How can this numpy 2D sorted array creation be optimized?

Question

I have a NxM matrix called coefficients that I want to sort:

import numpy
N = 10
M = 42
coefficients = numpy.random.uniform(size=(N, M))

I have an array called order with N elements that says the order that the rows of coefficients should be in:

order = numpy.random.choice(range(N), N, False)

I'm sorting coefficients by sorting order:

coefficients = numpy.array([mag for (orig, mag)
                            in sorted(zip(order, coefficients),
                                      key=lambda pair: pair[0])])

This works, but it's probably slower than it should be. If this was in 1D, I'd use fromiter, but I don't know how to tackle this since it's 2D. Is there an optimization I can make here?

Community · Accepted Answer · 2017-05-23 10:33:27Z

4

To answer your question, just coefficients[order.argsort()] is enough :)

5 Comments

user2357112 Over a year ago

This goes in the opposite direction from the original code - the OP's code has original[0] go to result[order[0]] rather than having result[0] come from original[order[0]]. This might be an acceptable difference, depending on how order needs to behave. If not, we need coefficients[order.argsort()].

user2357112 Over a year ago

(I don't know why the other guys deleted their answers. There are two deleted answers that already provide this information.)

rhombidodecahedron Over a year ago

Omg, so obvious. Thanks a bunch you two!

Jill-Jênn Vie Over a year ago

@JoeKington, he might have thought the same :D

user2357112 Over a year ago

It's also worth noting that if the sort gets expensive, result = numpy.empty_like(coefficients); result[order] = coefficients can avoid the sort. At the scale given in the question, it's not a problem, but with millions of rows, it becomes a big difference.

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