Sorting based on frequency and alphabetical order

问题

I am trying to sort based on frequency and display it in alphabetical order After freq counting , I have a list with (string, count) tuple E.g tmp = [("xyz", 1), ("foo", 2 ) , ("bar", 2)]

I then sort as sorted(tmp, reverse=True) This gives me [("foo", 2 ) , ("bar", 2), ("xyz", 1)]

How can I make them sort alphabetically in lowest order when frequency same, Trying to figure out the comparator function

expected output:[("bar", 2), ("foo", 2 ), ("xyz", 1)]

回答1:

You have to sort by multiple keys.

sorted(tmp, key=lambda x: (-x[1], x[0]))

Source: Sort a list by multiple attributes?.

回答2:

Use this code:

from operator import itemgetter

tmp = [('xyz',1), ('foo', 2 ) , ('bar', 2)]
print(sorted(tmp, key=itemgetter(0,1)))

This skips the usage of function call.

来源：https://stackoverflow.com/questions/65113602/sorting-based-on-frequency-and-alphabetical-order