How to append to a string?

Question

Here I am counting each character in a string and return that character with it's number of occurences in the string. e.g.: ab("aabbc") and return a string: a2b2c1

However I have a list here where instead of returning a2b2c1, it returns ['a',2,'b',2,'c',1]. I want it to return it in the form of a string, and not a list.

Here's what I did:

def ab(n):

    char = []
    
    for i in n:
        if i not in char:
            char.append(i)
            count = n.count(i)
            char.append(count)
    return(char)

Note: posted code does not work for general run-length encoding such as: `ab('aabcaa')` — DarrylG, Oct 22 '22 at 22:27
Welcome to Stack Overflow! Please take the [tour]. What have you already tried? For example, do you know how to [convert integers to strings](/q/961632/4518341) and [join strings together](https://docs.python.org/3/library/stdtypes.html#str.join)? For more tips, see [ask]. — wjandrea, Oct 22 '22 at 22:31
BTW, you might consider using a [`Counter`](https://docs.python.org/3/library/collections.html#collections.Counter) to do this more easily. — wjandrea, Oct 22 '22 at 22:33
If there's no constraint on _how_ to count the occurrences, I suggest looking at [collections.Counter](https://docs.python.org/3/library/collections.html#collections.Counter), it's designed for this purpose. The only work you'll need to do is format the output. — Michael Ruth, Oct 22 '22 at 22:34
I'm happy you were able to solve your problem. Please don't replace the question with a message which should be in the comments. This buries the context needed to understand any answers. — Michael Ruth, Oct 22 '22 at 22:39
You could post your *own* answer, so that you could also share the idea and *earn* the reputation too! — Daniel Hao, Oct 22 '22 at 22:42

Sash Sinha · Answer 1 · 2022-10-22T22:40:49.683

Consider recognizing that strings are immutable in python:

def ab(n):
    char = []
    i = 0
    while i < len(n):
        run_length = 1
        while i + 1 < len(n) and n[i] == n[i + 1]:
            run_length += 1
            i += 1
        char.append(str(run_length))
        char.append(n[i])
        i += 1
    return ''.join(char)

print(f'{ab("aabbc") = }')
print(f'{ab("aabcaa") = }')

Output:

ab("aabbc") = '2a2b1c'
ab("aabcaa") = '2a1b1c2a'

Alternatively, if you do not actually need to implement run-length encoding, you could utilize collections.Counter:

>>> from collections import Counter
>>> n = 'aabcaa'
>>> ''.join(f'{c}{k}' for k, c in Counter(n).items())
'4a1b1c'

score 0 · Answer 2 · answered Oct 22 '22 at 22:44

0

Considering strings are immutable in Python, you may wish to use re.sub with a lambda to generate a new string with the run length encoding.

import re

s = "aaaabbc"

re.sub(r"(\D)(\1*)", lambda m: f"{m.group(1)}{1+len(m.group(2))}", s)
# 'a4b2c1'

answered Oct 22 '22 at 22:44

Chris

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score 0 · Answer 3 · answered Oct 22 '22 at 22:57

There is dictionary in python for this kind of problem Here how it works:

words = "aabbc"

occurrences = {}

for char in words:
    char[i] =  char.get(i, 0) + 1

print(occurrences)

Output:

{'a': 2, 'b': 2, 'c': 1}

To get dictionary value:

print(occurrences["a"])

Output:

How to append to a string?

3 Answers3