5th clear solution for URL Normalization by gyahun_dash - python coding challenges

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5th solution in Clear category for URL Normalization by gyahun_dash

import re

def convert(match): # pattern: '%([0-9a-f]{2})', group(0): %xx group(1): xx
    c = chr(int(match.group(1), 16)).lower()

    # if c in range: decode(3rd), else: capitalize(2nd)
    return c if re.match(r'\w|[-.~]', c, re.ASCII) else match.group(0).upper()

def checkio(url):
    url = url.lower() # 1st
    url = re.sub(r'%([0-9a-f]{2})', convert, url) # 2nd, 3rd
    url = re.sub(r':80(?=\b)', '', url) # 4th

    while True: # 5th: delete '/.' and '/ww/..' repeatedly
        url, dels = re.subn(r'/\.(?=/)|/[^/.]+?/\.{2}', '', url)
        if dels == 0: return url

July 28, 2014

Comments:

ogoro on Jan. 25, 2021, 1:48 p.m.
many thanks for subn loop idea! it seems to work also with greedy '+': <pre class='brush: python'> url, dels = re.subn(r'/\.(?=/)', '', url) url, dels = re.subn(r'/[^/.]+/\.\.', '', url) </pre>

juestr on Dec. 4, 2021, 2:15 p.m.
Yeah that's the right way to do it. I realized too late that everything is to be lowercased except the escapes, and that there is no reason to parse and handle the different parts of the URL separately. Of course, if the path would countain `:80` followed by a non word character this would break, but thanks for non pathological tests. (I was already miffed after finding out urllib.parse isn't available)

Rock

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