Jean Jean - 18 days ago 5
Python Question

Vectorise Python code

I have coded a kriging algorithm but I find it quite slow. Especially, do you have an idea on how I could vectorise the piece of code in the cons function below:

import time
import numpy as np

B = np.zeros((200, 6))
P = np.zeros((len(B), len(B)))

def cons():
for i in range(len(B)):
for j in range(len(B)):
P[i,j] = corr(B[i], B[j])
return time2-time1

def corr(x,x_i):
return np.exp(-np.sum(np.abs(np.array(x) - np.array(x_i))))

time_av = 0.
for i in range(30):
print "Average=", time_av/100.

Edit: Bonus questions

  1. What happens to the broadcasting solution if I want
    corr(B[i], C[j])
    with C the same dimension than B

  2. What happens to the scipy solution if my p-norm orders are an array:

    def corr(x, x_i):
    return np.exp(-np.sum(np.abs(np.array(x) - np.array(x_i))**p))

    For 2., I tried
    P = np.exp(-cdist(B, C,'minkowski', p))
    but scipy is expecting a scalar.


Your problem seems very simple to vectorize. For each pair of rows of B you want to compute

P[i,j] = np.exp(-np.sum(np.abs(B[i,:] - B[j,:])))

You can make use of array broadcasting and introduce a third dimension, summing along the last one:

P2 = np.exp(-np.sum(np.abs(B[:,None,:] - B),axis=-1))

The idea is to reshape the first occurence of B to shape (N,1,M) while the second B is left with shape (N,M). With array broadcasting, the latter is equivalent to (1,N,M), so

B[:,None,:] - B

is of shape (N,N,M). Summing along the last index will then result in the (N,N)-shape correlation array you're looking for.

Note that if you were using scipy, you would be able to do this using scipy.spatial.distance.cdist (or, equivalently, a combination of scipy.spatial.distance.pdist and scipy.spatial.distance.squareform), without unnecessarily computing the lower triangular half of this symmetrix matrix. Using @Divakar's suggestion in comments for the simplest solution this way:

from scipy.spatial.distance import cdist
P3 = 1/np.exp(cdist(B, B, 'minkowski',1))

cdist will compute the Minkowski distance in 1-norm, which is exactly the sum of the absolute values of coordinate differences.