Sorting a record array in numpy

Question

I have an numpy structure array

import numpy as np
np.array([(0, 1, 1167606000), (0, 1, 1167606005), (0, 1, 1167606008),
       (0, 10, 1167606010), (0, 10, 1167606012), (1, 0, 1167606000),
       (1, 2, 1167606001), (1, 0, 1167606005), (1, 0, 1167606008),
       (2, 1, 1167606001), (2, 3, 1167606002), (3, 2, 1167606002),
       (3, 4, 1167606003), (4, 3, 1167606003), (4, 5, 1167606004),
       (5, 4, 1167606004), (5, 6, 1167606005), (6, 5, 1167606005),
       (6, 7, 1167606006), (7, 6, 1167606006), (7, 8, 1167606007),
       (8, 7, 1167606007), (8, 9, 1167606008), (9, 8, 1167606008),
       (9, 10, 1167606009), (10, 9, 1167606009), (10, 0, 1167606010),
       (10, 0, 1167606012)], 
      dtype=[('fr', '<i8'), ('to', '<i8'), ('time', '<i8')])

Is there a vectorized way to sort it first by the minimum of 'fr', 'to' and then by 'time'. Another thing is that I want to sort this without making any copies.

Edit: The sort is not by 'fr', 'to' and 'time', but first by the minimum of 'fr' and 'to' followed by 'time'. The expected answer in the above case is

(0, 1, 1167606000),
(1, 0, 1167606000),
(0, 1, 1167606005),
(1, 0, 1167606005), 
(0, 1, 1167606008),
(1, 0, 1167606008),
(0, 10, 1167606010), 
(0, 10, 1167606012), 
(1, 2, 1167606001), 
(2, 1, 1167606001), 
(2, 3, 1167606002), 
(3, 2, 1167606002),
(3, 4, 1167606003), 
(4, 3, 1167606003), 

...

Answer 1

You can give an order argument to sort :

a.sort(order=['fr', 'to', 'time'])

---

Sorting by minimum of two columns:

Using lexsort , you can sort by any set of keys. Here, give it a['time'] and np.minimum(a['to'], a['fr']) (sorts by last item first)

inds = np.lexsort((a['time'], np.minimum(a['to'], a['fr'])))
a = a[inds]

To avoid making a copy of a when you rearrange it, you can use take instead of a = a[inds]

np.take(a, inds, out=a)