how to take a matrix in python?

Question

i want to create a matrix of size 1234*5678 with it being filled with 1 to 5678 in row major order?>..!!

Answer 1

Here's a forum post that has some code examples of what you are trying to achieve.

Answer 2

Or just use Numerical Python if you want to do some mathematical stuff on matrix too (like multiplication, ...). If they use row major order for the matrix layout in memory I can't tell you but it gets coverd in their documentation

Answer 3

I think you will need to use numpy to hold such a big matrix efficiently , not just computation. You have ~5e6 items of 4/8 bytes means 20/40 Mb in pure C already, several times of that in python without an efficient data structure (a list of rows, each row a list).

Now, concerning your question:

import numpy as np
a = np.empty((1234, 5678), dtype=np.int)
a[:] = np.linspace(1, 5678, 5678)

You first create an array of the requested size, with type int (I assume you know you want 4 bytes integer, which is what np.int will give you on most platforms). The 3rd line uses broadcasting so that each row (a[0], a[1], ... a[1233]) is assigned the values of the np.linspace line (which gives you an array of [1, ....., 5678]). If you want F storage, that is column major:

a = np.empty((1234, 4567), dtype=np.int, order='F')
...

The matrix a will takes only a tiny amount of memory more than an array in C, and for computation at least, the indexing capabilities of arrays are much better than python lists.

A nitpick: numeric is the name of the old numerical package for python - the recommended name is numpy.