Pythonic way to sparsely randomly populate array?

Question

Problem: Populate a 10 x 10 array of zeros randomly with 10 1's, 20 2's, 30 3's.

I don't actually have to use an array, rather I just need coordinates for the positions where the values would be. It's just easier to think of in terms of an array.

I have written several solutions for this, but they all seem to be non-straight forward and non-pythonic. I am hoping someone can give me some insight. My method has been using a linear array of 0--99, choosing randomly (np.random.choice) 10 values, removing them from the array, then choosing 20 random values. After that, I convert the linear positions into (y,x) coordinates.

import numpy as np

dim = 10
grid = np.arange(dim**2)

n1 = 10
n2 = 20
n3 = 30

def populate(grid, n, dim):
    pos = np.random.choice(grid, size=n, replace=False)
    yx = np.zeros((n,2))
    for i in xrange(n):
        delPos = np.where(grid==pos[i])
        grid = np.delete(grid, delPos)
        yx[i,:] = [np.floor(pos[i]/dim), pos[i]%dim]
    return(yx, grid)

pos1, grid = populate(grid, n1, dim)
pos2, grid = populate(grid, n2, dim)
pos3, grid = populate(grid, n3, dim)

Extra Suppose when I populate the 1's, I want them all on one half of the "array." I can do it using my method (sampling from grid[dim**2/2:]), but I haven't figured out how to do the same with the other suggestions.

Answer 1

You can create a list of all coordinates, shuffle that list and take the first 60 of those (10 + 20 + 30):

>>> import random
>>> coordinates = [(i, j) for i in xrange(10) for j in xrange(10)]
>>> random.shuffle(coordinates)
>>> coordinates[:60]
[(9, 5), (6, 9), (1, 5), ..., (0, 2), (5, 9), (2, 6)]

You can then use the first 10 to insert the 10 values, the next 20 for the 20 values and the remaining for the 30 values.