How to sort a list according to another list? Python

Question

So if I have one list:

name = [\'Megan\', \'Harriet\', \'Henry\', \'Beth\', \'George\']

And I have another list where each value represents the

Answer 1

You can sort them at the same time as tuples by using zip. The sorting will be by name:

tuples = sorted(zip(name, score_list))

And then

name, score_list = [t[0] for t in tuples], [t[1] for t in tuples]

Answer 2

One obvious approach would be to zip both the list sort it and then unzip it

>>> name, score = zip(*sorted(zip(name, score_list)))
>>> name
('Beth', 'George', 'Harriet', 'Henry', 'Megan')
>>> score
(6, 10, 6, 5, 9)

Answer 3

Consider a third-party tool more_itertools.sort_together:

Given

import more_itertools as mit


name = ["Megan", "Harriet", "Henry", "Beth", "George"]
score_list = [9, 6, 5, 6, 10]

Code

mit.sort_together([name, score_list])
# [('Beth', 'George', 'Harriet', 'Henry', 'Megan'), (6, 10, 6, 5, 9)]

Install more_itertools via > pip install more_itertools.

Answer 4

The correct way if you have a dict is to sort the items by the key:

name = ['Megan', 'Harriet', 'Henry', 'Beth', 'George']

score_list = [9, 6, 5, 6, 10]
d = dict(zip(name, score_list))

from operator import itemgetter
print(sorted(d.items(), key=itemgetter(0)))
[('Beth', 6), ('George', 10), ('Harriet', 6), ('Henry', 5), ('Megan', 9)]