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1302 Discussions

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  • hzerryn
     + 0 comments
    if __name__ == '__main__':
    s = input()
    
    counts = {}
    
    for i in s:
        counts[i] = s.count(i)
        
    counts = sorted(counts.items(), key=lambda x: (-x[1], x[0]))
    
    for i in range(3):
        print(counts[i][0], counts[i][1])

  • piras_2019843
     + 0 comments
    import math
import os
import random
import re
import sys
from collections import Counter


if __name__ == '__main__':
    s = input()

count= Counter(s)
sorted_items = sorted(count.items(), key=lambda x: (-x[1], x[0]))
for char, freq in sorted_items[:3]:
    print(char, freq)
    

    ``

  • shaswathkumar
     + 0 comments

    Did a little digging and found the Counter.most_common(n) method, which returns the list you need to iterate print(*) through. Sorted the string beforehand because .most_common(n) will sort by order of elements encountered.

    from collections import Counter


if __name__ == '__main__':
    s = sorted(input())
    s_count = Counter(s)
    [print(*entry) for entry in s_count.most_common(3)]

  • dheenupriya2001
     + 0 comments
    from collections import Counter


if __name__ == '__main__':
    s = input()
    c = Counter(s)
    sort_val = sorted(c.values(), reverse=True) #nos. sorted in ascending order
    sort_counter = dict(sorted(c.items())) #sorted the counter
    i = 0
    while(i<3):
        for k,v in sort_counter.items():
            if v==sort_val[i]:
                print(k +" "+ str(v))
                sort_counter.pop(k)
                i+=1
                break

  • jonade_naeem
     + 0 comments

    I did this:

    s = input().lower()

# Create dictionary of each char count
dict = {char: s.count(char) for char in s}

# Return list of tuples by value count in descending order, and alphabetic for equal values
sorted_list = sorted(dict.items(), key = lambda x: (-x[1], x[0]))

# Print top three values of sorted_list
for key, value in sorted_list[:3]:
    print(key, value)