CODEWITHSUNDEEP
pythonnumpy
Luk
Luk

Reputation: 1149

How to compare numpy arrays in terms of similarity

I am given two numpy-arrays: One of dimensions i x mand the other of dimensions j x m. What I want to do is, loop through the FirstArray and compare each of its elements with each of the elements of the SecondArray. When I say 'compare', I mean: I want to compute the Euclidean distance between the elements of FirstArray and SecondArray. Then, I want to store the index of the element of SecondArray that is closest to the corresponding element of FirstArray, and I also want to store the index of the element of SecondArray that is second closest to the element of the FirstArray.

In code this would look somewhat similar to this: 

        smallest = None
        idx = 0
        for i in range(0, FirstArrayRows):
            for j in range(0, SecondArrayRows): 
                EuclideanDistance = np.sqrt(np.sum(np.square(FirstArray[i,:] - SecondArray[j,:]))) 
                if smallest is None or EuclideanDistance < smallest:
                    smallest = EuclideanDistance
                    idx_second = idx
                    idx = j
           Closest[i] = idx 
           SecondClosest[i] = idx_second

And I think this works. However, there are two cases when this code fails to give the correct index for the second closest element of SecondArray:

  1. when the element of SecondArray that is closest to the element of FirstArray is at j = 0.
  2. when the element of SecondArray that is closest to the element of FirstArray is at j = 1.

So I wonder: Is there a better way of implementing this? I know there is. Maybe someone can help me see it?

Upvotes: 0

Views: 261

Answers (1)

Mercury
Mercury

Reputation: 4146

You could use numpy's broadcasting to your advantage. Compute the Euclidean distance with all elements of the second array in a single operation. Then, you can find the two smallest distances using argpartition.

import numpy as np

i, j, m = 3, 4, 5
a = np.random.choice(10,(i,m))
b = np.random.choice(10,(j,m))
print('First array:\n',a)
print('Second array:\n',b)

closest, second_closest = np.zeros(i), np.zeros(i)
for i in range(a.shape[0]):
    dist = np.sqrt(((a[i,:] - b)**2).sum(axis=1))
    closest[i], second_closest[i] = np.argpartition(dist, 2)[:2]

print('Closest:', closest)
print('Second Closest:', second_closest)

Output:

First array:
 [[3 9 0 2 2]
 [1 2 9 9 7]
 [4 0 6 6 4]]
Second array:
 [[9 9 2 2 3]
 [9 9 0 2 3]
 [1 1 6 7 7]
 [5 7 0 4 4]]
Closest: [3. 2. 2.]
Second Closest: [1. 3. 3.]

Upvotes: 2

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