CODEWITHSUNDEEP
pythonnumpymatrixvectorscikit-learn
mel
mel

Reputation: 2790

Distance of an array of vector from it's own element

I've got an array of vector and I want to build a matrix that shows me the distance between its own vector. For example I've got that matrix with those 2 vectors:

[[a, b , c]
 [d, e , f]]

and I want to get that where dist is an euclidian distance for example:

[[dist(vect1,vect1), dist(vect1,vect2)]
 [dist(vect2,vect1), dist(vect2,vect2)]]

So obviously I'm expecting a symmetric matrix with null value on the diagonal. I tried something using scikit-learn.

#Create clusters containing the similar vectors from the clustering algo
labels = db.labels_
n_clusters_ = len(set(labels)) - (1 if -1 in labels else 0)
list_cluster = [[] for x in range(0,n_clusters_ + 1)]
for index, label in enumerate(labels):
    if label == -1:
        list_cluster[n_clusters_].append(sparse_matrix[index])
    else:
        list_cluster[label].append(sparse_matrix[index])
vector_rows = []
for cluster in list_cluster:
    for row in cluster:
         vector_rows.append(row)
#Create my array of vectors per cluster order
sim_matrix = np.array(vector_rows)
#Build my resulting matrix
sim_matrix = metrics.pairwise.pairwise_distances(sim_matrix, sim_matrix)

The problem is my resulting matrix is not symmetric so I guess there is something wrong in my code.

I add a little sample if you want to test, I did it with an euclidean distance vector per vector:

input_matrix = [[0, 0, 0, 3, 4, 1, 0, 2], 
                [0, 0, 0, 2, 5, 2, 0, 3], 
                [2, 1, 1, 0, 4, 0, 2, 3], 
                [3, 0, 2, 0, 5, 1, 1, 2]]
expected_result = [[0, 2, 4.58257569, 4.89897949], 
                   [2, 0, 4.35889894, 4.47213595], 
                   [4.58257569,  4.35889894, 0, 2.64575131], 
                   [4.89897949, 4.47213595, 2.64575131, 0]]

Upvotes: 1

Views: 443

Answers (1)

Tonechas
Tonechas

Reputation: 13723

The functions pdist and squareform will do the trick:

In [897]: import numpy as np
     ...: from scipy.spatial.distance import pdist, squareform

In [898]: input_matrix = np.asarray([[0, 0, 0, 3, 4, 1, 0, 2],
     ...:                            [0, 0, 0, 2, 5, 2, 0, 3],
     ...:                            [2, 1, 1, 0, 4, 0, 2, 3],
     ...:                            [3, 0, 2, 0, 5, 1, 1, 2]])

In [899]: squareform(pdist(input_matrix))
Out[899]: 
array([[0.        , 2.        , 4.58257569, 4.89897949],
       [2.        , 0.        , 4.35889894, 4.47213595],
       [4.58257569, 4.35889894, 0.        , 2.64575131],
       [4.89897949, 4.47213595, 2.64575131, 0.        ]])

As expected, the resulting distance matrix is a symmetric array.

By default pdist computes the euclidean distance. You can calculate a different distance by passing the proper value to parameter metric in the function call. For example:

In [900]: squareform(pdist(input_matrix, metric='jaccard'))
Out[900]: 
array([[0.        , 1.        , 0.875     , 0.71428571],
       [1.        , 0.        , 0.875     , 0.85714286],
       [0.875     , 0.875     , 0.        , 1.        ],
       [0.71428571, 0.85714286, 1.        , 0.        ]])

Upvotes: 2

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