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Is there a way to change a variable within a function if the variable is being passed as an argument?

If I had the code:
num = 3
def add(variable, number):
variable += number
add(num, 2)
print(num)
It would output 3 because inside the function, the temporary variable variable inside the function is being added to instead of the num variable. Is there a way to make it so when I pass num as the first argument, 3 would be added to num instead of variable and num would be changed to 5?

When you call a function with arguments, you pass the arguments value to the function. That function has no connection to the original variable and it cannot be reassigned to a new value.
The correct way to accomplish your goal here is to return a new value:
num = 3
def add(variable, number):
return variable + number
num = add(num, 2)
print(num)

You could write it like this:
num = 3
def add(n1, n2):
return n1 + n2
print(add(num, 2))
The original assignment is not going to update when printing num.

I think you need to explicitly assign the new value to the num variable. For example, you could do:
num = 3
def add(variable, number)
return variable + number
num = add(num, 2)
print(num) #Prints out 5

Your question touches on one of the more esotheric and easy to misunderstand features of Python: mutables and immutables.
Here is the official documentation on mutable and immutable sequence types;
https://docs.python.org/3/library/stdtypes.html#mutable-sequence-types
https://docs.python.org/3/library/stdtypes.html#immutable-sequence-types
As you can tell, only sequences and complex objects are mutable, and thus may be altered by the function.
Base types like integers and floats cannot be changes within the function as you describe, though strings can, as they count as sequences.
In your example, you're using an integer - so this will not work. However, with a string, array, dictionary or complex object (a class of your own design), this _will work.
If you're coming at it from another programming language like C++, think of mutables like references, and immutables like values. If you pass a value, any changes to the object referenced within the scope of the function do not live past the life of the function scope. Any changes to objects with references on them, however, do persist.
This post does an excellent job of explaining this in much more depth, and I would highly recommend skimming it;
https://medium.com/#meghamohan/mutable-and-immutable-side-of-python-c2145cf72747

The purpose of the add function is to add two numbers together which will result in a number. This is something completely different than doing operations on already existing variables. If you want to use variables both inside and outside function, you can use the keyword 'global'. This lets the function know that this is a globally defined variable and you will be able to do operations on it. So you can for example do:
num = 3
def add(number):
global num
num += number
add(2)
print(num)
Which will print 5
If you just want a function to add two numbers and assign this result to the already globally defined variable, you can use:
num = 3
def add(variable, number):
return variable + number
num = add(num, 2)
print(num)
This will also print 5

Why can't I change the default value of a function after it was defined?

i = 5
def f(arg=i):
print(arg)
i = 6
f()
I am learning Python from the official documentation. There I find the above piece of code which I am unable to understand as to why 5 is printed instead of 6. I am relatively new to Python. Can somebody help me understand the concept?

def f(arg=i) says "make me a function f where the default value for arg is whatever i is right now". At the time of defining the function, i=5.

i = 5
def f(arg=i)
print(arg)
The i is evaluated at the time of definition, so the code above has the same meaning as the code below:
def f(arg=5)
print(arg)
This means that, when the function is called without arguments, arg will have the value 5, no matter what the value of i is now.
In order to get what you want, just do the following:
def f(arg)
print(arg)
i = 6
f(i)

Because the function takes its default value on the first declaration of 'i'.
Change to i=6 on the first line if you want you code to print 6.
Hope I helped !

This is the difference between something being handled by reference vs by value. When you defined the function f you told it to set the argument's default value to i this is done by value, not by reference, so it took whatever the value of i was at that time and set the default for the function to that. Changing the value of i after that point does not change the value of arg. If you want it to work that way you could do this:
i = 5
def f(arg = None):
if (arg = None)
arg = i
print(arg)
i = 6
f()
This lets you pass a value for arg into the function as normal, but if you don't (or you explicitly pass None) it updates arg to the current value of i if arg is still None (Python's version of NULL if you're familiar with other languages)
Something similar can be done using the or operator, arg = arg or i,but that will check if arg is falsy, and when using integers like you are in your example, 0 will be caught by the check.

What others have said is true...the default is evaluated at the time of function creation, but it is not that it takes the "value of i" at the time of creation. The default is assigned the object referred to by "i" at the time of creation. This is an important point, because if that object is mutable, the default can be changed!
Here's what happens:
import inspect
i = 5 # name "i" refers to an immutable Python integer object of value 5.
print(f'i = {i} (id={id(i)})') # Note value and ID
# Create function "f" with a parameter whose default is the object
# referred to by name "i" *at this point*.
def f(arg=i):
print(f'arg = {arg} (id={id(arg)})')
# Use the inspect module to extract the defaults from the function.
# Note the value and ID
defaults = dict(inspect.getmembers(f))['__defaults__']
print(f'defaults = {defaults} (id={id(defaults[0])})')
# name "i" now refers to a different immutable Python integer object of value 6.
i = 6
print(f'i = {i} (id={id(i)})') # Note value and ID (changed!)
f() # default for function still referes to object 5.
f(i) # override the default with object currently referred to by name "i"
Output:
i = 5 (id=2731452426672) # Original object
defaults = (5,) (id=2731452426672) # default refers to same object
i = 6 (id=2731452426704) # Different object
arg = 5 (id=2731452426672) # f() default is the original object
arg = 6 (id=2731452426704) # f(i) parameter is different object
Now see the results of a mutable default:
import inspect
i = [5] # name "i" refers to an mutable Python list containing immutable integer object 5
print(f'i = {i} (id={id(i)})') # Note value and ID
# Create function "f" with a parameter whose default is the object
# referred to by name "i" *at this point*.
def f(arg=i):
print(f'arg = {arg} (id={id(arg)})')
# Use the inspect module to extract the defaults from the function.
# Note the value and ID
defaults = dict(inspect.getmembers(f))['__defaults__']
print(f'defaults = {defaults} (id={id(defaults[0])})')
# name "i" now refers to a different immutable Python integer object of value 6.
i[0] = 6 # MUTATE the content of the object "i" refers to.
print(f'i = {i} (id={id(i)})') # Note value and ID (UNCHANGED!)
f() # default for function still refers to original list object, but content changed!
i = [7] # Create a different list object
print(f'i = {i} (id={id(i)})') # Note value and ID (changed)
f(i) # override the default currently refered to by name "i"
Output:
i = [5] (id=2206901216704) # Original object
defaults = ([5],) (id=2206901216704) # default refers to original object
i = [6] (id=2206901216704) # Still original object, but content changed!
arg = [6] (id=2206901216704) # f() default refers to orginal object, but content changed!
i = [7] (id=2206901199296) # Create a new list object
arg = [7] (id=2206901199296) # f(i) parameter refers to new passed object.
This can have strange side effects if not understood well:
>>> def f(a,b=[]): # mutable default
... b.append(a)
... return b
...
>>> x = f(1)
>>> x
[1]
>>> y = f(2) # some would think this would return [2]
>>> y
[1, 2]
>>> x # x changed from [1] to [1,2] as well!
[1, 2]
Above, b refers to the original default list object. Appending to it mutates the default list. Returning it makes x refer to the same object. The default list now contains [1] so appending in the 2nd call make it [1,2]. y refers to the same default object as x so both names refer see the same object content.
To fix, make the default immutable and create a new list when the default is seen:
>>> def f(a,b=None):
... if b is None:
... b = []
... b.append(a)
... return b
...
>>> f(1)
[1]
>>> f(2)
[2]

This is because you are assigning the value when the function is created. arg at the time of creation will be defaulted to what i is in that moment. Since at the time of the function being created the value of i is 5 then that's what the default value of that argument becomes. After the initial creation of the function i in the function argument is no longer linked to i in the body.

var passed to a function by kwargs is not updated [duplicate]

This question already has answers here:
How do I pass a variable by reference?
(39 answers)
Closed 8 months ago.
In some languages you can pass a parameter by reference or value by using a special reserved word like ref or val. When you pass a parameter to a Python function it never alters the value of the parameter on leaving the function.The only way to do this is by using the global reserved word (or as i understand it currently).
Example 1:
k = 2
def foo (n):
n = n * n #clarity regarding comment below
square = n
return square
j = foo(k)
print j
print k
would show
>>4
>>2
showing k to be unchanged.
In this example the variable n is never changed
Example 2:
n = 0
def foo():
global n
n = n * n
return n
In this example the variable n is changed.
Is there any way in Python to call a function and tell Python that the parameter is either a value or reference parameter instead of using global?

There are essentially three kinds of 'function calls':
Pass by value
Pass by reference
Pass by object reference
Python is a PASS-BY-OBJECT-REFERENCE programming language.
Firstly, it is important to understand that a variable, and the value of the variable (the object) are two seperate things. The variable 'points to' the object. The variable is not the object. Again:
THE VARIABLE IS NOT THE OBJECT
Example: in the following line of code:
>>> x = []
[] is the empty list, x is a variable that points to the empty list, but x itself is not the empty list.
Consider the variable (x, in the above case) as a box, and 'the value' of the variable ([]) as the object inside the box.
PASS BY OBJECT REFERENCE (Case in python):
Here, "Object references are passed by value."
def append_one(li):
li.append(1)
x = [0]
append_one(x)
print x
Here, the statement x = [0] makes a variable x (box) that points towards the object [0].
On the function being called, a new box li is created. The contents of li are the SAME as the contents of the box x. Both the boxes contain the same object. That is, both the variables point to the same object in memory. Hence, any change to the object pointed at by li will also be reflected by the object pointed at by x.
In conclusion, the output of the above program will be:
[0, 1]
Note:
If the variable li is reassigned in the function, then li will point to a separate object in memory. x however, will continue pointing to the same object in memory it was pointing to earlier.
Example:
def append_one(li):
li = [0, 1]
x = [0]
append_one(x)
print x
The output of the program will be:
[0]
PASS BY REFERENCE:
The box from the calling function is passed on to the called function. Implicitly, the contents of the box (the value of the variable) is passed on to the called function. Hence, any change to the contents of the box in the called function will be reflected in the calling function.
PASS BY VALUE:
A new box is created in the called function, and copies of contents of the box from the calling function is stored into the new boxes.

You can not change an immutable object, like str or tuple, inside a function in Python, but you can do things like:
def foo(y):
y[0] = y[0]**2
x = [5]
foo(x)
print x[0] # prints 25
That is a weird way to go about it, however, unless you need to always square certain elements in an array.
Note that in Python, you can also return more than one value, making some of the use cases for pass by reference less important:
def foo(x, y):
return x**2, y**2
a = 2
b = 3
a, b = foo(a, b) # a == 4; b == 9
When you return values like that, they are being returned as a Tuple which is in turn unpacked.
edit:
Another way to think about this is that, while you can't explicitly pass variables by reference in Python, you can modify the properties of objects that were passed in. In my example (and others) you can modify members of the list that was passed in. You would not, however, be able to reassign the passed in variable entirely. For instance, see the following two pieces of code look like they might do something similar, but end up with different results:
def clear_a(x):
x = []
def clear_b(x):
while x: x.pop()
z = [1,2,3]
clear_a(z) # z will not be changed
clear_b(z) # z will be emptied

OK, I'll take a stab at this. Python passes by object reference, which is different from what you'd normally think of as "by reference" or "by value". Take this example:
def foo(x):
print x
bar = 'some value'
foo(bar)
So you're creating a string object with value 'some value' and "binding" it to a variable named bar. In C, that would be similar to bar being a pointer to 'some value'.
When you call foo(bar), you're not passing in bar itself. You're passing in bar's value: a pointer to 'some value'. At that point, there are two "pointers" to the same string object.
Now compare that to:
def foo(x):
x = 'another value'
print x
bar = 'some value'
foo(bar)
Here's where the difference lies. In the line:
x = 'another value'
you're not actually altering the contents of x. In fact, that's not even possible. Instead, you're creating a new string object with value 'another value'. That assignment operator? It isn't saying "overwrite the thing x is pointing at with the new value". It's saying "update x to point at the new object instead". After that line, there are two string objects: 'some value' (with bar pointing at it) and 'another value' (with x pointing at it).
This isn't clumsy. When you understand how it works, it's a beautifully elegant, efficient system.

Hope the following description sums it up well:
There are two things to consider here - variables and objects.
If you are passing a variable, then it's pass by value, which means the changes made to the variable within the function are local to that function and hence won't be reflected globally. This is more of a 'C' like behavior.
Example:
def changeval( myvar ):
myvar = 20;
print "values inside the function: ", myvar
return
myvar = 10;
changeval( myvar );
print "values outside the function: ", myvar
O/P:
values inside the function: 20
values outside the function: 10
If you are passing the variables packed inside a mutable object, like a list, then the changes made to the object are reflected globally as long as the object is not re-assigned.
Example:
def changelist( mylist ):
mylist2=['a'];
mylist.append(mylist2);
print "values inside the function: ", mylist
return
mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist
O/P:
values inside the function: [1, 2, 3, ['a']]
values outside the function: [1, 2, 3, ['a']]
Now consider the case where the object is re-assigned. In this case, the object refers to a new memory location which is local to the function in which this happens and hence not reflected globally.
Example:
def changelist( mylist ):
mylist=['a'];
print "values inside the function: ", mylist
return
mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist
O/P:
values inside the function: ['a']
values outside the function: [1, 2, 3]

Python is neither pass-by-value nor pass-by-reference. It's more of "object references are passed by value" as described here:
Here's why it's not pass-by-value. Because
def append(list):
list.append(1)
list = [0]
reassign(list)
append(list)
returns [0,1] showing that some kind of reference was clearly passed as pass-by-value does not allow a function to alter the parent scope at all.
Looks like pass-by-reference then, hu? Nope.
Here's why it's not pass-by-reference. Because
def reassign(list):
list = [0, 1]
list = [0]
reassign(list)
print list
returns [0] showing that the original reference was destroyed when list was reassigned. pass-by-reference would have returned [0,1].
For more information look here:
If you want your function to not manipulate outside scope, you need to make a copy of the input parameters that creates a new object.
from copy import copy
def append(list):
list2 = copy(list)
list2.append(1)
print list2
list = [0]
append(list)
print list

Technically python do not pass arguments by value: all by reference. But ... since python has two types of objects: immutable and mutable, here is what happens:
Immutable arguments are effectively passed by value: string, integer, tuple are all immutable object types. While they are technically "passed by reference" (like all parameters), since you can't change them in-place inside the function it looks/behaves as if it is passed by value.
Mutable arguments are effectively passed by reference: lists or dictionaries are passed by its pointers. Any in-place change inside the function like (append or del) will affect the original object.
This is how Python is designed: no copies and all are passed by reference. You can explicitly pass a copy.
def sort(array):
# do sort
return array
data = [1, 2, 3]
sort(data[:]) # here you passed a copy
Last point I would like to mention which is a function has its own scope.
def do_any_stuff_to_these_objects(a, b):
a = a * 2
del b['last_name']
number = 1 # immutable
hashmap = {'first_name' : 'john', 'last_name': 'legend'} # mutable
do_any_stuff_to_these_objects(number, hashmap)
print(number) # 1 , oh it should be 2 ! no a is changed inisde the function scope
print(hashmap) # {'first_name': 'john'}

So this is a little bit of a subtle point, because while Python only passes variables by value, every variable in Python is a reference. If you want to be able to change your values with a function call, what you need is a mutable object. For example:
l = [0]
def set_3(x):
x[0] = 3
set_3(l)
print(l[0])
In the above code, the function modifies the contents of a List object (which is mutable), and so the output is 3 instead of 0.
I write this answer only to illustrate what 'by value' means in Python. The above code is bad style, and if you really want to mutate your values you should write a class and call methods within that class, as MPX suggests.

Consider that the variable is a box and the value it points to is the "thing" inside the box:
1. Pass by reference : function shares the same box and thereby the thing inside also.
2. Pass by value : function creates a new box, a replica of the old one, including a copy of whatever thing is inside it. Eg. Java - functions create a copy of the box and the thing inside it which can be: a primitive / a reference to an object. (note that the copied reference in the new box and the original both still point to the same object, here the reference IS the thing inside the box, not the object it is pointing to)
3. Pass by object-reference: the function creates a box, but it encloses the same thing the initial box was enclosing. So in Python:
a) if the thing inside said box is mutable, changes made will reflect back in the original box (eg. lists)
b) if the thing is immutable (like python strings and numeric types), then the box inside the function will hold the same thing UNTIL you try to change its value. Once changed, the thing in the function's box is a totally new thing compared to the original one. Hence id() for that box will now give the identity of the new thing it encloses.

The answer given is
def set_4(x):
y = []
for i in x:
y.append(i)
y[0] = 4
return y
and
l = [0]
def set_3(x):
x[0] = 3
set_3(l)
print(l[0])
which is the best answer so far as it does what it says in the question. However,it does seem a very clumsy way compared to VB or Pascal.Is it the best method we have?
Not only is it clumsy, it involves mutating the original parameter in some way manually eg by changing the original parameter to a list: or copying it to another list rather than just saying: "use this parameter as a value " or "use this one as a reference". Could the simple answer be there is no reserved word for this but these are great work arounds?

class demoClass:
x = 4
y = 3
foo1 = demoClass()
foo1.x = 2
foo2 = demoClass()
foo2.y = 5
def mySquare(myObj):
myObj.x = myObj.x**2
myObj.y = myObj.y**2
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)
mySquare(foo1)
mySquare(foo2)
print('After square:')
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)

In Python the passing by reference or by value has to do with what are the actual objects you are passing.So,if you are passing a list for example,then you actually make this pass by reference,since the list is a mutable object.Thus,you are passing a pointer to the function and you can modify the object (list) in the function body.
When you are passing a string,this passing is done by value,so a new string object is being created and when the function terminates it is destroyed.
So it all has to do with mutable and immutable objects.

Python already call by ref..
let's take example:
def foo(var):
print(hex(id(var)))
x = 1 # any value
print(hex(id(x))) # I think the id() give the ref...
foo(x)
OutPut
0x50d43700 #with you might give another hex number deppend on your memory
0x50d43700

How to access a list passed a parameter to a function inside the function?

How to access the list sent as a parameter to function inside the function?
def my_fun(mylist):
mylist[0]=a
mylist[1]=b
mylist[2]=c
mylist[3]=d
e=a+b
f=c+d
return (e,f))
val = [1,2,3,4]
my_fun(val)
NameError: name 'a' is not defined

You dont want to assign your list index to a variable (mylist[0] = a) but the other way around (a = mylist[0]), so that you can store the values when a list is passed as an argument.
def my_fun(mylist):
a = mylist[0]
b = mylist[1]
c = mylist[2]
d = mylist[3]
e=a+b
f=c+d
return (e,f)
def main():
val = [1,2,3,4]
x = my_fun(val)
print(x)
main()

What you are looking is globals() scope dictionary. Python has 2 functions locals() that return local variable dictionary inside the scope, and globals() that return global variables outside the scope. Both functions return a dict which change during execution. So you can change, check for change or even change them artificially during your code.
Let's say you have this type of code:
a = 4;
b = 3;
c = 5;
d = 2;
# version 1
def my_fun(mylist):
global a,b,c,d; # mentioning them as global variables
mylist[0]=a
mylist[1]=b
mylist[2]=c
mylist[3]=d
e=a+b
f=c+d
return (e,f)
By declaring global a,b,c,d you are placing them temporarily in global namespace. This way you can access the global variables from inside the function.
# or version 2
def my_fun2(mylist):
mylist[0]=globals()['a']
mylist[1]=globals()['b']
mylist[2]=globals()['c']
mylist[3]=globals()['d']
e=globals()['a']+globals()['b']
f=globals()['c']+globals()['d']
return (e,f)
By calling globals() and calling the variable name ['a'] you are accessing directly global namespace and as any dictionary you can change it by your will. But be carefull, changing variables starting with and ending __ (underscores) [ like __name__ can mess up your code, sometimes even interpreter. So keep track what you are changing and what you are changing it to.
#version 3
def my_fun3(mylist):
mylist[0]=globals()['a']
mylist[1]=globals()['b']
mylist[2]=globals()['c']
mylist[3]=globals()['d']
# changing of the global variables.
globals()['a'] = mylist[0]+mylist[1];
globals()['b'] = mylist[2]+mylist[3];
Be careful with this information, and test them in interpreter(IDE) first before committing to your code.
To get more information about function ( name, argument names, constants and etc ), you can do next:
>>def a():return 0
>>a.__code__ #grabs code object
>>a.__code__.co_name # gets the name of the function you can search
And more information on what you can call from __code__ object is dir(f_name.__code__) where f_name is a in this case.

Hole-in-scope, dead code or why such output?

Code
def change1(list1):
list1[1] = list1[1] + 5
def change2(number):
number = number + 2
def main():
numbers = [4, 8, 12]
change1(numbers)
variable = 15
change2(variable)
i = 0
while i < 3:
print numbers[i]
i += 1
print variable
main()
When I read it, I thought it will output 4 8 12 15 but it outputs 4 13 12 15. I can see here that Python deals with integer and lists differently, I assumed that the last thing is impossible without global. I cannot understand the output, in such case, why would it not output 4 13 12 17?
You can see here almost identical code with different types and different reference:
$ python test2.py
4
13
12
15
$ python test3.py
4
13
12
17
$ cat test2.py test3.py
Pass-by-reference examples
test2.py: pass-by-reference and mutable data type -example. Table/list is not enough to affect the local variable in main, you need the Reference!
def change1(list1):
list1[1] = list1[1] + 5
def change2(number):
number = [x+2 for x in number]
def main():
numbers = [4, 8, 12]
change1(numbers)
variable = [15]
change2(variable)
i = 0
while i < 3:
print numbers[i]
i += 1
print variable[0]
main()
test3.py: pass-by-reference example, changing a mutable data type list/table outside the main function
def change1(list1):
list1[1] = list1[1] + 5
def change2(number):
number[0] += 2
def main():
numbers = [4, 8, 12]
change1(numbers)
variable = [15]
change2(variable)
i = 0
while i < 3:
print numbers[i]
i += 1
print variable[0]
main()
pass-by-value examples
test4.py: trying to find an example with pass-by-value, why it does not work?
$ cat test4.py
# Not yet a pass-by-value example!
global variable
variable = [15]
def change1(list1):
list1[1] = list1[1] + 5
def change2(number):
number = [x+2 for x in number]
def main():
numbers = [4, 8, 12]
change1(numbers)
#variable = 15
change2(variable)
i = 0
while i < 3:
print numbers[i]
i += 1
print variable[0]
main()
$ python test4.py
4
13
12
15 # I expected 17! Why no 17?

def change1(list1):
# `list1[1] =` means you are changing the object passed in
list1[1] = list1[1] + 5
def change2(number):
# `number = ` means you create a **new local variable**, number,
# based on the `number`you passed in
number = [x+2 for x in number]
So if you want to change existing objects, you have to referene them in some way, for example in
def change3(number):
# `number[:]` is the whole existing list and you overwrite it
number[:] = [x+2 for x in number]
Note the [ .. ] when changing a list.

Python parameters are passed by reference. You mutating only one object in change1.
However, numerical values and Strings are all immutable. You cannot change the value of a passed in immutable and see that value change in the caller. Dictionaries and Lists on the other hand are mutable, and changes made to them by a called function will be preserved when the function returns.
More: http://www.penzilla.net/tutorials/python/functions/

The definitive answer is that Python is actually "call by sharing", also known as "call by object" or "call by object reference".
This has been extensively discussed before. From that article:
From time to time, people who’ve read a little CS but not a lot CS (or too much of just one kind of CS) pop up on comp.lang.python and waste a lot of energy trying to tell everyone that Python’s using some calling model that it doesn’t really use. It always turns out that they don’t really understand Python’s model, and quite often, they don’t understand their favourite model either.
But nevermind, the only thing you need to know is that Python’s model is neither “call by value” nor “call by reference” (because any attempt to use those terms for Python requires you to use non-standard definitions of the words “-value” and “-reference”). The most accurate description is CLU’s “call by object” or “call by sharing“. Or, if you prefer, “call by object reference“.
You should also read this, if you haven’t done so already.
Python's semantics are most similar to the semantics of the language CLU. The CLU Reference Manual by Liskov et al describes the semantics like this:
"We call the argument passing technique call by sharing,
because the argument objects are shared between the
caller and the called routine. This technique does not
correspond to most traditional argument passing techniques
(it is similar to argument passing in LISP). In particular it
is not call by value because mutations of arguments per-
formed by the called routine will be visible to the caller.
And it is not call by reference because access is not given
to the variables of the caller, but merely to certain objects."

In change1 you exchange the value in the list with value + 5.
In change2 you add 5 to number. The result is a new object and is not just applied to the passed variable.
If you come from C++: No there is no int& var in Python.
You get the expected result when doing this:
def change2(number):
return number + 5
variable = 15
variable = change2(variable)
If you still don't want to return a value, you could create a MutableInt class.
class MutableInt(object):
def __init__(self, value = 0):
self._value = int(value)
def __add__(self, other):
self._value += int(other)
return self
def __sub__(self, other):
self._value -= int(other)
return self
...

All the examples show call-by-value. Python only has call-by-value. There is no call-by-reference. All values in python are references (it is not possible to have an "object" as the value). Hence it is references that are copied when passed to the function. Lists are mutable, so it is possible to mutate its contents through a shared reference. In change2 you are reassigning a local variable to point to another object, which, like all assignments to local variables, has no effect on any calling scope, since it is call-by-value.