Numpy: Efficient way of finding the index of the element in an array with the smallest value given an index array

Question

Say I have a numpy array a = np.array([1, 5, 3, 2, 4, 6, 7]). Now I have another numpy array b = np.array([-1, -2, 3, 2, -1, -3]). The length of b is smaller than or equal to a. I wanna find the index i of the smallest element in a such that b[i] > 0. So in the example above, the result will be 3 since according to b only indices 2, 3 are valid and a[2] == 3 and a[3] == 2, so index 3 is chosen.

My current solution is

    smallest = np.inf
    index = None
    for i in range(len(b)):
        if b[i] > 0:
            if(a[i] < smallest):
                smallest = a[i]
                index = i

I am not sure if I can use numpy to do it more efficiently. Any advice is appreciated. Thank you.

Divakar · Accepted Answer · 2019-02-28 19:49:17Z

1

Here's one vectorized way -

In [72]: idx = np.flatnonzero(b>0)

In [73]: idx[a[:len(b)][idx].argmin()]
Out[73]: 3

answered Feb 28, 2019 at 19:49

Divakar

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3 Comments

Paul Panzer Over a year ago

What's the purpose of that [:len(b)]?

Divakar Over a year ago

@PaulPanzer Was because a could be bigger than b, but then with idx being decided by b, I figured later, idx can't have indices larger than len(b), so isn't really needed for that specific case.

Paul Panzer Over a year ago

Ah, ok, I thought maybe it's some clever optimization I don't understand...

Paritosh Singh · Accepted Answer · 2019-02-28 19:56:58Z

0

You can use the intermediate results of indices from b to get the right index later, heres a way.

import numpy as np
a = np.array([1, 5, 3, 2, 4, 6, 7])
b = np.array([-1, -2, 3, 2, -1, -3])

indices_to_check = np.where(b > 0)[0]
result = indices_to_check[np.argmin(a[indices_to_check])]
#Output:
3

edited Feb 28, 2019 at 19:56

answered Feb 28, 2019 at 19:50

Paritosh Singh

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Comments

PiRK · Accepted Answer · 2019-02-28 19:59:24Z

0

one liner:

idx = np.argwhere(a==a[:len(b)][b>0].min())[0]

Understandable code:

shortened_a = a[:len(b)]
filtered_a = shortened_a[b>0]
smallest = filtered_a.min()
indices_of_smallest = np.argwhere(a==smallest)
first_idx = indices_of_smallest[0]

answered Feb 28, 2019 at 19:59

PiRK

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1 Comment

PiRK Over a year ago

It is code efficient. To figure out if it is computationaly efficient, you should always profile. But generally speaking, python for loops with dynamic type detection don't beat numpy C code with typed data.

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