how do I print one letter that is duplicated on string?

Question

In a string, I want to just print one letter if it happens 2 times, i tried using itertools but if the letter is written 4 times it just prints one, and I need to print 2. Example:

input = rruunnnniinngg
output = running

Thanks btw.

Answer 1

With re.sub() function (assuming only adjacent characters):

import re

s = 'rruunnnniinngg'
result = re.sub(r'(\S)\1', '\\1', s)

print(result)

The output:

running

---

• (\\S) - regex captured group containing a single non-whitespace character

• \\1 - the value of the 1st captured group (the immediate previous character repeated)

Answer 2

You can try this:

from itertools import groupby
input1 = "rruunnnniinngg"
final_string = ''.join(map(lambda x: x[:(len(x)/2)], [''.join(list(b)) for a, b in groupby(input1)]))

Output:

'running'

Answer 3

it can be simply implemented using a while loop that :
-adds the current character in the output string
- checks weather the next character is same as current character.
-if yes, then it skips the next character by increasing the counter by an extra 1 unit.

def func(s):
    i=0
    p=''
    while(i<len(s)-1):
         p+=s[i]
         if(s[i+1]==s[i]):
            i+=1
         i+=1
    return p