Work faster then (-items.count(x), items.index(x)) speedy solution for Sort Array by Element Frequency by Gi-gik - python coding challenges

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Work faster then (-items.count(x), items.index(x)) solution in Speedy category for Sort Array by Element Frequency by Gi-gik

from collections import Counter

def frequency_sort(items):
	groups = Counter(items)
	sort = sorted(groups.items(), key=lambda e: e[1], reverse=True)
	for key, value in sort:
		yield from [key]*value

March 14, 2020

Comments:

leggewie on May 31, 2021, 8:55 a.m.
nice

Tenfrow on Oct. 18, 2021, 10:43 p.m.
Can you (or somebody) explain what `yield from [key]*value` actually does?

msol294 on Nov. 13, 2021, 3:13 p.m.
Thanks to you i knew two new things!

andrey.sereda on June 22, 2022, 7:39 a.m.
Very cool! but why do you really need sorting (Counter should already do it for you)? doesn't just yielding from groups.elements() give the same result?

froogss on Aug. 11, 2022, 1:35 a.m.
Counter CAN do it for you, but the elements method on the Counter returns them in the order the Counter encountered them in the iterable you gave it. You can use the more suitable Counter.most_common() method to get them in frequency order. You can use this to get a pretty short solution like this <pre class='brush: python'>return [item for pair in Counter(items).most_common() for item in ([pair[0]]*pair[1])] </pre>

117370 on Feb. 3, 2023, 1:28 a.m.
Oh thanks I didn't know about yield before

List, list and list again

0 %

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