Sort in descending order python

Question

I will like to sort by position in descending order. Given:

name = ["Shawn", "Patrick", "Nancy", "Viola"]

position = [3,1,4,2]
   
l =[name,positions]    

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

Does this answer your question? [Sorting list based on values from another list?](https://stackoverflow.com/questions/6618515/sorting-list-based-on-values-from-another-list) — RichieV, Sep 03 '20 at 03:16
`list(zip(*sorted(zip(name, position), key=operator.itemgetter(1), reverse=True)))[0]` — inspectorG4dget, Sep 03 '20 at 03:17

deepyaman · Answer 1 · 2020-09-03T03:21:23.843

0

Change l = [name, positions] to l = list(zip(name,position)):

>>> name = ["Shawn", "Patrick", "Nancy", "Viola"]
>>> position = [3,1,4,2]
>>> l = list(zip(name,position))
>>> l.sort(key=lambda x: -x[1])
>>> l
[('Nancy', 4), ('Shawn', 3), ('Viola', 2), ('Patrick', 1)]

Also note the modified sort key for descending order.

edited Sep 03 '20 at 03:21

answered Sep 03 '20 at 03:15

deepyaman

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score 0 · Answer 2 · answered Sep 03 '20 at 03:21

Use zip to pair up items in corresponding positions. (This will produce an iterable, so either convert to list or use sorted rather than .sort.)
To reverse the order, either use -x[1] as the key, or pass reverse=True.
(typo) Name the position variable consistently.
(style) Avoid using l as a variable name, as it can be confused with 1 and I.

name = ["Shawn", "Patrick", "Nancy", "Viola"]
position = [3, 1, 4, 2]
l = zip(name, position)
l = sorted(l, key=lambda x: x[1], reverse=True)

Sort in descending order python

2 Answers2