CODEWITHSUNDEEP
pythonstringsortingdictionarycount
Luwit Roy
Luwit Roy

Reputation: 5

Sort according to alphabetical order with respect to count

from collections import *
def compress(s):
    res = Counter(s)

    for key,value in res.items():
        print(key, value)

compress("hhgoogle")

Output:

h 2
g 2
o 2
l 1
e 1

How to sort it in alphabetical order with respect to count, i.e.:

Required output:

g 2
h 2
o 2
e 1
l 1

Upvotes: 0

Views: 79

Answers (3)

Pedro Lobito
Pedro Lobito

Reputation: 98921

You can use:

from collections import *

print(sorted(Counter('hhgoogle').items(), key=lambda x: (-x[1], x[0])))
# [('g', 2), ('h', 2), ('o', 2), ('e', 1), ('l', 1)]

demo

Upvotes: 1

Red
Red

Reputation: 27557

Here is what you can do:

from collections import *
def compress(s):
    res= Counter(s)
    l1 = sorted(res.most_common(), key=lambda t: t[0])
    l2 = sorted(l1, key=lambda t: t[1], reverse=True)
    print('\n'.join([f"{t[0]} {t[1]}" for t in l2]))
compress("hhgoogle")

Output:

g 2
h 2
o 2
e 1
l 1

Upvotes: 0

Nico Müller
Nico Müller

Reputation: 1874

First sort it alphabetically, then sort the occurrences. The join is merely to output it in the format you have described.

from collections import *
def compress(s):
    res= Counter(s)
    alphabetic_res = sorted(res.most_common(), key=lambda tup: tup[0])
    final_res = sorted(alphabetic_res, key=lambda tup: tup[1], reverse=True)
    print("\n".join("%s,%s" % tup for tup in final_res))

compress("hhgoogle")

OUT: g,2
     h,2
     o,2
     e,1
     l,1

Upvotes: 0

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