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.

Ignacio Vazquez-Abrams · Accepted Answer · 2012-06-21 11:03:34Z

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, 2012 at 11:03

Ignacio Vazquez-Abrams

804k160 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Gareth Latty Over a year ago

Might need a little more explanation than this - then again, this is no doubt a dupe.

Vivek S Over a year ago

how does python determine when to rebind and when to mutate?

Ignacio Vazquez-Abrams Over a year ago

It rebinds when you assign to a bare name (e.g. foo =). Every other assignment operation is (usually) mutation.

Vivek S Over a year ago

Yeah! got it. But how can i replace all values of x with y.should i manually loop through each element?

Vivek S Over a year ago

@ignacio: great!...but one more silly question.slice assign just shallow copies..so how will i deep copy?

|

Ashwini Chaudhary · Accepted Answer · 2012-06-21 11:06:40Z

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, 2012 at 11:06

Ashwini Chaudhary

252k60 gold badges478 silver badges519 bronze badges

6 Comments

Vivek S Over a year ago

How come X suddenly becomes local variable.It should still be the reference even after assigning rit?

Ashwini Chaudhary Over a year ago

x is a local variable inside the functions because it is used in the arguments (inside the function header).

Lev Levitsky Over a year ago

It is a reference. Bit in func this reference is reassigned to point to a new object.

stensootla Over a year ago

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 ?

Vivek S Over a year ago

@geekid you just need to do a shallw copy using slice assign like this x=y[:]

|

Collectives™ on Stack Overflow

Python list reference in a function.

2 Answers 2

7 Comments

6 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

7 Comments

6 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related