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Convert 1d array to lower triangular matrix

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.

numpy.tril

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


3 Answers
3

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

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 -

5k x 5k

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


k

n*(n-1)

You could also use the numpy function np.tril_indices:

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]])


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)


[[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]]


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