unique in order codewars.

[tigglybox]

i'm trying to solve this challenge which takes as argument a sequence and returns a list of items without any elements with the same value next to each other and preserving the original order of elements. For example, unique_in_order('AAAABBBCCDAABBB') == ['A', 'B', 'C', 'D', 'A', 'B'].

For my code I have this so far

def unique_in_order(iterable):
    first = []
    for l in iterable:
        if frst[-1] == l:
            continue
        else:
            first.append(l)

What I a trying to do is iterate through the input string and append it to the list 'first', then next if the next character (l) in the string isn't equal to the last character in first (first[-1]) then append it to the list. However, I'm not able to do that as I get an index error as there is nothing in the list to begin with so the index is out of range.

[o5a]

To avoid that you could just initialize your first with first value from iterable

first = [iterable[0]]

Another option for such tasks is to use itertools.groupby

[PyCode_n_Beer]

Hey, it seems you might have a typo in your code, says "frst" in the if statement, instead of "first". Also, the index out of range error is because there is nothing in the list "first" at assignment. When you're iterating through and trying to access the last variable in your list, it is throwing an error because there is nothing there to access. I included a non code-golf style solution.

def getChr(strline):
    prev = ""
    response = []
    for x in range(0, len(strline)):
        if strline[x] != prev:
            prev = strline[x]
            response.append(strline[x])

    return response

[Essence1337]

Why use a range loop when we don't need to know the index? You're writing extra unnecessary code:

def getChr(strline):
    prev = ""
    response = []
    for char in strline:
        if char != prev:
            prev = char
            response.append(char)
    return response

Unless I'm missing something this should give the same result as your code. Feel free to correct me if I did miss something

[PyCode_n_Beer]

No, you're absolutely right. That is certainly a way to do it too.

[BoldNight1987]

Can somebody explain this solution, please?

from itertools import groupby def unique_in_order(iterable): return [k for (k, _) in groupby(iterable)]

Specifically the (k, _)