CODEWITHSUNDEEP
python-3.x
praveen nani
praveen nani

Reputation: 87

Sorting of list depending on frequency in descending order

I have a list of numbers list: [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]. I want to sort it by the frequency of elements to get this: [1, 1, 1, 1, 3, 3, 3, 2, 2, 2].

If several elements have the same frequency, sort them by descending value. Can you find any way to do this. I'm using mentioned method but getting my output as:[1, 1, 1, 1, 2, 2, 2, 3, 3, 3].

from collections import Counter 
list = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(list)
x = sorted(c.most_common(), key=lambda x: (-x[1], x[0])) 
y = [([v] * n) for (v, n) in x]
z = sum(y, [])
print(z)

Upvotes: 1

Views: 338

Answers (2)

hiro protagonist
hiro protagonist

Reputation: 46849

if your list is long, you may want to have to sort only the items that occur the same amount of times:

from collections import Counter
from itertools import groupby

lst = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(lst)

ret = []
for key, group in groupby(c.most_common(), key=lambda x: x[1]):
    items = sorted((item for item, _ in group), reverse=True)
    for item in items:
        ret.extend(key * [item])
print(ret)
# [1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

Upvotes: 1

Rakesh
Rakesh

Reputation: 82765

Looks like you need to use reverse=True

Ex:

from collections import Counter 

data = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(data)

data.sort(key=lambda x: (c[x], x), reverse=True)
print(data)

Output:

[1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

Upvotes: 3

Related Questions