Updating an 2d array created using 2 * operator produces unexpected side effect in python

Asked Feb 02 '23 at 03:21

Active Feb 02 '23 at 03:27

Viewed 18 times

I tried making a 2d array using 2 approaches in python. Why does approach 1 provide an unexpected side effect where all the rows get updated?

dp1 = [ [0] * 5 ] * 8
dp1[2][2] = dp1[1][1] + 1

print("DP1")
for dp in dp1: print(dp)

dp2 = [[0] * 5 for _ in range(8)]
dp2[2][2] = dp2[1][1] + 1

print("DP2")
for dp in dp2: print(dp)

Output:

DP1
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 1, 0, 0]


DP2
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 1, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]
[0, 0, 0, 0, 0]

edited Feb 02 '23 at 03:27

juanpa.arrivillaga

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asked Feb 02 '23 at 03:21

Parth

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read this answer https://stackoverflow.com/a/67842423/7552340 – Damzaky Feb 02 '23 at 03:24
1

So, that is not an array, that is a *list* which contains references to lists. In fact, what it contains is **8 references to the same list**, so first `[0] * 5` list is created, let's call it `x`, then `[x]*8` creates another list, with 8 references *to the same list* – juanpa.arrivillaga Feb 02 '23 at 03:25
1

On the other hand `[[0] * 5 for _ in range(8)]` creates a new list, `[0]*5` 8 times. – juanpa.arrivillaga Feb 02 '23 at 03:25
@Damzaky that answer is just wrong. Python *never* uses call by reference – juanpa.arrivillaga Feb 02 '23 at 03:47

Updating an 2d array created using 2 * operator produces unexpected side effect in python

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