Question

My function finds in string hex notation (hexadecimal CSS colors) and replaces with the short notation.
For example: #000000 can be represented as #000

import re

def to_short_hex (string):
    match = re.findall(r'#[\w\d]{6}\b', string)

    for i in match:
        if not re.findall(r'#' + i[1] + '{6}', i):
            match.pop(match.index(i))

    for i in match:
        string = string.replace(i, i[:-3])

    return string;

to_short_hex('text #FFFFFF text #000000 #08088')

Out:

text #FFF text #000 #08088

Is there any way to optimize my code using list comprehension etc..?

Answer 1

This is what re.sub is for! It’s not a great idea to use a regex to find something and then do a further sequence of search-and-replace operations to change it. For one thing, it’s easy to accidentally replace things you didn’t mean to, and for another it does a lot of redundant work.

Also, you might want to shorten ‘#aaccee’ to ‘#ace’. This example does that too:

def to_short_hex(s):
    def shorten_match(match):
        hex_string = match.group(0)
        if hex_string[1::2]==hex_string[2::2]:
            return '#'+hex_string[1::2]
        return hex_string
    return re.sub(r"#[\da-fA-F]{6}\b", shorten_match, s)

Explanation

re.sub can take a function to apply to each match. It receives the match object and returns the string to substitute at that point.

Slice notation allows you to apply a stride. hex_string[1::2] takes every second character from the string, starting at index 1 and running to the end of the string. hex_string[2::2] takes every second character from the string, starting at index 2 and running to the end. So for the string “#aaccee”, we get “ace” and “ace”, which match. For the string “#123456”, we get “135” and “246”, which don’t match.