Python multiple dimension arrays [duplicate]

Question

Possible Duplicate:
How to initialize a two-dimensional array in Python?

In solving a simple problem regarding a two dimensional array I came across a solution on this site that explained how to declare one in Python using the overload operator.

Example:

Myarray = [[0]*3]*3

this would produce the following array (list)

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

This seems fine until you use it:

if you assign an element for example:

Myarray [0][0] = 1

you get the unexpected output:

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

In effect assigning Myarray [1][0] and Myarray[2][0] at the same time

My solution:

Myarray = [[][][]]
for i in range(0,3):
  for j in range (0,3):
     Myarray[i].append(0)

This solution works as intended:

Marray[0][1] = 1

gives you

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

Is there a simpler way to do this? This was a solution to an A level Cambridge question and seems too long winded for students compared to other languages.

When you do Myarray = [[0]*3]*3, you're actually multiplying the references. List comprehension is a one-line way to do it. — Makoto
– Makoto, Commented Nov 9, 2012 at 0:03
I like [x[:] for x in [[0]*3]*3] for non-numpy 2D arrays but I'm in the minority on that one. — DSM
– DSM, Commented Nov 9, 2012 at 0:05
@ScottHunter: There are a few Python-like languages where it is just [[0]*3]*3, because they don't do references right. Of course staying in Python, assuming you've done from numpy import *: zeros((3,3)) is actually shorter, as well as better. — abarnert
– abarnert, Commented Nov 9, 2012 at 0:24

Glorfindel · Accepted Answer · 2023-04-06 18:00:58Z

6

With vanilla Python, you could use this, a nested list comprehension

>>> m = [[0 for y in range(3)] for x in range(3)]
>>> m
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Unlike the multiplied list you showed in your example, it has the desired behavior

>>> m[1][0] = 99
>>> m
[[0, 0, 0], [99, 0, 0], [0, 0, 0]]

However, for serious use of multidimensional arrays and/or numerical programming, I'd suggest you use Numpy arrays.

edited Apr 6, 2023 at 18:00

Glorfindel

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answered Nov 9, 2012 at 0:01

Junuxx

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2 Comments

Scott Hunter Over a year ago

Better yet: [[0]*3 for x in range(3)]

Timothy Lawman Over a year ago

Both the answer above and Scotts work just fine and I thank you both. For the purpose of A level Junuxx is more readable so gets the vote!

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