the same value (given an appropriate environment). If this is not possible, a string of the form <...some useful description...> should be returned. The return value must be a string object. If a class defines __repr__() but not __str__(), then __repr__() is also used when an “informal” string representation of instances of that class is required.
In your case for print_class() , you have to specifically call the method when printing the object. But in case of __repr__() , it gets internally called by print .
This is especially useful, when you are mixing different classes/types . For Example lets take a list which can have numbers and objects of your point class, now you want to print the elements of the list.
If you do not define the __repr__() or __str__() , you would have to first check the instance , whether its of type Point if so call print_class() , or if not directly print the number.
But when your class defines the __repr__() or __str__() , you can just directly call print on all the elements of the list, print statement would internally take care of printing the correct values.
Example , Lets assume a class which has print_class() method, but no __repr__() or __str__() , code -
>>> class CA:
... def __init__(self,x):
... self.x = x
... def print_class(self):
... return self.x
...
>>> l = [1,2,3,CA(4),CA(5)]
>>> for i in l:
... print(i)
...
1
2
3
<__main__.CA object at 0x00590F10>
<__main__.CA object at 0x005A5070>
SyntaxError: invalid syntax
>>> for i in l:
... if isinstance(i, CA):
... print(i.print_class())
... else:
... print(i)
...
1
2
3
4
5
As you can see, when we mix numbers and objects of type CA in the list, and then when we just did print(i) , it did not print what we wanted. For this to work correctly, we had to check the type of i and call the appropriate method (as done in second case).
Now lets assume a class that implements __repr__() instead of print_class() -
>>> class CA:
... def __init__(self,x):
... self.x = x
... def __repr__(self):
... return str(self.x)
...
>>>
>>> l = [1,2,3,CA(4),CA(5)]
>>> for i in l:
... print(i)
...
1
2
3
4
5
As you can see in second case, simply printing worked, since print internally calls __str__() first, and as that did not exist fell back to __repr__() .
And not just this, when we do str(list) , internally each list's element's __repr__() is called. Example -
First case (without __repr__() ) -
>>> str(l)
'[1, 2, 3, <__main__.CA object at 0x005AB3D0>, <__main__.CA object at 0x005AB410>]'
Second case (with __repr__() ) -
>>> str(l)
'[1, 2, 3, 4, 5]'
Also, in interactive interpreter, when you are directly using the object, it shows you the output of repr() function, Example -
>>> class CA:
... def __repr__(self):
... return "CA instance"
...
>>>
>>> c = CA()
>>> c
CA instance

The difference is that the __repr__ function is automatically called by Python in certain contexts, and is part of a predefined API with specific requirements. For instance, if you enter p by itself(not print p) in the interactive shell after creating your p object, its __repr__ will be called. It will also be used for print p if you don't define a __str__on p. (That is, you had to write print p.print_class(), but you didn't have to write print p.__repr__(); Python called __repr__ automatically for you.) The requirements for __repr__ are described in the documentation:
Called by the repr() built-in function and by string conversions (reverse quotes) to compute the “official” string representation of an object. If at all possible, this should look like a valid Python expression that could be used to recreate an object with the same value (given an appropriate environment). If this is not possible, a string of the form <...some useful description...> should be returned.
In short, if you write your own method called print_class you can make it do whatever you want and tell people how to use it, because it's your API. If you use __repr__ you're supposed to follow the conventions of Python's API. Either one may make sense depending on the context.

It helps you do more efficient coding work. even though you get same result using user define method like 'print_class()' as repr, but you don't need to type in '.print_class()' by repr method.

Check if class object

Is it possible in python to check if an object is a class object. IE if you have
class Foo(object):
pass
How could you check if o is Foo (or some other class) or an instance of Foo (or any other class instance)? In Java this would be a simple matter. Just check if the object is an instance of Class. Is there something similar in python or are you supposed to just not care?
Slight clarification: I'm trying to make a function that prints information about the parameter its given. So if you pass in o, where o = Foo() it prints out information about Foo. If you pass in Foo it should print out the exact same information. Not information about Type.

Use the isinstance builtin function.
>>> o = Foo()
>>> isinstance(o, Foo)
True
>>> isinstance(13, Foo)
False
This also works for subclasses:
>>> class Bar(Foo): pass
>>> b = Bar()
>>> isinstance(b, Foo)
True
>>> isinstance(b, Bar)
True
Yes, normally, you are supposed to not particularly care what type the object is. Instead, you just call the method you want on o so that people can plug in arbitrary objects that conform to your interface. This wouldn't be possible if you were to aggressively check the types of objects that you're using. This principle is called duck typing, and allows you a bit more freedom in how you choose to write your code.
Python is pragmatic though, so feel free to use isinstance if it makes sense for your particular program.
Edit:
To check if some variable is a class vs an instance, you can do this:
>>> isinstance(Foo, type) # returns true if the variable is a type.
True
>>> isinstance(o, type)
False

My end goal is to make a function that prints out information about an object if its an instance and print something different if its a class. So this time I do care.
First, understand that classes are instances — they're instances of type:
>>> class Foo(object):
... pass
...
>>> isinstance(Foo, type)
True
So, you can pick out classes that way, but keep in mind that classes are instances too. (And thus, you can pass classes to functions, return them from functions store them in lists, create the on the fly…)

the isinstance() function
isinstance(o, Foo)
and you can also use it to compare o to object
In [18]: class Foo(object): pass
In [20]: o_instance = Foo()
In [21]: o_class = Foo
In [22]: isinstance(o_instance, Foo)
Out[22]: True
In [23]: isinstance(o_class, Foo)
Out[23]: False
In [24]: isinstance(o_instance, object)
Out[24]: True
In [25]: isinstance(o_class, object)
Out[25]: True

I had to do like Thanatos said and check
isinstance(Foo, type)
But in the case of old class types you have to also do
isinstance(Foo, types.ClassType)