How to order an array and count it in Python?

Question

I want to find only the top 3 distinct items in descending order. If there's a tiebreaker, sort by alphabetical order. If there are 3 items or fewer, returning the distinct list of items is sufficient.

So if I have input of: ["a","a","b","b","c","c","c","d","d","d","d"]

The output will be ["d","c","a"]

Because d has 4 counts, c 3 counts, a and b have the same frequency, but a is alphabetically first.

In MySQL, I would usually use this:

SELECT id, COUNT(*) as frequency FROM mylist GROUP BY id ORDER BY frequency, id

How can I do that in Python?

I use this code based on SAI SANTOH CHIRAG's solution:

def main(output):
    arr = sorted(output,key=lambda i:[output.count(i),-ord(i)],reverse=True)
    out = []
    for i in arr: 
        if i not in out: out.append(i) 
        print(out[:3])

but why is the result like this:

Input (stdin) = a a a b b c d d d d
output = ['d']
['d']
['d']
['d']
['d', 'a']
['d', 'a']
['d', 'a']
['d', 'a', 'b']
['d', 'a', 'b']
['d', 'a', 'b']

instead of what I want, which would be:

['d','a','b']

Answer 1

You use sorted and key for that. Try in this way:

arr = sorted(x,key=lambda i:[x.count(i),-ord(i)],reverse=True)

With this you get all the elements in the sorted order in the increase of count and then alphabetical order. Then do this to get all elements only once:

out = []
for i in arr:
    if i not in out:
        out.append(i)
print(out[:3])

Answer 2

collections.Counter will do:

the_list = ["a","a","b","b","c","c","c","d","d","d","d"]
counter = Counter(sorted(the_list))
top_3 = counter.most_common(3)

at this point, top_3 is of the form [(<entry>, <freq>)] e.g.

[('d', 4), ('c', 3), ('a', 2)]

Take out the first elements from it via list comprehension:

result = [item for item, freq in top_3]

and we get

['d', 'c', 'a']

Notes:

1. We pass the sorted list to the Counter because otherwise it will break the ties according to the insertion order; sorting forces the insertion order to be the alphabetical order in a way.

2. .most_common(3) will return at most 3 elements so we are fine e.g. even if only 2 unique entries are there. E.g. if the_list = ["b", "a"], result will be ["a", "b"] even though number of unique elements is less than 3.

Answer 3

you can use Counter from collections

from collections import Counter

inputs = ["a","a","b","b","c","c","c","d","d","d","d"]
counts = Counter(x)
counts.most_common(3)

final = [i[0] for i in counts.most_common]

output for counts.most_common()

[('d', 4), ('c', 3), ('a', 2), ('b', 2)]