Reputation: 25
Suppose I have a NumPy array (5×5) with distances between points. The matrix is square and symmetrical.
distances = np.array([
[0, 3, 2, 1, 4],
[3, 0, 1, 3, 5],
[2, 1, 0, 7, 6],
[1, 3, 7, 0, 9],
[4, 5, 6, 9, 0],
])
I also have a points
list of 5 elements as following:
points = ["A", "D", "A", "D", "F"]
so that, for example, the distance between the first "A"
and the first "D"
of the points list is distances[0, 1]
, which is 3
.
Is it possible, using the np.where()
function to get the indices where the distance is less than 2
, the first point is "A"
and the second point is "D"
? That is, to get [0, 4]
and [2, 1]
?
If I use np.where((distances <= 2) & (distances > 0))
I get the indices of the array for all values <=2
. I want to get the indices for those values <= 2
, where the first point is "A"
and the second point is "D"
.
Upvotes: 1
Views: 128
Reputation: 13733
Something like this?
In [152]: import numpy as np
In [153]: points = np.array(['A', 'D', 'A', 'D', 'F'], dtype='str')
In [154]: distances = np.array(
...: [[0, 3, 2, 1, 4],
...: [3, 0, 1, 3, 5],
...: [2, 1, 0, 7, 6],
...: [1, 3, 7, 0, 9],
...: [4, 5, 6, 9, 0]])
In [155]: first = 'A'
In [156]: second = 'D'
In [157]: threshold = 2
In [158]: rows = np.flatnonzero(points == first)
In [159]: cols = np.flatnonzero(points == second)
In [160]: ixgrid = np.ix_(rows, cols)
In [161]: idx1, idx2 = np.where(distances[ixgrid] <= threshold)
In [162]: np.stack([rows[idx1], cols[idx2]]).T
Out[162]:
array([[0, 3],
[2, 1]], dtype=int64)
Upvotes: 1