Python .sort() command works as pointer?

Question

i encountered a weird thing about the .sort() command in python 2.7.

to explain i'll use a simple code -

A = [1,3,2]
B = A
B.sort()

print A

in this code i create a non sorted list A, i then copy that list into B, sort B and print A. for some reason when i print A i get a sorted list [1,2,3] even though i used the sort command on B.

on the other hand if i'll write the following code -

A = [1,3,2]
B = A
B[1] = 123

print A

it'll print A as it should - [1,3,2]

if someone can suggest an explanation it'll be great thanks

You didn't make a copy; you only created an additional reference to the same object. This has little to do with sorting; `list.sort()` sorts the list in-place so all references to the list will see the changes made. — Martijn Pieters, Oct 17 '15 at 13:40
Your second example prints `[1, 123, 2]`, as it should. It'll never print `[1, 3, 2]`. — Martijn Pieters, Oct 17 '15 at 13:41

score 0 · Accepted Answer · answered Oct 17 '15 at 13:18

0

This has nothing to do with sort().

B = A

Now B and A are two different names for, or references to, the same list object.

You can confirm this by calling id() and seeing that it reutns the same value for both names.

If you want a copy of the list, do:

B = list(A)

answered Oct 17 '15 at 13:18

Jonathon Reinhart

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does the same apply to any variable or only lists? – Ido GD Oct 17 '15 at 13:31
1

@IdoGD: it applies to any mutable object that you alter. Numbers and strings are not mutable, so you end up *rebinding* the name. `a = 5`. `a = a + 5` assigns a new object to `a`, the `int` instance is never itself altered. – Martijn Pieters Oct 17 '15 at 13:42
got it, thanks alot! – Ido GD Oct 17 '15 at 13:47

Python .sort() command works as pointer?

1 Answers1