Python: How do I sort on the second item of a tuple? [duplicate]

Question

I have this piece of code

relations = dict()
relations[('a','b','c')] = 'first'
relations[('a','c','c')] = 'third'
relations[('d','b','c')] = 'second'

relations_sorted = sorted(relations.keys(), key=lambda tup: tup[1])

for key in relations_sorted:
  print( key, relations[key])

This prints the dictionary sorted on the second element of the key tuple (solution found on StackOverflow).

(('d', 'b', 'c'), 'second')
(('a', 'b', 'c'), 'first')
(('a', 'c', 'c'), 'third')

How do I extend this to sorting on the combination of second, then first, then third element in the tuple? E.g.

(('a', 'b', 'c'), 'first')
(('d', 'b', 'c'), 'second')
(('a', 'c', 'c'), 'third')

MSeifert · Accepted Answer · 2017-01-31 10:28:54Z

4

Just use key=lambda tup: (tup[1], tup[0], tup[2])

or even faster/easier/better with operator.itemgetter:

from operator import itemgetter

relations_sorted = sorted(relations.keys(), key=itemgetter(1, 0, 2))

answered Jan 31, 2017 at 10:28

MSeifert

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Andrey · Accepted Answer · 2017-01-31 10:29:00Z

1

Change to key=lambda tup: (tup[1], tup[0], tup[2])

answered Jan 31, 2017 at 10:29

Andrey

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