Numpy：在给定索引数组的情况下找到具有最小值的数组中元素索引的有效方法

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.

Answer 1

Here's one vectorized way -

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

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

Answer 2

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

Answer 3

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]