Sorting numpy array in the same manner as sorted(, key=)

Question

For example, to sort an array of complex numbers first by real part, then by magnitude of imaginary part, and then with negative imaginary parts first:

def order(a):
    return a.real, abs(a.imag), sign(a.imag)

z = array(sorted(z, key=order))

So

array([ 1.+2.j, 5.+0.j, 1.+0.j, 1.+1.j, 1.+1.j, 1.-1.j, 6.+0.j, 1.-1.j, 1.-2.j])

becomes

array([ 1.+0.j, 1.-1.j, 1.-1.j, 1.+1.j, 1.+1.j, 1.-2.j, 1.+2.j, 5.+0.j, 6.+0.j])

I think there's a way to do the same thing using numpy's argsort, which is probably faster, but I can't figure it out:

In [2]: argsort((a.real, abs(a.imag), sign(a.imag)))
Out[2]: 
array([[0, 2, 3, 4, 5, 7, 8, 1, 6],
       [1, 2, 6, 3, 4, 5, 7, 0, 8],
       [5, 7, 8, 1, 2, 6, 0, 3, 4]])

Jaime · Accepted Answer · 2013-01-18 13:58:41Z

4

You can use np.lexsort :

import numpy as np
a = np.array([ 1.+2.j, 5.+0.j, 1.+0.j, 1.+1.j, 1.+1.j, 1.-1.j, 6.+0.j,
               1.-1.j, 1.-2.j])
sorted_idx = np.lexsort((np.sign(a.imag), np.abs(a.imag), a.real))
>>> a[sorted_idx]
array([ 1.+0.j,  1.-1.j,  1.-1.j,  1.+1.j,  1.+1.j,  1.-2.j,  1.+2.j,
        5.+0.j,  6.+0.j])

Notice that the sorting keys are in reversed ordered, i.e. last is principal.

edited Jan 18, 2013 at 13:58

answered Jan 18, 2013 at 5:34

Jaime

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