Closest element to a value (Elementwise, numpy array)

Question

I used to use min([a, b], key=lambda x:abs(x-x0)) to find which of a and b are the closest to x0.

a = 1
b = 2
x0 = 1.49
print(min([a, b], key=lambda x:abs(x-x0)))
# >>> 1

Now, a and b are numpy arrays with an arbitrary number of dimensions. I would like to build an array composed of the closest values to x0 between both arrays, element by element.

import numpy as np

a = np.array([[1, 2], [3, 5]])
b = np.array([[6, 2], [6, 2]])

## case 1
x0 = 4
# >>> should return np.array([[6, 2], [3, 5]])

## case 2
x0 = np.array([[1, 2], [3, 4]])
# >>> should return np.array([[1, 2], [3, 5]])

To find the elementwise minimum between two arrays, we can use numpy.minimum. Unfortunately, it does not take lambda functions as arguments.

How should I do ?

arbitrary number of dimensions is somewhat arbitrary. Are their dimensions equal, and how do they relate to the dimension of x0? Do you have a sample in hand? — Quang Hoang
– Quang Hoang, Commented Oct 28, 2020 at 16:39
In my case, x0 is a single float. The dimensions of a and b should not play in this problem, as soon as a.shape == b.shape, right ? — Liris
– Liris, Commented Oct 28, 2020 at 16:41

Marat · Accepted Answer · 2020-10-28 17:36:47Z

1

Is this what you are looking for?

np.where(np.abs(a - x0) < np.abs(b - x0), a, b)

edited Oct 28, 2020 at 17:36

answered Oct 28, 2020 at 16:48

Marat

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2 Comments

Liris Over a year ago

I think there is a mistake: np.abs(a - x0) < np.abs(a - x0) should be np.abs(a - x0) < np.abs(b - x0) right ?

mathfux Over a year ago

This works only for two points a and b and my solution works for any list of points.

mathfux · Accepted Answer · 2020-10-28 18:16:59Z

0

numpy doesn't support a key function. A possible fix is storing distances in a separate array and determining index of smallest distance with np.argmin. An example with 4D points:

points = np.array([[0, 0.1, 0.2, 0.3], [3, 4, 5, 5], [6, 4.5, 1, 1]])
x0 = [3, 0.5, 5, 5]
distances = np.sum((points - x0)**2, axis=1)
output = points[np.argmin(distances)]

Output:

array([3., 4., 5., 5.])

answered Oct 28, 2020 at 18:16

mathfux

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