I need to create a function called compress that compresses a string by replacing any repeated letters with a letter and number. Can someone suggest a better way to do this?

s=input("Enter the string:")

temp={}

result=" "

for x in s:

    if x in temp:

        temp[x]=temp[x]+1

    else:

        temp[x]=1

for key,value in temp.items():

    result+=str(key)+str(value)

        print(result)

Encapsulate your code into functions

Your code is neither reusable nor testable, wrap it into a function and call it from under an if __name__ == '__main__' guard. This will allow you to test it more easily. You will also be able to return values instead of printing them, this will make the code more reusable:

def compress(string):

    temp={}

    result=" "

    for x in string:

        if x in temp:

            temp[x] = temp[x]+1

        else:

            temp[x] = 1

    for key, value in temp.items():

        result += str(key) + str(value)



    return result





if __name__ == '__main__':

    s = input("Enter the string:")

    print(compress(s))

You can then jump into an interactive shell and type:

>>> from your_file_name import compress

>>> compress('aaabbccccddefg')

 a3b2c4d2e1f1g1

>>> compress('b'*42)

 b42

Use existing data structures

collections.Counter offers simplified ways of counting elements of an iterable:

from collections import Counter





def compress(string):

    temp = Counter()

    result = " "

    for x in string:

        temp[x] += 1



    for key, value in temp.items():

        result += str(key) + str(value)

    return result





if __name__ == '__main__':

    s = input("Enter the string:")

    print(compress(s))

You can even simplify further as Counter can take any iterable in its constructor. You can also use str.join to simplify even further:

from collections import Counter





def compress(string):

    counts = Counter(string)

    return ''.join(letter+str(count) for letter, count in counts.items())





if __name__ == '__main__':

    print(compress(input("Enter the string: ")))

Possible bug

To me, compressing a string mean having a mean to decompress it back. Using a dictionnary keyed by each letter as you do, you lost an important information: which letter is next to which one. Also 'aabaabaabaa' will be compressed to 'a8b3' which, to me, doesn't make sense and would be better as 'a2b1a2b1a2b1a2'. But I might be wrong. In this case, itertools.groupby is much more usefull as it will keep the ordering and avoid aggregating separate groups of letters:

import itertools





def compress(string):

    return ''.join(

            letter + str(len(list(group)))

            for letter, group in itertools.groupby(string))





if __name__ == '__main__':

    print(compress(input("Enter the string: ")))

In the itertools module there is the groupby function that groups together runs of the same values.

You can use it like this here:

from itertools import groupby



def compress(s):

    out = 

    for name, group in groupby(s):

        length_of_run = len(list(group))

        if length_of_run == 1:

            out.append(name)

        else:

            out.append(f"{name}{length_of_run}")

    return "".join(out)

This also uses the more modern f-strings instead of manually building the string with str calls and + and puts everything into a function that you can reuse.

It also has the advantage that it directly iterates over the input, instead of over its indices (have a look at Loop like a Native!). This makes it work also for a generator, which does not have a length:

from itertools import islice, cycle



compress(islice(cycle('a'), 10))

# 'a10'

This could be one of the way to do it:

count = 1

for i in range(1, len(input) + 1):

    if i == len(input):

        print(input[i - 1] + str(count), end="")

        break

    else:

        if input[i - 1] == input[i]:

            count += 1

    else:

            print(input[i - 1] + str(count), end="")

            count = 1