Update value in Multidimensional list in Python

Question

I have an list like the following

line_37_data = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

When I print line_37_data[0][0] , the value 0 is printed.

When I update the list as line_37_data[0][0] = 5, the list gets modified like below

[[5, 0, 0], [5, 0, 0], [5, 0, 0]]

How can I can update the value in the list based on the index ?

Note :- I don't use NumPy. This is pure plain Python without any libraries. I am using 2.7 and not Python 3

Python 3.6.1 (default, Dec 2015, 13:05:11) [GCC 4.8.2] on linux [[5, 0, 0], [0, 0, 0], [0, 0, 0]] (this also happens in 2.7) — Max Deepfield, Aug 18 '17 at 21:25
What you are claiming is not the case. Problem seems somewhere else. — haccks, Aug 18 '17 at 21:26
@MaxDeepfield , thanks for the reply. I am using Python 2 and not 3 — RandomWanderer, Aug 18 '17 at 21:26
It looks like you're doing something like `x = [0, 0, 0]; y = [x, x, x]` and that each item in the outer is a copy of the same inner list. — Brett Beatty, Aug 18 '17 at 21:27
@BrettBeatty . You are correct. I am doing the same. Initializing the list like this [[0]*3]*3 . Is this wrong ? — RandomWanderer, Aug 18 '17 at 21:28
@BrettBeatty . Thanks for correcting me. This works now. line_37_data = [[0 for i in range(3)] for j in range(3)]. — RandomWanderer, Aug 18 '17 at 21:37

score 3 · Accepted Answer · answered Aug 18 '17 at 21:37

If you pass in the same list as each element of your outer list, manipulating it will show in each place it appears. If you're just looking to fill a 2d list with zeros, list comprehension would be easy:

def generate_2d(h, w):
    return [[0 for x in range(w)] for y in range(h)]

array = generate_2d(3, 3)

# Format is array[y][x] based on names in function
array[0][0] = 5
array[1][2] = 7

assert array == [
    [5, 0, 0],
    [0, 0, 7],
    [0, 0, 0]]

Update value in Multidimensional list in Python

1 Answers1