What's the difference between these two ways of initializing an array?

Question

matrix = [[0]*1005]*1005

and

matrix = [[0 for _ in range(1005)] for _ in range(1005)]

I found if i using the former to initial array, and run matrix[1][1] = 1 it will set the second column in every row to 1. And the second performed as i wished. I can't figure out why?

possible duplicate of ["Least Astonishment" in Python: The Mutable Default Argument](http://stackoverflow.com/questions/1132941/least-astonishment-in-python-the-mutable-default-argument) — Joel Cornett, Jul 05 '12 at 00:31

score 8 · Accepted Answer · answered Jul 05 '12 at 00:12

8

[[0] * 1005] * 1005 will create an array of 1005 references to an array with 1005 zeroes. They're all the same reference, and so changing one will change all the others. You can shorten your working code to this, though:

matrix = [[0] * 1005 for _ in range(1005)]

_{This is pretty much the only warning anybody ever gives you about the list multiplication syntax, actually :)}

answered Jul 05 '12 at 00:12

Ry-

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Can you give me a link about this in python reference. Actually, what does it means about `[0]*1005`. Thanks. – Frank Cheng Jul 05 '12 at 00:23
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@user674199: http://docs.python.org/library/stdtypes.html#typesseq. See s * n: "n shallow copies of s concatenated". – Ry- Jul 05 '12 at 00:28
`[0] * 1005` is ok since integers are immutable. – John La Rooy Jul 05 '12 at 02:42

score 3 · Answer 2 · answered Jul 05 '12 at 00:20

3

Just to expand on whats already been said

a = [[0]] * 5
print a # [[0], [0], [0], [0], [0]]
a[0][0] = 1
print a # [[1], [1], [1], [1], [1]]

answered Jul 05 '12 at 00:20

Jakob Bowyer

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What's the difference between these two ways of initializing an array?

2 Answers2