I have a condensed distance matrix from scipy that I need to pass to a C function that requires the matrix be converted to the lower triangle read by rows. For example:
0 1 2 3
0 4 5
0 6
0
The condensed form of this is: [1,2,3,4,5,6] but I need to convert it to
0
1 0
2 4 0
3 5 6 0
The lower triangle read by rows is: [1,2,4,3,5,6].
I was hoping to convert the compact distance matrix to this form without creating a redundant matrix.
Here's a quick implementation--but it creates the square redundant distance matrix as an intermediate step:
In [128]: import numpy as np
In [129]: from scipy.spatial.distance import squareform
c is the condensed form of the distance matrix:
In [130]: c = np.array([1, 2, 3, 4, 5, 6])
d is the redundant square distance matrix:
In [131]: d = squareform(c)
Here's your condensed lower triangle distances:
In [132]: d[np.tril_indices(d.shape[0], -1)]
Out[132]: array([1, 2, 4, 3, 5, 6])
Here's a method that avoids forming the redundant distance matrix. The function condensed_index(i, j, n) takes the row i and column j of the redundant distance matrix, with j > i, and returns the corresponding index in the condensed distance array.
In [169]: def condensed_index(i, j, n):
...: return n*i - i*(i+1)//2 + j - i - 1
...:
As above, c is the condensed distance array.
In [170]: c
Out[170]: array([1, 2, 3, 4, 5, 6])
In [171]: n = 4
In [172]: i, j = np.tril_indices(n, -1)
Note that the arguments are reversed in the following call:
In [173]: indices = condensed_index(j, i, n)
indices gives the desired permutation of the condensed distance array.
In [174]: c[indices]
Out[174]: array([1, 2, 4, 3, 5, 6])
(Basically the same function as condensed_index(i, j, n) was given in several answers to this question.)
