Question

I have an object roughly like so:

class Hand(object):
    def __init__(self, finger_names, finger_lengths, nail_sizes):
        self.finger_names = finger_names
        self.finger_lengths = finger_lengths
        self.nail_sizes = nail_sizes

   def _sort_by_finger_lengths(self):
        ????

finger_names, finger_lengths and nail_sizes are all either equally long lists or empty (for example, if the person hasn’t measured their nail_sizes yet). The goal is to sort the attributes of the object by finger_lengths. So if you start with a Hand object where the lists are ordered by left-to-right names (pinky, ring, middle, pointer, thumb), you end up with a Hand object with all attributes sorted by finger_lengths.

Like so:

finger_names = [pinky, thumb, pointer, ring, middle]
finger_lengths = [6, 7, 12, 13, 15]
nail_sizes = []

EDITED TO ADD: Hand is just an example class. The real code has good reasons for having lists for each attribute.

Answer 1

One simple approach is to do this:

>>> finger_names, finger_lengths
(('pointer', 'ring', 'pinky', 'middle', 'thumb'), (12, 13, 6, 15, 7))
>>> s_tuples = sorted(zip(finger_names, finger_lengths), key=lambda x: x[1])
>>> finger_names, finger_lengths = zip(*s_tuples)
>>> finger_names, finger_lengths
(('pinky', 'thumb', 'pointer', 'ring', 'middle'), (6, 7, 12, 13, 15))

However, Nick is right that you should probably link these up in a data structure, instead of depending on their order to associate them.

The above strategy (using key) still works in that case, but you don’t need to use zip(*s_tuples) to disassociate them.

On the other hand, if you want to disassociate them, there’s a one-liner solution that I forgot about before.

finger_lengths, finger_names = zip(*sorted(zip(finger_lengths, finger_names)))

Or, if you want to sort in place, saving one copy operation:

s_tuples = zip(finger_lengths, finger_names)
s_tuples.sort()
finger_lengths, finger_names = zip(*temp)