CODEWITHSUNDEEP
pythonmatrix
termlim
termlim

Reputation: 1175

Python: How to create a zero-indexed 2-dimensional matrix with a size of n

Given a value n, what is the most efficient way to create a zero-indexed matrix with the column and row size equal to the value n?

I know the following command will create a 2d matrix with 0 as all of the cell values:

[[0 for x in range(n)] for y in range(n)]

So when the n = 4, the matrix would look as follows:

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

But is there any way to create the matrix as follows?

[[ 0,  1,  2,  3],
 [ 4,  5,  6,  7],
 [ 8,  9, 10, 11],
 [12, 13, 14, 15]]

Upvotes: 0

Views: 640

Answers (3)

hiro protagonist
hiro protagonist

Reputation: 46901

you could simply use the current coordinates x and y to calculate the value x + y*n:

n = 4
print([[x + y*n for x in range(n)] for y in range(n)])

our you could use itertools.count: 

from itertools import count

n = 4
counter = count()
print([[next(counter) for x in range(n)] for y in range(n)])

Upvotes: 1

Abhishek Chomal
Abhishek Chomal

Reputation: 13

Something like this would work.

[[i for i in range(j, j + 4)] for j in range(0, 20, 4)]

[[0, 1, 2, 3],
 [4, 5, 6, 7],
 [8, 9, 10, 11],
 [12, 13, 14, 15],
 [16, 17, 18, 19]]

Upvotes: 0

sacuL
sacuL

Reputation: 51395

Since you are talking about 2-d matrices, and you're specifically asking about efficiency, it might be best to turn to numpy:

import numpy as np

n = 4    
my_matrix = np.arange(n*n).reshape(n,n)

>>> my_matrix
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])

Or you could use a list comprehension such as the following (or one of the ones suggested by @hiroprotagonist):

n = 4  
my_ListOfLists = [list(range(i,i+n)) for i in range(0,n*n,n)]

>>> my_ListOfLists
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]

Upvotes: 1

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