how to separate alternating digits and characters in string into dict or list?

‘L134e2t1C1o1d1e1’

the original string is "LeetCode"

but I need to separate strings from digits, digits can be not only single-digit but also 3-4 digits numbers like 345.

My code needs to separate into dict of key values; keys are characters and numbers is the digit right after the character. Also create 2 lists of separate digits, letters only.

expected output:

letters = ['L', 'e', 't', 'C', 'o', 'd', 'e']
digits = [134,2,1,1,1,1,1]

This code is not properly processing this.

def f(s):

    d = dict()
    letters = list()
    # letters = list(filter(lambda x: not x.isdigit(), s))

    i = 0
    while i < len(s):
        print('----------------------')
        if not s[i].isdigit():

            letters.append(s[i])

        else:
            j = i
            temp = ''
            while j < len(s) and s[j].isdigit():
                j += 1

            substr = s[i:j]
            print(substr)
            
        i += 1

    print('----END -')
    print(letters)

>Solution :

You edited your question and I got a bit confused, so here is a really exhaustive code giving you a list of letters, list of the numbers, the dict with the number associated with the number, and finally the sentence with corresponding number of characters …

def f(s):
    letters = [c for c in s if c.isalpha()]
    numbers = [c for c in s if c.isdigit()]

    mydict = {}
    currentKey = ""
    for c in s:
        print(c)
        if c.isalpha():
            mydict[c] = [] if c not in mydict.keys() else mydict[c]
            currentKey = c
        elif c.isdigit():
            mydict[currentKey].append(c)

    sentence = ""
    for i in range(len(letters)):
        count = int(numbers[i])
        while count > 0:
            sentence += letters[i]
            count -= 1

    print(letters)
    print(numbers)
    print(mydict)
    print(sentence)