Given a 2-d numpy array, X, of shape [m,m], I wish to apply a function and obtain a new 2-d numpy matrix P, also of shape [m,m], whose [i,j]th element is obtained as follows:
P[i][j] = exp (-|| X[i] - x[j] ||**2)
where ||.|| represents the standard L-2 norm of a vector. Is there any way faster than a simple nested for loop?
For example,
X = [[1,1,1],[2,3,4],[5,6,7]]
Then, at diagonal entries the rows accessed will be the same and the norm/magnitude of their difference will be 0. Hence,
P[0][0] = P[1][1] = P[2][2] = exp (0) = 1.0
Also,
P[0][1] = exp (- || X[0] - X[1] ||**2) = exp (- || [-1,-2,-3] || ** 2) = exp (-14)
etc.
The most trivial solution using a nested for loop is as follows:
import numpy as np
X = np.array([[1,2,3],[4,5,6],[7,8,9]])
P = np.zeros (shape=[len(X),len(X)])
for i in range (len(X)):
for j in range (len(X)):
P[i][j] = np.exp (- np.linalg.norm (X[i]-X[j])**2)
print (P)
This prints:
P = [[1.00000000e+00 1.87952882e-12 1.24794646e-47]
[1.87952882e-12 1.00000000e+00 1.87952882e-12]
[1.24794646e-47 1.87952882e-12 1.00000000e+00]]
Here, m is of the order of 5e4.
As hpaulj mentioned cdist does it better. Try the following.
from scipy.spatial.distance import cdist
import numpy as np
np.exp(-cdist(X,X,'sqeuclidean'))
Notice the sqeuclidean. This means that scipy does not take the square root so you don't have to square like you did above with the norm.