Python: double sort

Question

Is there a quick and easy way to sort a list of tuples containing two items? (a short list of the tuples I am trying to sort:

[('this', 4), ('in', 4), ('dedicated', 4), ('who', 3), ('us', 3), ('they', 3), ('so', 3), ('shall', 3), ('people', 3), ('is', 3), ('great', 3), ('dead', 3), ('are', 3), ('It', 3), ('which', 2), ('what', 2)]

I am trying to sort them first by frequency (largest first), so the number, and then by alphabetical order.

This is what I have so far:

word_list.sort(key=itemgetter(1,0), reverse = True)

This sorts the list by frequency in descending order.

So what is the problem actually? Your input list already seems to be sorted. — Marcin
– Marcin, Commented Mar 27, 2015 at 2:09

Marcin · Accepted Answer · 2015-03-27 02:17:53Z

3

I think I understand what you want to do. Have order of frequencies different than words. For this you need to sort twice:

from operator import itemgetter

word_list = [('this', 4), ('in', 4), ('dedicated', 4), 
             ('who', 3), ('us', 3), ('they', 3), ('so', 3), ('shall', 3), ('people', 3), 
             ('is', 3), ('great', 3), ('dead', 3), ('are', 3), ('It', 3), 
             ('which', 2), ('what', 2)]


#first we set words in alphabetical order
word_list2 = sorted(word_list, key=lambda l: l[0].lower())

# then we sort them by frequency
word_list2 = sorted(word_list2, key=itemgetter(1), reverse = True)

print(word_list2)

The result is:

[('dedicated', 4), ('in', 4), ('this', 4), ('are', 3), ('dead', 3), ('great', 3), ('is', 3), ('It', 3), ('people', 3), ('shall', 3), ('so', 3), ('they', 3), ('us', 3), ('who', 3), ('what', 2), ('which', 2)]

This is so called a complex sort. More here. And it works, because sort operations in python are stable. It means that:

when multiple records have the same key, their original order is preserved.

answered Mar 27, 2015 at 2:17

Marcin

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3 Comments

ACx Over a year ago

yep, I actually just figured it out myself as well. Thank you, you are correct sir. I do have one question, why did you include 'key=lambda' in the first sort?

Marcin Over a year ago

I did lambda to sort using lower letters, as you have mixture of uppercase and lowercase in your words. Without this, sorting is not as it should be, unless you exactly want that.

DSM Over a year ago

If we're using Python 3 we can use casefold instead of lower. (And in this case we don't really need to sort twice, because we could sort by a tuple and flip the sign of the frequency, but that's hard to pull off in the general case.)

hjonasson · Accepted Answer · 2015-03-27 02:20:21Z

1

I would use two sorting functions. One for sorting as you do already. Then sort the results of that by alphabetical order. Does that make sense to you?

word_list = [('this', 4), ('in', 4), ('dedicated', 4), ('who', 3), ('us', 3), ('they', 3), ('so', 3), ('shall', 3), ('people', 3), ('is', 3), ('great', 3), ('dead', 3), ('are', 3), ('It', 3), ('which', 2), ('what', 2)]

word_list.sort(key=lambda i:i[1], reverse = True)
word_list.sort(key=lambda i:i[0].lower())

answered Mar 27, 2015 at 2:20

hjonasson

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