Item assignment in a nested list

Question

I'm trying to change an item in a nested list. I thought this would be very straightforward. I have the following:

temp = [1, 2, 3, 4]
my_list = [temp for i in xrange(4)]
print "my_list = ", my_list

out: my_list =  [[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]

So just a normal list of lists. I want to access one item:

print my_list[0][1]

out: 2

As expected. The problem comes when changing the item. I just want to change item in my_list[0][1], but I get the following:

my_list[0][1]= "A"
print my_list

out: [[1, 'A', 3, 4], [1, 'A', 3, 4], [1, 'A', 3, 4], [1, 'A', 3, 4]]

Why is it changing the four positions, instead of just one? How to avoid it?

@ajcr I just reopened the question because the suggested question was not exactly about this problem. I will be happy if you find a correct dup. — Mazdak, Oct 27 '15 at 16:10
@Kasramvd: Hmm... I think http://stackoverflow.com/questions/240178/python-list-of-lists-changes-reflected-across-sublists-unexpectedly is the right duplicate for explaining the problem, but I agree that the solution isn't immediately obvious from those answers (so I guess the question could be left open). — Alex Riley, Oct 27 '15 at 16:15

score 2 · Accepted Answer · edited Oct 27 '15 at 16:15

2

Since lists are mutable objects when you repeat a list inside another one you just made a list of same objects (all of them point to one memory address),for getting rid of this problem You need to copy the nested lists in each iteration :

>>> temp = [1, 2, 3, 4]
>>> my_list = [temp[:] for i in xrange(4)]
>>> my_list[0][1]= "A"
>>> print my_list
[[1, 'A', 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]
>>>

edited Oct 27 '15 at 16:15

Chad S.

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answered Oct 27 '15 at 16:04

Mazdak

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This isn't caused by the mutability of lists, it's because everything in python is a reference. – Chad S. Oct 27 '15 at 16:16
1

@ChadS. but you don't see the same *outcome* with immutable objects, which are also referenced. – jonrsharpe Oct 27 '15 at 16:17
By the way... so every time I slice a list, a copy is created instead of access to the same place in memory? – Alejandro Oct 27 '15 at 16:18
@Alejandro It actually create a shallow copy of your lists read more https://docs.python.org/2/library/copy.html – Mazdak Oct 27 '15 at 16:21
@alejandro, yes. one way in which it's often used is to copy a list: `l2 = l[:]` – acushner Oct 27 '15 at 16:21
1

First time I post in the community, and feel overwhelmed with the fast responses, clarity and friendly vibe. Thanks for the answers. Very clear now – Alejandro Oct 27 '15 at 16:26
You can also make a copy of the list with `list(temp)` – Chad S. Oct 27 '15 at 16:33

Item assignment in a nested list

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