Question

I have experienced some problem by using a nested list in Python in the code shown bleow.

Basically, I have a 2D list contains all 0 values, I want to update the list value in a loop.

However, Python does not produce the result I want. Is there something that I misunderstand about range() and Python list indices?

some_list = 4 * [(4 * [0])]
for i in range(3):
    for j in range(3):
        some_list[i+1][j+1] = 1
for i in range(4):
    print(some_list[i])

The results I expected are:

[0, 0, 0, 0]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]

But the actual results from Python are:

[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]
[0, 1, 1, 1]

What’s going on here?

Answer 1

The problem is caused by the fact that python chooses to pass lists around by reference.

Normally variables are passed “by value”, so they operate independently:

>>> a = 1
>>> b = a
>>> a = 2
>>> print b
1

But since lists might get pretty large, rather than shifting the whole list around memory, Python chooses to just use a reference (‘pointer’ in C terms). If you assign one to another variable, you assign just the reference to it. This means that you can have two variables pointing to the same list in memory:

>>> a = [1]
>>> b = a
>>> a[0] = 2
>>> print b
[2]

So, in your first line of code you have 4 * [0]. Now [0] is a pointer to the value 0 in memory, and when you multiply it, you get four pointers to the same place in memory. BUT when you change one of the values then Python knows that the pointer needs to change to point to the new value:

>>> a = 4 * [0]
>>> a
[0, 0, 0, 0]
>>> [id(v) for v in a]
[33302480, 33302480, 33302480, 33302480]
>>> a[0] = 1
>>> a
[1, 0, 0, 0]

The problem comes when you multiply this list – you get four copies of the list pointer. Now when you change one of the values in one list, all four change together:

>>> a[0][0] = 1
>>> a
[[1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0]]

The solution is to avoid the second multiplication. A loop does the job:

>>> some_list = [(4 * [0]) for _ in range(4)]