[FIXED] How to create a list which only adds letters that are unique to adjacent indices in python?

Issue

I have created a function which randomly generates a list of the letters "a", "b", "c", and "d". I would like to create a new list which is the same as the first list but with any letters/items which are the same as the previous letter/item removed. Where I am having problems is referring to the previous letter in the list.

For example, if :

letterlist = ['a','a','a','b','b','a,',b']

then the output should be,

nondupelist = ['a','b','a','b']

The problem is that nodupeletterlist is the same as letterlist - meaning it's not removing items which are the same as the last - because I am getting the function to refer to the previous item in letterlist wrong. I have tried using index and enumerate, but I am obviously using them wrong because I'm not getting the correct results. Below is my current attempt.

import random

def rdmlist(letterlist, nodupeletterlist):
    for item in range(20):
        rng = random.random()
        if rng < 0.25:
            letterlist.append("a")
        elif 0.25 <= rng and rng < 0.5:
            letterlist.append("b")
        elif 0.5 <= rng and rng < 0.75:
            letterlist.append("c")
        else:
            letterlist.append("d")
    for letter in letterlist:
        if letter != letterlist[letterlist.index(letter)-1]:
            nodupeletterlist.append(letter)
        else:
            pass
    return

letterlist1 = []
nodupeletterlist1 = []
rdmlist(letterlist1, nodupeletterlist1)

EDIT: This is what I ended up using. I used this solution simply because I understand how it works. The answers below may provide more succinct or pythonic solutions.

    for index, letter in enumerate(letterlist, start=0):
        if 0 == index:
            nodupeletterlist.append(letter)
        else:
            pass
    for index, letter in enumerate(letterlist[1:], start = 1):
        if letter != letterlist[index-1]:
            nodupeletterlist.append(letter)
        else:
            pass

Solution

You can use itertools.groupby:

import itertools

nodupeletterlist = [k for k, _ in itertools.groupby(letterlist)]

---

Solution without using itertools, as requested in the comments:

def nodupe(letters):
    if not letters:
        return []
    r = [letters[0]]
    for ch in letters[1:]:
        if ch != r[-1]:
            r.append(ch)
    return r

nodupeletterlist = nodupe(letterlist)

---

A fixed version of the proposed "working solution":

def nodupe(letters):
    if not letters:
        return []
    r = [letters[0]]
    r += [l for i, l in enumerate(letters[1:]) if l != letters[i]]
    return r

nodupeletterlist = nodupe(letterlist)

---

You can also simplify your random generator a bit, by using random.choices:

import random

chars = 'abcd'
letterlist = random.choices(chars, k=20)

or by using random.randint:

import random

start, end = ord('a'), ord('d')
letterlist = [chr(random.randint(start, end)) for _ in range(20)]

Answered By - Jan Christoph Terasa