Python - removing repeated letters in a string

Question

Say I have a string in alphabetical order, based on the amount of times that a letter repeats.

• Example: "BBBAADDC".

There are 3 B's, so they go at the start, 2 A's and 2 D's, so the A's go in front of the D's because they are in alphabetical order, and 1 C. Another example would be CCCCAAABBDDAB.

Note that there can be 4 letters in the middle somewhere (i.e. CCCC), as there could be 2 pairs of 2 letters.

However, let's say I can only have n letters in a row. For example, if n = 3 in the second example, then I would have to omit one "C" from the first substring of 4 C's, because there can only be a maximum of 3 of the same letters in a row.

Another example would be the string "CCCDDDAABC"; if n = 2, I would have to remove one C and one D to get the string CCDDAABC

Example input/output:

1. n=2: Input: AAABBCCCCDE, Output: AABBCCDE
2. n=4: Input: EEEEEFFFFGGG, Output: EEEEFFFFGGG
3. n=1: Input: XXYYZZ, Output: XYZ

How can I do this with Python? Thanks in advance!

This is what I have right now, although I'm not sure if it's on the right track. Here, z is the length of the string.

for k in range(z+1):
        if final_string[k] == final_string[k+1] == final_string[k+2] == final_string[k+3]: 
            final_string = final_string.translate({ord(final_string[k]): None})
return final_string

Answer 1

Here's my solution:

def snip_string(string, n):
    list_string = list(string)
    list_string.sort()
    chars = set(string)
    for char in chars:
        while list_string.count(char) > n:
            list_string.remove(char)
    return ''.join(list_string)

Calling the function with various values for n gives the following output:

>>> string = "AAAABBBCCCDDD"
>>> snip_string(string, 1)
'ABCD'
>>> snip_string(string, 2)
'AABBCCDD'
>>> snip_string(string, 3)
'AAABBBCCCDDD'
>>> 

Edit

Here is the updated version of my solution, which only removes characters if the group of repeated characters exceeds n.

import itertools

def snip_string(string, n):
    groups = [list(g) for k, g in itertools.groupby(string)]
    string_list = []
    for group in groups:
        while len(group) > n:
            del group[-1]
        string_list.extend(group)
    return ''.join(string_list)

Output:

>>> string = "DDDAABBBBCCABCDE"
>>> snip_string(string, 3)
'DDDAABBBCCABCDE'

Answer 2

hello = "hello frrriend"


def replacing() -> str:
    global hello
    j = 0
    for i in hello:
        if j == 0:
            pass
        else:
            if i == prev:
                hello = hello.replace(i, "")
                prev = i
        prev = i
        j += 1
    return hello

replacing()

looks a bit primal but i think it works, thats what i came up with on the go anyways , hope it helps :D

Answer 3

Ok, based on your comment, you're either pre-sorting the string or it doesn't need to be sorted by the function you're trying to create. You can do this more easily with itertools.groupby():

import itertools

def max_seq(text, n=1):
    result = []
    for k, g in itertools.groupby(text):
        result.extend(list(g)[:n])
    return ''.join(result)


max_seq('AAABBCCCCDE', 2)
# 'AABBCCDE'
max_seq('EEEEEFFFFGGG', 4)
# 'EEEEFFFFGGG'
max_seq('XXYYZZ')
# 'XYZ'
max_seq('CCCDDDAABC', 2)
# 'CCDDAABC'

In each group g, it's expanded and then sliced until n elements (the [:n] part) so you get each letter at most n times in a row. If the same letter appears elsewhere, it's treated as an independent sequence when counting n in a row.

---

Edit: Here's a shorter version, which may also perform better for very long strings. And while we're using itertools, this one additionally utilises itertools.chain.from_iterable() to create the flattened list of letters. And since each of these is a generator, it's only evaluated/expanded at the last line:

import itertools

def max_seq(text, n=1):
    sequences = (list(g)[:n] for _, g in itertools.groupby(text))
    letters = itertools.chain.from_iterable(sequences)
    return ''.join(letters)

Answer 4

from itertools import groupby
n = 2
def rem(string):
    out = "".join(["".join(list(g)[:n]) for _, g in groupby(string)])
    print(out)

So this is the entire code for your question.

s = "AABBCCDDEEE"
s2 = "AAAABBBDDDDDDD"
s3 = "CCCCAAABBDDABBB"
s4 = "AAAAAAAA"
z = "AAABBCCCCDE"

With following test:

AABBCCDDEE
AABBDD
CCAABBDDABB
AA
AABBCCDE