Python variable weirdness?

Question

What's going on with my Python variable? old_pos seems to be linked to pos:

Code:

pos = [7, 7]
direction = [1, 1]
old_pos = pos
print 'pos     = '+str(pos)
print 'old_pos = '+str(old_pos)
pos[0] += direction[0]
pos[1] += direction[1]
print 'pos     = '+str(pos)
print 'old_pos = '+str(old_pos)

Output:

pos     = [7, 7]
old_pos = [7, 7]
pos     = [8, 8]
old_pos = [8, 8]

However, if I replace old_pos = pos with old_pos = tuple(pos) or even old_pos = list(pos), I don't get this problem:

pos     = [7, 7]
old_pos = [7, 7]
pos     = [8, 8]
old_pos = [7, 7]

@recursive: It depends on how you define "variables", but it's usually easier to think about if you don't consider Python names to be the same thing as variables. See: python.net/crew/mwh/hacks/objectthink.html — Daniel Pryden
– Daniel Pryden, Commented Sep 1, 2010 at 15:45

recursive · Accepted Answer · 2010-09-01 14:55:56Z

13

When you say old_pos = pos, you are not creating a copy of pos, but just making another reference to the same list. If you want two lists that behave independently, you'll need to make a copy, like using the list(pos) function as you mention, or using the slice notation pos[:].

answered Sep 1, 2010 at 14:55

recursive

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Comments

Steven Rumbalski · Accepted Answer · 2010-09-01 15:36:07Z

old_pos = pos does not create a copy of the object refered to by the name pos, rather it creates a second reference called old_pos to the very same object. Actions done to pos affect the same object referred to by old_pos. In the same way, the names "Steven" and "Mr. Rumbalski" both refer to me. If you punch Steven in the face, Mr. Rumbalski will be injured, because the two names refer to the same object -- me.

Here are 3 ways to make an actual copy instead of a second reference:

Using slice notation

old_pos = pos[:]

Using the list constructor

old_pos = list(pos)

Using the copy module

from copy import copy
old_pos = copy(pos)

Note that these copies are all shallow copies, which in this case is fine. To learn about the difference between shallow copy and deep copy read the documentation of the copy module.

Richard Fearn · Accepted Answer · 2010-09-01 14:56:08Z

3

old_pos seems to be linked to pos

Correct - this:

old_pos = pos

makes old_pos and pos point to the same list. It doesn't create a new copy of pos.

answered Sep 1, 2010 at 14:56

Richard Fearn

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Comments

c_harm · Accepted Answer · 2010-09-01 14:59:58Z

3

recursive is right about the cause. You can see that they have identical memory addresses:

>>> pos = [7, 7]
>>> old_pos = pos
>>> id(pos)
4299304472
>>> id(old_pos)
4299304472

This is called passing by reference, versus passing by value. You can also remedy this situation by using the copy module.

>>> from copy import copy
>>> pos = [7, 7]
>>> old_pos = pos
>>> id(pos)
4299304472
>>> id(old_pos)
4299304472
>>> old_pos = copy(pos)
>>> id(old_pos)
4299349240

answered Sep 1, 2010 at 14:59

c_harm

1 Comment

Daniel Pryden Over a year ago

Um... it's not really passing by reference because you're not passing anything in the example. It's just that Python names have reference semantics in how they refer to Python objects.

PieterJanS · Accepted Answer · 2016-01-25 18:12:46Z

0

In addition to the aforementioned comments, some additional steps must be taken in case of multi dimensional arrays. For instance, when you have a two dimensional array a = [[0,1,2],[3,4],[5,6,7,8]], the code b = a will not create an independent copy. However, it will create another reference to the references of the same one dimensional lists within the two dimensional list. To solve this, you must use one of the aforementioned methods for every list within the two dimensional list. For example, b = [i[:] for i in a] will create an independent copy of the list.

answered Jan 25, 2016 at 18:12

PieterJanS

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