https://www.codewars.com/kata/reverse-or-rotate/train/python

Task

The input is a string str of digits. Cut the string into chunks (a chunk here is a substring of the initial string) of size sz (ignore the last chunk if its size is less than sz).

If a chunk represents an integer such as the sum of the cubes of its digits is divisible by 2, reverse that chunk; otherwise rotate it to the left by one position. Put together these modified chunks and return the result as a string.

If

• sz is <= 0 or if str is empty return ""
• sz is greater (>) than the length of str it is impossible to take a chunk of size sz hence return "".

Examples:

revrot("123456987654", 6) --> "234561876549"
revrot("123456987653", 6) --> "234561356789"
revrot("66443875", 4) --> "44668753"
revrot("66443875", 8) --> "64438756"
revrot("664438769", 8) --> "67834466"
revrot("123456779", 8) --> "23456771"
revrot("", 8) --> ""
revrot("123456779", 0) --> "" 
revrot("563000655734469485", 4) --> "0365065073456944"

Best Practices

Py First:

def revrot(s, n, res=""):
    if not s or n < 1 or n > len(s):
        return ""

    while len(s) >= n:
        group = s[:n]
        if sum([int(d)**3 for d in group]) % 2 == 0:
            res += group[::-1]
        else:
            res += group[1:] + group[0]
        s = s[n:]

    return res

Py Second:

def revrot(strng, sz):
    func = lambda x : x[1:] + x[0] if sum(int(i) for i in x) % 2 else x[::-1]
    return "" if sz <= 0 or sz > len(strng) else "".join(func(strng[i:i+sz]) for i in xrange(0, len(strng) - sz + 1, sz))

Py Third:

def revrot(strng, sz):

    def chunks():
        if sz <= 0:
            raise StopIteration

        string = strng
        while len(string) >= sz:
            chunk, string = string[:sz], string[sz:]
            yield chunk

    def rev_or_rotate(chunk):
        if sum(int(digit)**3 for digit in chunk) % 2 == 0:
            return chunk[::-1]

        else:
            return chunk[1:] + chunk[:1]

    return "".join(map(rev_or_rotate, chunks()))