Adapting the same sorting order from a list

Question

I have a list containing numbers and one containing users. The first element of the first list belongs to the first element of the second list and so on. I then want to sort the number list using sort() and want the usernames to still match with the numbers they had before.

from:

users = ["max", "david", "freddy"]

numbers = [9, 3, 10]

to:

users = ["david", "max", "freddy"]

numbers = [3, 9, 10]

Why not just use a dict (key value pair) to map these numbers to users? — plum 0, Dec 29 '20 at 15:31
Does this answer your question? [How to sort two lists (which reference each other) in the exact same way](https://stackoverflow.com/questions/9764298/how-to-sort-two-lists-which-reference-each-other-in-the-exact-same-way) — Georgy, Dec 30 '20 at 15:00

Sash Sinha · Accepted Answer · 2020-12-29T17:47:37.067

2

You could use a list comprehension along with zip and sorted:

>>> users = ["max", "david", "Freddy"]
>>> numbers = [9, 3, 10]
>>> sorted_users = [u for _, u in sorted(zip(numbers, users)]
>>> sorted_users
['david', 'max', 'freddy']

edited Dec 29 '20 at 17:47

answered Dec 29 '20 at 15:32

Sash Sinha

18,743
3
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The `lambda` is unnecessary. It will sort by the first element first by default. – SethMMorton Dec 29 '20 at 16:04
True, removed it – Sash Sinha Dec 29 '20 at 17:48

dwb · Answer 2 · 2020-12-29T15:43:14.770

I would suggest grouping your values together in a list of tuples or dictionaries:

data = [
    ('max',    9),
    ('david',  3),
    ('freddy', 10)
]

and then you can use custom sorting to sort this list:

data.sort(key=lambda x: x[1])

This way your data is stored in a way which keeps their relationship.

To use dictionaries instead requires this simple change:

data = [
    {'name': 'max',    'number': 9},
    {'name': 'david',  'number': 3},
    {'name': 'freddy', 'number': 10}
]

and then change your sort code to

data.sort(key=lambda x: x['number'])

score 1 · Answer 3 · answered Dec 29 '20 at 15:33

1

You can zip the two into one list, sort based on the number, then unzip them:

users = ["max", "david", "freddy"]
numbers = [9, 3, 10]

zipped = sorted(zip(numbers, users), key=lambda x: x[0])

numbers, users = zip(*zipped)

answered Dec 29 '20 at 15:33

Aplet123

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nsh1998 · Answer 4 · 2021-01-04T21:05:31.360

0

You could try a dictionary for user numbers and user names

users = ["max", "david", "freddy"]

numbers = [9, 3, 10]

dictionary = {users[i]: numbers[i] for i in range(len(numbers))} 

{'max': 9, 'david': 3, 'freddy': 10}

and then sort the dictionary and convert back to lists. \

edited Jan 04 '21 at 21:05

answered Dec 29 '20 at 15:36

nsh1998

106
9

score 0 · Answer 5 · answered Dec 30 '20 at 00:50

Using the right combination of zip and map, you can do this in one line:

users   = ["max", "david", "freddy"]
numbers = [9, 3, 10]

numbers,users = map(list,zip(*sorted(zip(numbers,users))))

print(users)   # ['david', 'max', 'freddy'] 
print(numbers) # [3, 9, 10]

Adapting the same sorting order from a list

5 Answers5