Develop Reference

Why can't I modify a list (as a whole) inside a function? [duplicate]

I moved from using Matlab to Python and the variable assignment while using functions is confusing me.
I have a code as follows:
a = [1,1,1]
def keeps(x):
y = x[:]
y[1] = 2
return y
def changes(x):
y = x
y[1] = 2
return y
aout = keeps(a)
print(a, aout)
aout = changes(a)
print(a, aout)
The first print statement gives [1, 1, 1] [1, 2, 1], while
the second one gives [1, 2, 1] [1, 2, 1].
I had a understanding (coming from Matlab) that the operations on a variable within a function are local. But here, if I don't make a copy of the variable inside a function, the values change outside the function as well. It's almost as if the variable is defined as global.
It will be very helpful if someone can explain how the variables are allocated differently in both the methods and what are the best practices if one wants to send a variable to the function without affecting it's value outside the function? Thanks.

Argument passing is done by assignment. In changes, the first thing that happens implicitly is
x = a when you call changes(a). Since assingment NEVER copies data you mutate a.
In keeps you are not mutating the argument list because x[:] is creating a (shallow) copy which then the name y is assigned to.
I highly recommend watching Facts and Myths about Python names and values.

Let's look at your code, but first, we will mode the function declarations to the top, so that the order of execution becomes clearer.
def keeps(x):
y = x[:] #Here you are creating a modifiable copy of the original x list and referencing it with y
y[1] = 2
return y
def changes(x):
y = x # Here you are just referencing x itself with a new name y
y[1] = 2
return y
a = [1,1,1]
aout = keeps(a)
print(a, aout)
aout = changes(a)
print(a, aout)
Basically if you just assign another variable name to a list, you are giving two names to the same object, so any changes in the contents may affect both "lists". When you use y = x[:]you are in fact creating a new copy of the x list in memory, through list slicing, and assigning the new variable name y to that new copy of the list.

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

Python3 - use function parameter as name of variable

I have a function that creates different forms of arrays and I don't know how to differentiate them. Is something similar to this possible?
def array_creator(nameArray):
nameArray = [0,1,2]
array_creator(a)
print(a) # prints [0,1,2]
At the moment I always run the function and then assign manually variables to store the arrays.
Thanks!

In Python you do this by returning a value from the function and binding this value to a local name, ie:
def array_creator():
return [0, 1, 2]
a = array_creator()

For your example to work you need to define your variable a before you use it. E.g. a = []. However your example won't work the way you want to. The reason for this is that you assign a new object ([1, 2, 3]) to your nameArray variable in line 2. This way you lose the reference to your object a. However it is possible to change your object a from inside the function.
def array_creator(nameArray):
nameArray.extend([0,1,2])
a = []
array_creator(a)
print(a) # prints [0,1,2]
This will work. Have a look at How to write functions with output parameters for further information.

Python does not have "output parameters", hence a plain assignment will only change the binding of the local variable, but will not modify any value, nor change bindings of variables outside the function.
However lists are mutable, so if you want to modify the argument just do so:
nameArray[:] = [0,1,2]
This will replace the contents of nameArray with 0,1,2 (works if nameArray is a list).
An alternative is to have your function simply return the value you want to assign:
def array_creator():
values = [0, 1, 2]
return values
my_arr = array_creator()
Finally, if the function wants to modify a global/nonlocal variable you have to declare it as such:
a = [1,2,3]
def array_creator():
global a
a = [0,1,2]
print(a) # [1,2,3]
array_creator()
print(a) # [0,1,2]
Or:
def wrapper():
a = [1,2,3]
def array_creator():
nonlocal a
a = [0,1,2]
return a, array_creator
a, creator = wrapper()
print(a) # [1,2,3]
creator()
print(a) # [0,1,2]
Note however that it is generally bad practice to use global variables in this way, so try to avoid it.

Python 3.4: Function unexpectedly redefines its input [duplicate]

I'm trying to understand Python's approach to variable scope. In this example, why is f() able to alter the value of x, as perceived within main(), but not the value of n?
def f(n, x):
n = 2
x.append(4)
print('In f():', n, x)
def main():
n = 1
x = [0,1,2,3]
print('Before:', n, x)
f(n, x)
print('After: ', n, x)
main()
Output:
Before: 1 [0, 1, 2, 3]
In f(): 2 [0, 1, 2, 3, 4]
After: 1 [0, 1, 2, 3, 4]
See also: How do I pass a variable by reference?

Some answers contain the word "copy" in the context of a function call. I find it confusing.
Python doesn't copy objects you pass during a function call ever.
Function parameters are names. When you call a function, Python binds these parameters to whatever objects you pass (via names in a caller scope).
Objects can be mutable (like lists) or immutable (like integers and strings in Python). A mutable object you can change. You can't change a name, you just can bind it to another object.
Your example is not about scopes or namespaces, it is about naming and binding and mutability of an object in Python.
def f(n, x): # these `n`, `x` have nothing to do with `n` and `x` from main()
n = 2 # put `n` label on `2` balloon
x.append(4) # call `append` method of whatever object `x` is referring to.
print('In f():', n, x)
x = [] # put `x` label on `[]` ballon
# x = [] has no effect on the original list that is passed into the function
Here are nice pictures on the difference between variables in other languages and names in Python.

You've got a number of answers already, and I broadly agree with J.F. Sebastian, but you might find this useful as a shortcut:
Any time you see varname =, you're creating a new name binding within the function's scope. Whatever value varname was bound to before is lost within this scope.
Any time you see varname.foo() you're calling a method on varname. The method may alter varname (e.g. list.append). varname (or, rather, the object that varname names) may exist in more than one scope, and since it's the same object, any changes will be visible in all scopes.
[note that the global keyword creates an exception to the first case]

f doesn't actually alter the value of x (which is always the same reference to an instance of a list). Rather, it alters the contents of this list.
In both cases, a copy of a reference is passed to the function. Inside the function,
n gets assigned a new value. Only the reference inside the function is modified, not the one outside it.
x does not get assigned a new value: neither the reference inside nor outside the function are modified. Instead, x’s value is modified.
Since both the x inside the function and outside it refer to the same value, both see the modification. By contrast, the n inside the function and outside it refer to different values after n was reassigned inside the function.

I will rename variables to reduce confusion. n -> nf or nmain. x -> xf or xmain:
def f(nf, xf):
nf = 2
xf.append(4)
print 'In f():', nf, xf
def main():
nmain = 1
xmain = [0,1,2,3]
print 'Before:', nmain, xmain
f(nmain, xmain)
print 'After: ', nmain, xmain
main()
When you call the function f, the Python runtime makes a copy of xmain and assigns it to xf, and similarly assigns a copy of nmain to nf.
In the case of n, the value that is copied is 1.
In the case of x the value that is copied is not the literal list [0, 1, 2, 3]. It is a reference to that list. xf and xmain are pointing at the same list, so when you modify xf you are also modifying xmain.
If, however, you were to write something like:
xf = ["foo", "bar"]
xf.append(4)
you would find that xmain has not changed. This is because, in the line xf = ["foo", "bar"] you have change xf to point to a new list. Any changes you make to this new list will have no effects on the list that xmain still points to.
Hope that helps. :-)

If the functions are re-written with completely different variables and we call id on them, it then illustrates the point well. I didn't get this at first and read jfs' post with the great explanation, so I tried to understand/convince myself:
def f(y, z):
y = 2
z.append(4)
print ('In f(): ', id(y), id(z))
def main():
n = 1
x = [0,1,2,3]
print ('Before in main:', n, x,id(n),id(x))
f(n, x)
print ('After in main:', n, x,id(n),id(x))
main()
Before in main: 1 [0, 1, 2, 3] 94635800628352 139808499830024
In f(): 94635800628384 139808499830024
After in main: 1 [0, 1, 2, 3, 4] 94635800628352 139808499830024
z and x have the same id. Just different tags for the same underlying structure as the article says.

My general understanding is that any object variable (such as a list or a dict, among others) can be modified through its functions. What I believe you are not able to do is reassign the parameter - i.e., assign it by reference within a callable function.
That is consistent with many other languages.
Run the following short script to see how it works:
def func1(x, l1):
x = 5
l1.append("nonsense")
y = 10
list1 = ["meaning"]
func1(y, list1)
print(y)
print(list1)

It´s because a list is a mutable object. You´re not setting x to the value of [0,1,2,3], you´re defining a label to the object [0,1,2,3].
You should declare your function f() like this:
def f(n, x=None):
if x is None:
x = []
...

n is an int (immutable), and a copy is passed to the function, so in the function you are changing the copy.
X is a list (mutable), and a copy of the pointer is passed o the function so x.append(4) changes the contents of the list. However, you you said x = [0,1,2,3,4] in your function, you would not change the contents of x in main().

Python is copy by value of reference. An object occupies a field in memory, and a reference is associated with that object, but itself occupies a field in memory. And name/value is associated with a reference. In python function, it always copy the value of the reference, so in your code, n is copied to be a new name, when you assign that, it has a new space in caller stack. But for the list, the name also got copied, but it refer to the same memory(since you never assign the list a new value). That is a magic in python!

When you are passing the command n = 2 inside the function, it finds a memory space and label it as 2. But if you call the method append, you are basically refrencing to location x (whatever the value is) and do some operation on that.

Python is a pure pass-by-value language if you think about it the right way. A python variable stores the location of an object in memory. The Python variable does not store the object itself. When you pass a variable to a function, you are passing a copy of the address of the object being pointed to by the variable.
Contrast these two functions
def foo(x):
x[0] = 5
def goo(x):
x = []
Now, when you type into the shell
>>> cow = [3,4,5]
>>> foo(cow)
>>> cow
[5,4,5]
Compare this to goo.
>>> cow = [3,4,5]
>>> goo(cow)
>>> goo
[3,4,5]
In the first case, we pass a copy the address of cow to foo and foo modified the state of the object residing there. The object gets modified.
In the second case you pass a copy of the address of cow to goo. Then goo proceeds to change that copy. Effect: none.
I call this the pink house principle. If you make a copy of your address and tell a
painter to paint the house at that address pink, you will wind up with a pink house.
If you give the painter a copy of your address and tell him to change it to a new address,
the address of your house does not change.
The explanation eliminates a lot of confusion. Python passes the addresses variables store by value.

As jouell said. It's a matter of what points to what and i'd add that it's also a matter of the difference between what = does and what the .append method does.
When you define n and x in main, you tell them to point at 2 objects, namely 1 and [1,2,3]. That is what = does : it tells what your variable should point to.
When you call the function f(n,x), you tell two new local variables nf and xf to point at the same two objects as n and x.
When you use "something"="anything_new", you change what "something" points to. When you use .append, you change the object itself.
Somehow, even though you gave them the same names, n in the main() and the n in f() are not the same entity, they only originally point to the same object (same goes for x actually). A change to what one of them points to won't affect the other. However, if you instead make a change to the object itself, that will affect both variables as they both point to this same, now modified, object.
Lets illustrate the difference between the method .append and the = without defining a new function :
compare
m = [1,2,3]
n = m # this tells n to point at the same object as m does at the moment
m = [1,2,3,4] # writing m = m + [4] would also do the same
print('n = ', n,'m = ',m)
to
m = [1,2,3]
n = m
m.append(4)
print('n = ', n,'m = ',m)
In the first code, it will print n = [1, 2, 3] m = [1, 2, 3, 4], since in the 3rd line, you didnt change the object [1,2,3], but rather you told m to point to a new, different, object (using '='), while n still pointed at the original object.
In the second code, it will print n = [1, 2, 3, 4] m = [1, 2, 3, 4]. This is because here both m and n still point to the same object throughout the code, but you modified the object itself (that m is pointing to) using the .append method... Note that the result of the second code will be the same regardless of wether you write m.append(4) or n.append(4) on the 3rd line.
Once you understand that, the only confusion that remains is really to understand that, as I said, the n and x inside your f() function and the ones in your main() are NOT the same, they only initially point to the same object when you call f().

Please allow me to edit again. These concepts are my experience from learning python by try error and internet, mostly stackoverflow. There are mistakes and there are helps.
Python variables use references, I think reference as relation links from name, memory adress and value.
When we do B = A, we actually create a nickname of A, and now the A has 2 names, A and B. When we call B, we actually are calling the A. we create a ink to the value of other variable, instead of create a new same value, this is what we call reference. And this thought would lead to 2 porblems.
when we do
A = [1]
B = A # Now B is an alias of A
A.append(2) # Now the value of A had been changes
print(B)
>>> [1, 2]
# B is still an alias of A
# Which means when we call B, the real name we are calling is A
# When we do something to B, the real name of our object is A
B.append(3)
print(A)
>>> [1, 2, 3]
This is what happens when we pass arguments to functions
def test(B):
print('My name is B')
print(f'My value is {B}')
print(' I am just a nickname, My real name is A')
B.append(2)
A = [1]
test(A)
print(A)
>>> [1, 2]
We pass A as an argument of a function, but the name of this argument in that function is B.
Same one with different names.
So when we do B.append, we are doing A.append
When we pass an argument to a function, we are not passing a variable , we are passing an alias.
And here comes the 2 problems.
the equal sign always creates a new name
A = [1]
B = A
B.append(2)
A = A[0] # Now the A is a brand new name, and has nothing todo with the old A from now on.
B.append(3)
print(A)
>>> 1
# the relation of A and B is removed when we assign the name A to something else
# Now B is a independent variable of hisown.
the Equal sign is a statesment of clear brand new name,
this was the concused part of mine
A = [1, 2, 3]
# No equal sign, we are working on the origial object,
A.append(4)
>>> [1, 2, 3, 4]
# This would create a new A
A = A + [4]
>>> [1, 2, 3, 4]
and the function
def test(B):
B = [1, 2, 3] # B is a new name now, not an alias of A anymore
B.append(4) # so this operation won't effect A
A = [1, 2, 3]
test(A)
print(A)
>>> [1, 2, 3]
# ---------------------------
def test(B):
B.append(4) # B is a nickname of A, we are doing A
A = [1, 2, 3]
test(A)
print(A)
>>> [1, 2, 3, 4]
the first problem is
the left side of and equation is always a brand new name, new variable,
unless the right side is a name, like B = A, this create an alias only
The second problem, there are something would never be changed, we cannot modify the original, can only create a new one.
This is what we call immutable.
When we do A= 123 , we create a dict which contains name, value, and adress.
When we do B = A, we copy the adress and value from A to B, all operation to B effect the same adress of the value of A.
When it comes to string, numbers, and tuple. the pair of value and adress could never be change. When we put a str to some adress, it was locked right away, the result of all modifications would be put into other adress.
A = 'string' would create a protected value and adess to storage the string 'string' . Currently, there is no built-in functions or method cound modify a string with the syntax like list.append, because this code modify the original value of a adress.
the value and adress of a string, a number, or a tuple is protected, locked, immutable.
All we can work on a string is by the syntax of A = B.method , we have to create a new name to storage the new string value.
please extend this discussion if you still get confused.
this discussion help me to figure out mutable / immutable / refetence / argument / variable / name once for all, hopely this could do some help to someone too.
##############################
had modified my answer tons of times and realized i don't have to say anything, python had explained itself already.
a = 'string'
a.replace('t', '_')
print(a)
>>> 'string'
a = a.replace('t', '_')
print(a)
>>> 's_ring'
b = 100
b + 1
print(b)
>>> 100
b = b + 1
print(b)
>>> 101
def test_id(arg):
c = id(arg)
arg = 123
d = id(arg)
return
a = 'test ids'
b = id(a)
test_id(a)
e = id(a)
# b = c = e != d
# this function do change original value
del change_like_mutable(arg):
arg.append(1)
arg.insert(0, 9)
arg.remove(2)
return
test_1 = [1, 2, 3]
change_like_mutable(test_1)
# this function doesn't
def wont_change_like_str(arg):
arg = [1, 2, 3]
return
test_2 = [1, 1, 1]
wont_change_like_str(test_2)
print("Doesn't change like a imutable", test_2)
This devil is not the reference / value / mutable or not / instance, name space or variable / list or str, IT IS THE SYNTAX, EQUAL SIGN.

How Does Calling Work In Python? [duplicate]

This question already has answers here:
Does Python make a copy of objects on assignment?
(5 answers)
How do I pass a variable by reference?
(39 answers)
Why can a function modify some arguments as perceived by the caller, but not others?
(13 answers)
Closed last month.
For a project I'm working on, I'm implementing a linked-list data-structure, which is based on the idea of a pair, which I define as:
class Pair:
def __init__(self, name, prefs, score):
self.name = name
self.score = score
self.preferences = prefs
self.next_pair = 0
self.prev_pair = 0
where self.next_pair and self.prev_pair are pointers to the previous and next links, respectively.
To set up the linked-list, I have an install function that looks like this.
def install(i, pair):
flag = 0
try:
old_pair = pair_array[i]
while old_pair.next_pair != 0:
if old_pair == pair:
#if pair in remainders: remainders.remove(pair)
return 0
if old_pair.score < pair.score:
flag = 1
if old_pair.prev_pair == 0: # we are at the beginning
old_pair.prev_pair = pair
pair.next_pair = old_pair
pair_array[i] = pair
break
else: # we are not at the beginning
pair.prev_pair = old_pair.prev_pair
pair.next_pair = old_pair
old_pair.prev_pair = pair
pair.prev_pair.next_pair = pair
break
else:
old_pair = old_pair.next_pair
if flag==0:
if old_pair == pair:
#if pair in remainders: remainders.remove(pair)
return 0
if old_pair.score < pair.score:
if old_pair.prev_pair==0:
old_pair.prev_pair = pair
pair.next_pair = old_pair
pair_array[i] = pair
else:
pair.prev_pair = old_pair.prev_pair
pair.next_pair = old_pair
old_pair.prev_pair = pair
pair.prev_pair.next_pair = pair
else:
old_pair.next_pair = pair
pair.prev_pair = old_pair
except KeyError:
pair_array[i] = pair
pair.prev_pair = 0
pair.next_pair = 0
Over the course of the program, I am building up a dictionary of these linked-lists, and taking links off of some and adding them in others. Between being pruned and re-installed, the links are stored in an intermediate array.
Over the course of debugging this program, I have come to realize that my understanding of the way Python passes arguments to functions is flawed. Consider this test case I wrote:
def test_install():
p = Pair(20000, [3, 1, 2, 50], 45)
print p.next_pair
print p.prev_pair
parse_and_get(g)
first_run()
rat = len(juggler_array)/len(circuit_array)
pref_size = get_pref_size()
print pref_size
print install(3, p)
print p.next_pair.name
print p.prev_pair
When I run this test, I get the following result.
0
0
10
None
10108
0
What I don't understand is why the second call to p.next_pair produces a different result (10108) than the first call (0). install does not return a Pair object that can overwrite the one passed in (it returns None), and it's not as though I'm passing install a pointer.
My understanding of call-by-value is that the interpreter copies the values passed into a function, leaving the caller's variables unchanged. For example, if I say
def foo(x):
x = x+1
return x
baz = 2
y = foo(baz)
print y
print baz
Then 3 and 2 should be printed, respectively. And indeed, when I test that out in the Python interpreter, that's what happens.
I'd really appreciate it if anyone can point me in the right direction here.

In Python, everything is an object. Simple assignment stores a reference to the assigned object in the assigned-to name. As a result, it is more straightforward to think of Python variables as names that are assigned to objects, rather than objects that are stored in named locations.
For example:
baz = 2
... stores in baz a pointer, or reference, to the integer object 2 which is stored elsewhere. (Since the type int is immutable, Python actually has a pool of small integers and reuses the same 2 object everywhere, but this is an implementation detail that need not concern us much.)
When you call foo(baz), foo()'s local variable x also points to the integer object 2 at first. That is, the foo()-local name x and the global name baz are names for the same object, 2. Then x = x + 1 is executed. This changes x to point to a different object: 3.
It is important to understand: x is not a box that holds 2, and 2 is then incremented to 3. No, x initially points to 2 and that pointer is then changed to point to 3. Naturally, since we did not change what object baz points to, it still points to 2.
Another way to explain it is that in Python, all argument passing is by value, but all values are references to objects.
A counter-intuitive result of this is that if an object is mutable, it can be modified through any reference and all references will "see" the change. For example, consider this:
baz = [1, 2, 3]
def foo(x):
x[0] = x[0] + 1
foo(baz)
print baz
>>> [2, 2, 3]
This seems very different from our first example. But in reality, the argument is passed the same way. foo() receives a pointer to baz under the name x and then performs an operation on it that changes it (in this case, the first element of the list is pointed to a different int object). The difference is that the name x is never pointed to a new object; it is x[0] that is modified to point to a different object. x itself still points to the same object as baz. (In fact, under the hood the assignment to x[0] becomes a method call: x.__setitem__().) Therefore baz "sees" the modification to the list. How could it not?
You don't see this behavior with integers and strings because you can't change integers or strings; they are immutable types, and when you modify them (e.g. x = x + 1) you are not actually modifying them but binding your variable name to a completely different object. If you change baz to a tuple, e.g. baz = (1, 2, 3), you will find that foo() gives you an error because you can`t assign to elements of a tuple; tuples are another immutable type. "Changing" a tuple requires creating a new one, and assignment then points the variable to the new object.
Objects of classes you define are mutable and so your Pair instance can be modified by any function it is passed into -- that is, attributes may be added, deleted, or reassigned to other objects. None of these things will re-bind any of the names pointing to your object, so all the names that currently point to it will "see" the changes.

Python does not copy anything when passing variables to a function. It is neither call-by-value nor call-by-reference, but of those two it is more similar to call-by-reference. You could think of it as "call-by-value, but the value is a reference".
If you pass a mutable object to a function, then modifying that object inside the function will affect the object everywhere it appears. (If you pass an immutable object to a function, like a string or an integer, then by definition you can't modify the object at all.)
The reason this isn't technically pass-by-reference is that you can rebind a name so that the name refers to something else entirely. (For names of immutable objects, this is the only thing you can do to them.) Rebinding a name that exists only inside a function doesn't affect any names that might exist outside the function.
In your first example with the Pair objects, you are modifying an object, so you see the effects outside of the function.
In your second example, you are not modifying any objects, you are just rebinding names to other objects (other integers in this case). baz is a name that points to an integer object (in Python, everything is an object, even integers) with a value of 2. When you pass baz to foo(x), the name x is created locally inside the foo function on the stack, and x is set to the pointer that was passed into the function -- the same pointer as baz. But x and baz are not the same thing, they only contain pointers to the same object. On the x = x+1 line, x is rebound to point to an integer object with a value of 3, and that pointer is what is returned from the function and used to bind the integer object to y.
If you rewrote your first example to explicitly create a new Pair object inside your function based on the information from the Pair object passed into it (whether this is a copy you then modify, or if you make a constructor that modifies the data on construction) then your function would not have the side-effect of modifying the object that was passed in.
Edit: By the way, in Python you shouldn't use 0 as a placeholder to mean "I don't have a value" -- use None. And likewise you shouldn't use 0 to mean False, like you seem to be doing in flag. But all of 0, None and False evaluate to False in boolean expressions, so no matter which of those you use, you can say things like if not flag instead of if flag == 0.

I suggest that you forget about implementing a linked list, and simply use an instance of a Python list. If you need something other than the default Python list, maybe you can use something from a Python module such as collections.
A Python loop to follow the links in a linked list will run at Python interpreter speed, which is to say, slowly. If you simply use the built-in list class, your list operations will happen in Python's C code, and you will gain speed.
If you need something like a list but with fast insertion and fast deletion, can you make a dict work? If there is some sort of ID value (string or integer or whatever) that can be used to impose an ordering on your values, you could just use that as a key value and gain lightning fast insert and delete of values. Then if you need to extract values in order, you can use the dict.keys() method function to get a list of key values and use that.
But if you really need linked lists, I suggest you find code written and debugged by someone else, and adapt it to your needs. Google search for "python linked list recipe" or "python linked list module".

I'm going to throw in a slightly complicating factor:
>>> def foo(x):
... x *= 2
... return x
...
Define a slightly different function using a method I know is supported for numbers, lists, and strings.
First, call it with strings:
>>> baz = "hello"
>>> y = foo(baz)
>>> y
'hellohello'
>>> baz
'hello'
Next, call it with lists:
>>> baz=[1,2,2]
>>> y = foo(baz)
>>> y
[1, 2, 2, 1, 2, 2]
>>> baz
[1, 2, 2, 1, 2, 2]
>>>
With strings, the argument isn't modified. With lists, the argument is modified.
If it were me, I'd avoid modifying arguments within methods.