Getting nearest index in Numpy

Question

import numpy as np

I have a given array (a):

a = np.array([[99,2,3,4,99],
              [6,7,8,99,10]])

I have 3 reference arrays (b,c,and d):

b = np.array([[99,12,13,14,99],
              [16,17,99,99,20]])

c = np.array([[21,22,23,24,99],
              [26,27,99,99,30]])

d = np.array([[31,32,33,34,35],
              [36,37,99,99,40]])

The reference arrays are given together in this form:

references = np.array([b,c,d])

I have to replace the value '99' in a given array 'a' using the nearest index value from the reference arrays if 'non 99' value is available.

The expected answer is:

answer = np.array([[21,2,3,4,35],
              [6,7,8,99,10]])

What is the fastest way of doing it?

unutbu · Accepted Answer · 2014-12-21 14:33:04Z

1

You could use np.select:

import numpy as np

a = np.array([[99,2,3,4,99],
              [6,7,8,99,10]])

b = np.array([[99,12,13,14,99],
              [16,17,99,99,20]])

c = np.array([[21,22,23,24,99],
              [26,27,99,99,30]])

d = np.array([[31,32,33,34,35],
              [36,37,99,99,40]])

references = np.array([b,c,d])

choices = np.concatenate([a[None, ...], references])
conditions = (choices != 99)

print(np.select(conditions, choices, default=99))

yields

[[21  2  3  4 35]
 [ 6  7  8 99 10]]

edited Dec 21, 2014 at 14:33

answered Dec 21, 2014 at 14:12

unutbu

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