create cosine similarity matrix numpy

Sal picture Sal · Jan 28, 2017 · Viewed 13.2k times · Source

Suppose I have a numpy matrix like the following:

array([array([ 0.0072427 ,  0.00669255,  0.00785213,  0.00845336,  0.01042869]),
   array([ 0.00710799,  0.00668831,  0.00772334,  0.00777796,  0.01049965]),
   array([ 0.00741872,  0.00650899,  0.00772273,  0.00729002,  0.00919407]),
   array([ 0.00717589,  0.00627021,  0.0069514 ,  0.0079332 ,  0.01069545]),
   array([ 0.00617369,  0.00590539,  0.00738468,  0.00761699,  0.00886915])], dtype=object)

How can I generate a 5 x 5 matrix where each index of the matrix is the cosine similarity of two corresponding rows in my original matrix?

e.g. row 0 column 2's value would be the cosine similarity between row 1 and row 3 in the original matrix.

Here's what I've tried:

from sklearn.metrics import pairwise_distances
from scipy.spatial.distance import cosine
import numpy as np

#features is a column in my artist_meta data frame
#where each value is a numpy array of 5 floating point values, similar to the
#form of the matrix referenced above but larger in volume

items_mat = np.array(artist_meta['features'].values)

dist_out = 1-pairwise_distances(items_mat, metric="cosine")

The above code gives me the following error:

ValueError: setting an array element with a sequence.

Not sure why I'm getting this because each array is of the same length (5), which I've verified.

Answer

piRSquared picture piRSquared · Jan 28, 2017

let m be the array

m = np.array([
        [ 0.0072427 ,  0.00669255,  0.00785213,  0.00845336,  0.01042869],
        [ 0.00710799,  0.00668831,  0.00772334,  0.00777796,  0.01049965],
        [ 0.00741872,  0.00650899,  0.00772273,  0.00729002,  0.00919407],
        [ 0.00717589,  0.00627021,  0.0069514 ,  0.0079332 ,  0.01069545],
        [ 0.00617369,  0.00590539,  0.00738468,  0.00761699,  0.00886915]
    ])

per wikipedia: Cosine_Similarity
enter image description here

We can calculate our numerator with

d = m.T @ m

Our ‖A‖ is

norm = (m * m).sum(0, keepdims=True) ** .5

Then the similarities are

d / norm / norm.T

[[ 1.      0.9994  0.9979  0.9973  0.9977]
 [ 0.9994  1.      0.9993  0.9985  0.9981]
 [ 0.9979  0.9993  1.      0.998   0.9958]
 [ 0.9973  0.9985  0.998   1.      0.9985]
 [ 0.9977  0.9981  0.9958  0.9985  1.    ]]

The distances are

1 - d / norm / norm.T

[[ 0.      0.0006  0.0021  0.0027  0.0023]
 [ 0.0006  0.      0.0007  0.0015  0.0019]
 [ 0.0021  0.0007  0.      0.002   0.0042]
 [ 0.0027  0.0015  0.002   0.      0.0015]
 [ 0.0023  0.0019  0.0042  0.0015  0.    ]]