How to settle a value in a list that contains a list?

Question

I want to make a simple array b, and set b[0][0] to be 1. However, b[1][0], b[2][0] and b[3][0] changed as well. How to solve this?

>>> a=[0]*5
>>> b=[a]*4
>>> b
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
>>> b[0][0]
0
>>> b[0][0]=1
>>> b
[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]

score 1 · Answer 1 · answered Jul 05 '16 at 12:11

1

List b was created with list a, and all rows in b reference the same list object as a:

You can avoid this behavior by using a copy of a:

b = [a.copy()]

And to avoid each row in b pointing to the same object:

b = [a.copy() for _ in range(4)]

answered Jul 05 '16 at 12:11

Moses Koledoye

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Then, how could I create a 4*5 array with default value 0? thanks – ss1234 Jul 05 '16 at 12:13
http://stackoverflow.com/a/240205/3125566 – Moses Koledoye Jul 05 '16 at 12:15
If in this way, how could I change the value in the position I want. – ss1234 Jul 05 '16 at 12:23
The same way you would in the first case: `b[0][0]=1` – Moses Koledoye Jul 05 '16 at 12:24

How to settle a value in a list that contains a list?

1 Answers1