How can I return the index of a numpy array?

Question

I have a numpy array that I am using to complete a nearest neighbor calculation:

def all_distance_compute(matrix, vector):
    diff = matrix[0:] - matrix[vector]
    distances = np.sqrt(np.sum(diff**2, axis=1))

    for i in range(len(distances)):        
        print i
        print distances[i]
return  distances

It seems to be working based on the result distances that is returned, however, I don't know how to look at all of the values in distances and return which element in the array is the minimum.

The for loop that I have in my function is purely for diagnostics, but I was thinking I could iterate thru this way and perhaps determine the minimum this way, but I also figured numpy probably has a better means of doing this. EDIT: So as I was typing out the question, I figured I would try my suggestion of iterating to find the minimum, and I changed my function to be this: code

for i in range(len(distances)):
        if distances[i] < min and distances[i] > 0:
            min = distances[i]
            mindex = i

return  min, mindex

What is vector? Could you add a representative sample case? — Divakar
– Divakar, Commented May 11, 2017 at 16:24

Grr · Accepted Answer · 2017-05-11 16:24:56Z

1

numpy.argsort will return you the array index sorted in ascending order.

For example:

In [1]: import numpy as np

In [2]: arr = np.array([5,3,8,2,1,9])

In [3]: np.argsort(arr)
Out [3]: array([4, 3, 1, 0, 2, 5])

In [4]: arr[np.argsort(arr)]
Out [4]: array([1, 2, 3, 5, 8, 9])

answered May 11, 2017 at 16:24

Grr

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2 Comments

Michael Melillo Over a year ago

Thank you! This also works. If I do np.argsort(distances) i get the same result as my mindex variable above.

Grr Over a year ago

Indeed. The numpy solution will be at least an order of magnitude faster though.

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