Convert 1d array to lower triangular matrix

Question

I would like to convert a 1 dimensional array into a lower, zero diagonal matrix while keeping all the digits.

I am aware of numpy.tril function, but it replaces some of the elements with zeros. I need to expand the matrix to contain all the original digits.

For example:

[10,20,40,46,33,14,12,46,52,30,59,18,11,22,30,2,11,58,22,72,12]

Should be

0
10 0
20 40 0
46 33 14 0
12 46 52 30 0
59 18 11 22 30 0
2 11 58 22 72 12 0

Divakar · Accepted Answer · 2018-07-20T09:58:08.820

11

With the input array holding all the values as required to fill up the lower diagonal places, here's one approach with masking -

def fill_lower_diag(a):
    n = int(np.sqrt(len(a)*2))+1
    mask = np.tri(n,dtype=bool, k=-1) # or np.arange(n)[:,None] > np.arange(n)
    out = np.zeros((n,n),dtype=int)
    out[mask] = a
    return out

Sample run -

In [82]: a
Out[82]: 
array([10, 20, 40, 46, 33, 14, 12, 46, 52, 30, 59, 18, 11, 22, 30,  2, 11,
       58, 22, 72, 12])

In [83]: fill_lower_diag(a)
Out[83]: 
array([[ 0,  0,  0,  0,  0,  0,  0],
       [10,  0,  0,  0,  0,  0,  0],
       [20, 40,  0,  0,  0,  0,  0],
       [46, 33, 14,  0,  0,  0,  0],
       [12, 46, 52, 30,  0,  0,  0],
       [59, 18, 11, 22, 30,  0,  0],
       [ 2, 11, 58, 22, 72, 12,  0]])

Timings on large array with 5k x 5k shape -

In [146]: np.random.seed(0)

In [147]: n = 5000

In [148]: a = np.random.randint(0,9,n*(n+1)/2)

In [149]: %timeit tril_indices_app(a) #@Brenlla's solution
1 loop, best of 3: 218 ms per loop

In [151]: %timeit fill_lower_diag(a) # From this post
10 loops, best of 3: 43.1 ms per loop

edited Jul 20 '18 at 09:58

answered Jul 20 '18 at 09:33

Divakar

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2

Yep, mask indexing can be much faster and more memory efficient (provided you are indexing [enough elements](https://stackoverflow.com/a/46043964/6091318)). My aproach would only be competitive for a small `k` relative to the triangular matrix size – Brenlla Jul 20 '18 at 10:19
@Brenlla I think its also about generating those indices vs getting the mask that showing the difference, specially with decent to large sized arrays. – Divakar Jul 20 '18 at 10:29
Yes, that's true. In your timing code shoudn't it be `n*(n-1)`? – Brenlla Jul 20 '18 at 10:39
@Brenlla Ah yes thanks. It would generate 5001 x 5001 shaped array. Timings shouldn't change much. – Divakar Jul 20 '18 at 10:42

score 10 · Answer 2 · answered Jul 20 '18 at 09:39

You could also use the numpy function np.tril_indices:

arr = np.array([10,20,40,46,33,14,12,46,52,30,59,18,11,22,30,2,11,58,22,72,12])
n = int(np.sqrt(len(arr)*2))+1

Generate zeros matrix and fill the lower part with your values:

idx = np.tril_indices(n, k=-1, m=n)
matrix = np.zeros((n,n)).astype(int)
matrix[idx] = arr
array([[ 0,  0,  0,  0,  0,  0,  0],
       [10,  0,  0,  0,  0,  0,  0],
       [20, 40,  0,  0,  0,  0,  0],
       [46, 33, 14,  0,  0,  0,  0],
       [12, 46, 52, 30,  0,  0,  0],
       [59, 18, 11, 22, 30,  0,  0],
       [ 2, 11, 58, 22, 72, 12,  0]])

hygull · Answer 3 · 2018-07-20T12:57:29.053

You can also try the below function to get the list.

import pprint

def get_triangled_list(l, rows, typ='lower'):
    if type(l) is not list:
        print 'First parameter should be a list'
        return None

    if type(rows) is not int:
        print 'Second parameter should be a list'
        return None

    if not(typ == 'lower' or typ == 'upper'):
        print 'ERROR:', typ, 'is not allowed type'
        return None

    new_l = []
    length = len(l)
    num_items = ((rows-1) * rows)/ 2

    if length != num_items:
        print 'ERROR: ', 'There should be exactly', num_items, 'items for ', rows, 'rows, found', length, 'items.'
        return None

    if typ == 'upper':
        for i in range(rows):
            temp_l = [0]*(i+1) + [l.pop(0) for j in range(7-(i+1))] 
            new_l.append(temp_l)
    elif typ=='lower':
        for i in range(rows):
            temp_l = [l.pop(0) for j in range(i)] + [0]*(rows-i)
            new_l.append(temp_l)

    return new_l

if __name__ == '__main__':
    l = [10,20,40,46,33,14,12,46,52,30,59,18,11,22,30,2,11,58,22,72,12]

    # TEST CASE 1 (LOWER TRIANGLE, default)
    new_lower = get_triangled_list(l, 7)
    pprint.pprint(new_lower)
    print('\n')

    # TEST CASE 2 (UPPER TRIANGLE, passing one more parameter)
    l = [10,20,40,46,33,14,12,46,52,30,59,18,11,22,30,2,11,58,22,72,12]
    new_upper = get_triangled_list(l, 7, 'upper')
    pprint.pprint(new_upper)

Output »

[[0, 0, 0, 0, 0, 0, 0],
 [10, 0, 0, 0, 0, 0, 0],
 [20, 40, 0, 0, 0, 0, 0],
 [46, 33, 14, 0, 0, 0, 0],
 [12, 46, 52, 30, 0, 0, 0],
 [59, 18, 11, 22, 30, 0, 0],
 [2, 11, 58, 22, 72, 12, 0]]


[[0, 10, 20, 40, 46, 33, 14],
 [0, 0, 12, 46, 52, 30, 59],
 [0, 0, 0, 18, 11, 22, 30],
 [0, 0, 0, 0, 2, 11, 58],
 [0, 0, 0, 0, 0, 22, 72],
 [0, 0, 0, 0, 0, 0, 12],
 [0, 0, 0, 0, 0, 0, 0]]

Convert 1d array to lower triangular matrix

3 Answers3

Output »

Linked