Most efficient way to initialize a matrix in python

Question

Whats the most efficient way to initialize a matrix in Python without using numpy. For example I'd like to create:

matrix = [
           [[0,0],[0,0],[0,0]]
           [[0,0],[0,0],[0,0]]
           [[0,0],[0,0],[0,0]]
         ]

I was creating one using:

dpcols = [[0,0]] * len(matrix[0])
matrix = [dpcols] * len(matrix)

But with this, if i change

matrix[1][1] = [1,1]

Then indexes [0][1], [1][1], and [2][1] all gets changed to [1,1], due to them all being linked together.

why the requirement of not using numpy?

James R
– James R

2017-03-15 20:28:14 +00:00
Commented Mar 15, 2017 at 20:28 — James R
– James R, Commented Mar 15, 2017 at 20:28
Its just a restriction, cant use certain libraries

user1179317
– user1179317

2017-03-15 21:16:12 +00:00
Commented Mar 15, 2017 at 21:16 — user1179317
– user1179317, Commented Mar 15, 2017 at 21:16

mitoRibo · Accepted Answer · 2017-03-15 20:35:32Z

4

I don't think this is as efficient as numpy since it's a double for loop, but it's easy to write:

>>> rows,cols = 2,3
>>> a = [[[0,0] for c in range(cols)] for r in range(rows)]
>>> a
[[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]]
>>> a[1][1] = [1,1]
>>> a
[[[0, 0], [0, 0], [0, 0]], [[0, 0], [1, 1], [0, 0]]]

edited Mar 15, 2017 at 20:35

answered Mar 15, 2017 at 20:32

mitoRibo

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2 Comments

TemporalWolf Over a year ago

matrix = [[[0, 0] for _ in range(cols)] for _ in range(rows)] in his use-case

user1179317 Over a year ago

Was wondering if there are other ways without using 2 loops, but i guess this is the best i can do. Thanks for the help

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