Python list reference in a function.

Question

I'm having trouble understanding the odd behaviour in python functions if i pass in a list. I made the following functions:

def func(x):
    y = [4, 5, 6]
    x = y

def funcsecond(x):
    y = [4, 5, 6]
    x[1] = y[1]

x = [1, 2, 3]

When i call func(x) and then print out x , it prints out [1, 2, 3], just the way x was before, it doesnt assign the list y to x. However, if i call funcsecond(x), it assigns 5 to the second position of x. Why is that so? When i assign the whole list, it doesn't do anything, but when i assign only one element, it changes the list i originally called it with. Thank you very much and i hope you understand what im intending to say , i'm having a hard time expressing myself in English.

score 9 · Answer 1 · answered Jun 21 '12 at 11:03

9

The former rebinds the name, the latter mutates the object. Changes to the name only exist in local scope, whereas a mutated object remains mutated after the scope is exited.

answered Jun 21 '12 at 11:03

Ignacio Vazquez-Abrams

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1

Might need a little more explanation than this - then again, this is no doubt a dupe. – Gareth Latty Jun 21 '12 at 11:04
how does python determine when to rebind and when to mutate? – Vivek S Jun 21 '12 at 11:06
2

It rebinds when you assign to a bare name (e.g. `foo =`). Every other assignment operation is (usually) mutation. – Ignacio Vazquez-Abrams Jun 21 '12 at 11:07
Yeah! got it. But how can i replace all values of x with y.should i manually loop through each element? – Vivek S Jun 21 '12 at 11:10
@ignacio: great!...but one more silly question.slice assign just shallow copies..so how will i deep copy? – Vivek S Jun 21 '12 at 11:14
You need to perform the deep copy yourself before slice-assigning. – Ignacio Vazquez-Abrams Jun 21 '12 at 11:15
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/12850/discussion-between-internal-server-error-and-ignacio-vazquez-abrams) – Vivek S Jun 21 '12 at 11:17

score 4 · Accepted Answer · answered Jun 21 '12 at 11:06

4

this is happening beacuse x points to a object which is mutable.

def func(x): # here x is a local variable which refers to the object[1,2,3]
    y = [4, 5, 6]
    x = y  #now the local variable x refers to the object [4,5,6]

def funcsecond(x): # here x is a local variable which refers to the object[1,2,3]
    y = [4, 5, 6]
    x[1] = y[1]  # it means [1,2,3][1]=5 , means you changed the object x was pointing to

x = [1, 2, 3]

answered Jun 21 '12 at 11:06

Ashwini Chaudhary

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How come X suddenly becomes local variable.It should still be the reference even after assigning rit? – Vivek S Jun 21 '12 at 11:09
2

`x` is a local variable inside the functions because it is used in the arguments (inside the function header). – Ashwini Chaudhary Jun 21 '12 at 11:11
It is a reference. Bit in `func` this reference is reassigned to point to a new object. – Lev Levitsky Jun 21 '12 at 11:12
Thank you , i understand it a little bit better now. So , if I want the first function to change x to y , i just have to do x[:] = y , right ? – stensootla Jun 21 '12 at 11:12
@geekid you just need to do a shallw copy using slice assign like this x=y[:] – Vivek S Jun 21 '12 at 11:16
@InternalServerError No, that will be an assignment again. geekkid, I think you are right, that's what Ignacio's link above suggests. – Lev Levitsky Jun 21 '12 at 11:17

Python list reference in a function.

2 Answers2