Using Python, how to print output string as -> aaa3bb2c1ddddd5 when Input string is aaabbcddddd

Using Python, how to print output string as -> aaa3bb2c1ddddd5 when Input string is aaabbcddddd
I want to concatenate actual character value and number of times a character is repeated in a string

def mycode(myString):
    lenstr = len(myString)
    print('length of string is '+str(lenstr));
    for ele in myString:
        count=0
        for character in myString:
            if character == ele:
                count = count+1
        totalstr = ele+str(count)
        
    return totalstr

If the string is always sorted and grouped together like that, then you can use a collections.Counter to do it.

from collections import Counter
inp = "aaabbcddddd"
counter = Counter(inp)
out = "".join(k * v + str(v) for k,v in counter.items())

Or in one line:

print(''.join(k * v + str(v) for k,v in Counter(inp).items()))

Output:

aaa3bb2c1ddddd5

Or you can do it manually:

inp = "aaabbcddddd"
last = inp[0]
out = inp[0]
count = 1
for i in inp[1:]:
    if i == last:
        count += 1
    else:
        out += str(count)
        count = 1
        last = i
    out += i
out += str(count)
print(out)
        

Here is a one line solution using a regex replacement with callback:

<!-- language: python -->

inp = "aaabbcddddd"
output = re.sub(r'((\w)*)', lambda m: m.group(1) + str(len(m.group(1))), inp)
print(output)  # aaa3bb2c1ddddd5

Another one-liner:

import itertools

test = 'aaabbcddddd'
out = ''.join(f"{(g := ''.join(ig))}{len(g)}" for _, ig in itertools.groupby(test))
assert out == 'aaa3bb2c1ddddd5'

you can do like..

Code:

Time Complexity: O(n)

input_string="aaabbcddddd"
res=""
count=1

for i in range(1, len(input_string)):
    if input_string[i] == input_string[i-1]:
        count += 1
    else:
        res+=input_string[i-1]*count + str(count)
        count = 1

res+=input_string[-1]*count + str(count)
print(res) #aaa3bb2c1ddddd5

def char_counter_string(string):
    prev_char = None
    char_counter = 0
    output = ''
    for char_index in range(len(string)+1):
        if char_index == len(string):
            output += str(char_counter)
            break
        if string[char_index] != prev_char and prev_char is not None:
            output += str(char_counter)
            char_counter = 0
        output += string[char_index]
        char_counter += 1
        prev_char = string[char_index]
    return output


if __name__ == '__main__':
    print(char_counter_string('aaabbcddddd'))

Here's another way, ...

Full disclosure: ... as long as the run of characters is 10 or less, it will work. I.e., if there are 11 of anything in row, this won't work (the count will be wrong).

It's just a function wrapping a reduce.

from functools import reduce

def char_rep_count(in_string):
    return reduce(
             lambda acc, inp:
               (acc[:-1]+inp+str(int(acc[-1])+1))
                 if (inp==acc[-2])
                 else (acc+inp+"1"),
             in_string[1:],
             in_string[0]+"1"
           )

And here's some sample output:

print(char_rep_count("aaabbcdddd"))
aaa3bb2c1dddd4

I think this fulfils the brief and is also very fast:

s = 'aaabbcddddd'

def mycode(myString):
    if myString:
        count = 1
        rs = [prev := myString[0]]
        for c in myString[1:]:
            if c != prev:
                rs.append(f'{count}')
                count = 1
            else:
                count += 1
            rs.append(prev := c)
        rs.append(f'{count}')
        return ''.join(rs)
    return myString