Replace values in subarray based upon dynamic condition in Numpy

Question

I have a Python Numpy array that is a 2D array where the second dimension is a subarray of 3 elements of integers. For example:

[ [2, 3, 4], [9, 8, 7], ... [15, 14, 16] ]

For each subarray I want to replace the lowest number with a 1 and all other numbers with a 0. So the desired output from the above example would be:

[ [1, 0, 0], [0, 0, 1], ... [0, 1, 0] ]

This is a large array, so I want to exploit Numpy performance. I know about using conditions to operate on array elements, but how do I do this when the condition is dynamic? In this instance the condition needs to be something like:

newarray = (a == min(a)).astype(int)

But how do I do this across each subarray?

akuiper · Accepted Answer · 2017-03-28 17:55:40Z

1

You can specify the axis parameter to calculate a 2d array of mins(if you keep the dimension of the result), then when you do a == a.minbyrow, you will get trues at the minimum position for each sub array:

(a == a.min(1, keepdims=True)).astype(int)
#array([[1, 0, 0],
#       [0, 0, 1],
#       [0, 1, 0]])

answered Mar 28, 2017 at 17:55

akuiper

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Paul Panzer · Accepted Answer · 2017-03-28 18:30:44Z

1

How about this?

import numpy as np

a = np.random.random((4,3))

i = np.argmin(a, axis=-1)
out = np.zeros(a.shape, int)
out[np.arange(out.shape[0]), i] = 1

print(a)
print(out)

Sample output:

# [[ 0.58321885  0.18757452  0.92700724]
#  [ 0.58082897  0.12929637  0.96686648]
#  [ 0.26037634  0.55997658  0.29486454]
#  [ 0.60398426  0.72253012  0.22812904]]
# [[0 1 0]
#  [0 1 0]
#  [1 0 0]
#  [0 0 1]]

It appears to be marginally faster than the direct approach:

from timeit import timeit

def dense():
    return (a == a.min(1, keepdims=True)).astype(int)

def sparse():
    i = np.argmin(a, axis=-1)
    out = np.zeros(a.shape, int)
    out[np.arange(out.shape[0]), i] = 1
    return out

for shp in ((4,3), (10000,3), (100,10), (100000,1000)):
    a = np.random.random(shp)
    d = timeit(dense, number=40)/40
    s = timeit(sparse, number=40)/40
    print('shape, dense, sparse, ratio', '({:6d},{:6d}) {:9.6g} {:9.6g} {:9.6g}'.format(*shp, d, s, d/s))

Sample run:

# shape, dense, sparse, ratio (     4,     3) 4.22172e-06 3.1274e-06   1.34992
# shape, dense, sparse, ratio ( 10000,     3) 0.000332396 0.000245348   1.35479
# shape, dense, sparse, ratio (   100,    10) 9.8944e-06 5.63165e-06   1.75693
# shape, dense, sparse, ratio (100000,  1000)  0.344177  0.189913   1.81229

edited Mar 28, 2017 at 18:30

answered Mar 28, 2017 at 18:08

Paul Panzer

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1 Comment

Neil Kimber Over a year ago

Many thanks - this is a good answer. Numpy sure is an interesting beast. Your solution works.

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