Sort list of lists ascending and then descending

Question

If I have a list that contains a list that looks like this

['a',1] ['a',2] ['a',3] ['b',1] ['b',2] ['b',3]

How can I sort them so that element 0 is sorted descending and element 1 sorted ascending so the result would look like

['b',1] ['b',2] ['b',3] ['a',1] ['a',2] ['a',3]

Using itemgetter I can pass in reverse on element 0 but I then resort against element to of course it ruins the previous sort. I can’t do a combined key since it needs to first sort descending and then ascending.

Asked By: Ominus

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Source

Answer 1

L = [['a',1], ['a',2], ['a',3], ['b',1], ['b',2], ['b',3]]
L.sort(key=lambda k: (k[0], -k[1]), reverse=True)

L now contains:

[['b', 1], ['b', 2], ['b', 3], ['a', 1], ['a', 2], ['a', 3]]

Answered By: Steven Rumbalski

Answer 2

Something like

def mycmp(a, b):

  res = cmp(a[0], b[0])
  if res == 0:
     return cmp(a[1], b[1])
  return res

newlist = sorted(input_list, cmp=mycmp)

The comparison method first checks the first item of each element. If they are equal it will check the second items of each element. The return value inside the mycmp() implementation may be negated in order to implemented a different sorting behavior.

Answered By: Andreas Jung

Answer 3

You can do successive rounds of sorting as python’s sort is stable. You need to first sort on the secondary key though. See also the official HOW TO.

from operator import itemgetter
l = [['a',2], ['a',1], ['b', 2], ['a',3], ['b',1], ['b',3]]
l.sort(key=itemgetter(1))
l.sort(key=itemgetter(0), reverse=True)
# [['b', 1], ['b', 2], ['b', 3], ['a', 1], ['a', 2], ['a', 3]]

Answered By: mhyfritz

Sort list of lists ascending and then descending

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