Why Python 2D array access modifies the whole column

Question

I had a confusion where I initialize a 2D array in this way:

>>> a = [[0] * 3] * 3
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0] = 1
>>> a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]

while I was expecting it to be

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

I also tried

>>> a = [[0 for _ in range(3)] for _ in range(3)]
>>> a
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> a[0][0] = 1
>>> a
[[1, 0, 0], [0, 0, 0], [0, 0, 0]]

Which worked as expected

Wonder what is the reason caused this?

Because in the first code, the x3 is making 3 references of the same list — BlueSheepToken
– BlueSheepToken, Commented Jul 31, 2019 at 6:23

BlueSheepToken · Accepted Answer · 2019-07-31 07:59:37Z

2

With the outter *3 you are making a shallow copy of a list.

This can be easily verifier by printing ids. An id is unique to each item.

a = [[0] * 3] * 3
print(*map(id, a))
# Same ID

Instead, you absolutely need to generate new lists

a = [[0] * 3 for _ in range(3)]
print(*map(id, a))
# Different ids

answered Jul 31, 2019 at 7:59

BlueSheepToken

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1 Comment

Always thought they are equivalent, thanks for your answer.