The way a 2D list is initialized changes its behavior

Asked Apr 26 '18 at 15:23

Active Apr 26 '18 at 15:23

Viewed 10 times

square = [[0 for i in range(3)] for j in range(3)]
square
>> [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
square[0][0] = 1
>> [[1, 0, 0], [0, 0, 0], [0, 0, 0]]

This is the expected behavior.

But then:

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

It seems like each list within the list is passed by reference to a list stored in memory.

I found this to be weird behaviour since I didn't explicitly create the sublist as an independent variable, and I wondered if anybody had thoughts to share on this...

asked Apr 26 '18 at 15:23

Antoine Guenet

2

*"I didn't explicitly create the sublist as an independent variable"* Exactly. That's why they're all the same list. – Aran-Fey Apr 26 '18 at 15:26
1

Note that you _can_ safely do `[[0] * 3 for j in range(3)]`. – PM 2Ring Apr 26 '18 at 15:29
Right. But how is `[for in]` different than `[] *` on that level? In neither case I explicitly create a variable. – Antoine Guenet Apr 26 '18 at 15:40
Oh great the question was asked before, thanks for finding it. – Antoine Guenet Apr 26 '18 at 15:52

The way a 2D list is initialized changes its behavior

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