Create list of items repeated N times without repeating itself

Question

I would like to repeat a list N times but without repeating an item itself. For example,

from itertools import *

items = [ _ for _ in range(3)]
# repeat each item twice
row = sorted(list(chain(*repeat(items, 2))))
row

[0, 0, 1, 1, 2, 2]

BUT, I would like to create another list (col), which also has 6 items:

col = [1, 2, 0, 2, 0, 1]

The goal of these lists is to create an adjacency matrix with diagonal items = 0. (COO format is my final goal so random.shuffle does not meet my need)

row = [0,0,1,1,2,2]
col = [1,2,0,2,0,1]
value = [1,1,1,1,1,1]

import scipy
mtx = scipy.sparse.coo_matrix((value, (row, col)), shape=(3, 3))
mtx.todense()

matrix([[0, 1, 1],
        [1, 0, 1],
        [1, 1, 0]])

I need the information into a COO format. Any suggestions? Thanks in advance!

Answer 1

If you want to fill all the non-diagonal indices of a matrix of size (n, m), you could do:

n, m = 3, 3
value = [1,3,7,2,1,4]

row, col = np.where(np.arange(m)[:,None] != np.arange(n))
a = np.zeros((n, m), dtype=int)
a[row, col] = value

>>> a
array([[0, 1, 3],
       [7, 0, 2],
       [1, 4, 0]])

If you just want to make a dense array from a COO spec (with specified row, col indices and corresponding values):

n, m = 3, 3
# or, if you only have row, col: n, m = np.max(np.c_[row, col], 0) + 1

row = [0,1,2,2]
col = [1,2,0,1]
value = [1,2,3,4]

a = np.zeros((n, m), dtype=int)
a[row, col] = value

>>> a
array([[0, 1, 0],
       [0, 0, 2],
       [3, 4, 0]])

Answer 2

Since you want to create a matrix of 1s with a diagonal of 0s, use numpy.identity:

N = 3

out = 1 - np.identity(N, dtype=int)

Output:

array([[0, 1, 1],
       [1, 0, 1],
       [1, 1, 0]])

If you just want to create the row/col indices:

a = np.arange(N)
row, col = np.where(a[:,None]!=a)

Output:

# row
array([0, 0, 1, 1, 2, 2])

# col
array([1, 2, 0, 2, 0, 1])