Sorting a tuple of strings alphabetically and reverse alaphabetically'

Question

Lets say you have a list of tuples (last_name, first_name) and you want to sort them reverse alphabetically by last_name and then if two people have the same name for them to be sorted alphabetically by first name. Is there a way to do this with one line?

e.g.

Sorted(names, key = ?) = [('Zane', 'Albert'), ('Zane', 'Bart'), ('Python', 'Alex'), ('Python', 'Monty')]

With two lines I guess you could rely on sort stability and first sort by first name and then sort by last name reversed.

yes... that is default sort ... you dont need anything... `sorted(names)` — Joran Beasley, Mar 21 '17 at 00:12
i rpobably read too fast ... but `(A,B) vs (A,C)` sorts exactly how you would expect which is what it sounds like what he wants (first on last name, and in event of tie on first name) — Joran Beasley, Mar 21 '17 at 00:23
@JoranBeasley What do you think is the negative of a *string*? — Stefan Pochmann, Mar 21 '17 at 00:31
@JoranBeasley doesn't work for strings `Traceback (most recent call last):File "", line 1, in names2.sort(key= lambda x:(-x[0],x[1])) File "", line 1, in names2.sort(key= lambda x:(-x[0],x[1])) TypeError: bad operand type for unary -: 'str'` — Seraphya, Mar 21 '17 at 00:31

score 2 · Answer 1 · edited Jun 20 '20 at 09:12

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Is there a way to do this with one line?

With two lines I guess you could rely on sort stability and first sort by first name and then sort by last name reversed.

Well you can do exactly that in one line:

sorted(sorted(names, key=lambda name: name[1]), key=lambda name: name[0], reverse=True)

Doing two sorts for exactly this problem in the Sorting Mini-HOW TO ("to sort the student data by descending grade and then ascending age, do the age sort first and then sort again using grade").

A silly alternative doing it with just one sorted and key:

sorted(names, key=lambda name: ([-ord(c) for c in name[0]], name[1])) == correct

A Python 2 solution with one sorted and cmp instead of key:

sorted(names, lambda a, b: cmp(b[0], a[0]) or cmp(a, b))

edited Jun 20 '20 at 09:12

Community

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answered Mar 21 '17 at 00:25

Stefan Pochmann

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Thanks; I was halfway to entering this when you had the job done. My response is no longer useful, except for the [link](https://wiki.python.org/moin/HowTo/Sorting). – Prune Mar 21 '17 at 00:28
@Seraphya Ok I just added a single-sorted solution. – Stefan Pochmann Mar 21 '17 at 00:29
@Prune Thanks, included that. – Stefan Pochmann Mar 21 '17 at 00:36

score 1 · Accepted Answer · edited Jul 15 '20 at 17:14

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Here's an itertools.groupby solution:

from itertools import groupby
from operator import itemgetter

li = [('x', 'y'), ('s', 'e'), ('s', 'a'), ('x', 'z')]

[p for k, g in groupby(sorted(li, reverse=True), itemgetter(0)) for p in reversed(list(g))]
# [('x', 'y'), ('x', 'z'), ('s', 'a'), ('s', 'e')]

edited Jul 15 '20 at 17:14

Augustin

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answered Mar 21 '17 at 00:31

Paul Panzer

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score 0 · Answer 3 · answered Mar 21 '17 at 00:31

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Yes, you can rely on the stability of the sorting:

names = [('Zane', 'Albert'), ('Zane', 'Bart'), ('Python', 'Alex'), ('Python', 'Monty')]

names.sort()
names.sort(reverse=True, key=lambda i: i[0])

answered Mar 21 '17 at 00:31

Peter Gerber

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FYI: Sort stability was added and Python 2.2 – Prune Mar 21 '17 at 00:44

Sorting a tuple of strings alphabetically and reverse alaphabetically'

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