Fairly new to Python, and I have been doing a few Edabit challenges, to better help with my problem-solving. I have just completed a some what semi-difficult challenge, and I was hoping for some feed back.

The challenge itself:

Create a function that takes an imgur link (as a string) and extracts the unique id and type. Return an object containing the unique id, and a string indicating what type of link it is.

The link could be pointing to:


I came up with the following.

import re

def imgurUrlParser(url):

    url_regex          =    "^[http:\/\/www\.|https:\/\/www\.|http:\/\/|https:\/\/|www\.]*[imgur|i.imgur]*\.com"
    url = re.match(url_regex, url).string

    gallery_regex     =     re.match(url_regex + "(\/gallery\/)(\w+)", url)
    album_regex       =     re.match(url_regex + "(\/a\/)(\w+)", url)
    image_regex       =     re.match(url_regex + "\/(\w+)", url)
    direct_link_regex =     re.match(url_regex + "(\w+)(\.\w+)", url)

    if gallery_regex:
        return { "id" : gallery_regex.group(2), "type" : "gallery" } 
    elif album_regex:
        return { "id" : album_regex.group(2), "type" : "album" }
    elif image_regex:
        return { "id" : image_regex.group(1), "type" : "image" } 
    elif direct_link_regex:
        return { "id" : direct_link_regex.group(1), "type" : "image"}
  • \$\begingroup\$ This code doesn't run. direct_link_regex is a string and strings don't have a group() method. Also, the parentheses for that regex are not balanced properly. Perhaps you copied an old version of your code? \$\endgroup\$
    – JakeD
    Commented Sep 25, 2018 at 2:42

3 Answers 3


According to PEP 8, the official Python style guide, function names should be lower_case_with_underscores. Furthermore, this function parses the URL, rather than creating a parser object, so the function name should be a verb phrase rather than a noun phrase.

By RFC 1738, the scheme and host portions of URLs are case-insensitive. Also, it is allowable to include a redundant port number in the URL.

Imgur also also partners with certain other websites. For instance, when you upload an image through the question editor Stack Exchange site, it will end up on https://i.stack.imgur.com.

There is a lot of commonality in the various regexes. Consider combining them all into a single regex. Use named capture groups to avoid the magic group numbers.

A docstring with doctests would be very beneficial for this function.

import re

def parse_imgur_url(url):
    Extract the type and id from an Imgur URL.

    >>> parse_imgur_url('http://imgur.com/a/cjh4E')
    {'id': 'cjh4E', 'type': 'album'}
    >>> parse_imgur_url('HtTP://imgur.COM:80/gallery/59npG')
    {'id': '59npG', 'type': 'gallery'}
    >>> parse_imgur_url('https://i.imgur.com/altd8Ld.png')
    {'id': 'altd8Ld', 'type': 'image'}
    >>> parse_imgur_url('https://i.sstatic.net/ELmEk.png')
    {'id': 'ELmEk', 'type': 'image'}
    >>> parse_imgur_url('http://not-imgur.com/altd8Ld.png') is None
    Traceback (most recent call last):
    ValueError: "http://not-imgur.com/altd8Ld.png" is not a valid imgur URL
    >>> parse_imgur_url('tftp://imgur.com/gallery/59npG') is None
    Traceback (most recent call last):
    ValueError: "tftp://imgur.com/gallery/59npG" is not a valid imgur URL
    >>> parse_imgur_url('Blah') is None
    Traceback (most recent call last):
    ValueError: "Blah" is not a valid imgur URL
    match = re.match(
    if not match:
        raise ValueError('"{}" is not a valid imgur URL'.format(url))
    return {
        'id': match.group('id'),
        'type': 'album' if match.group('album') else
                'gallery' if match.group('gallery') else

Note that the regex above relies on the (?aiLmsux-imsx:...) feature of Python 3.6, and the doctests rely on the predictable order of dictionary keys in Python 3.6 / 3.7.

  • \$\begingroup\$ Can you comment on why you chose to name the function parse_imgr_url instead of imgr_url_parser? \$\endgroup\$ Commented Sep 25, 2018 at 15:48
  • \$\begingroup\$ @MatthewLeingang Thanks for reminding me. Added the recommendation in Rev 3. \$\endgroup\$ Commented Sep 25, 2018 at 16:09
  • \$\begingroup\$ @MatthewLeingang function (method/action/procedure, etc.) should start with a verb. They do something. A noun is more suitable on a variable or an object. \$\endgroup\$
    – aloisdg
    Commented Sep 26, 2018 at 6:56

For a first pass, not bad! Your code is pretty easy to follow.


  1. Don't use [] to match different strings. [] matches any set of characters, so [imgur|i.imgur]* will match ``, g, mgi, etc. You probably wanted a non-capturing group, which is specified with (?: ...), re Docs

  2. Name functions with snake_case, as recommended by PEP 8.

  3. The challenge as stated doesn't specify what should happen if the string passed in doesn't match the link format. Right now your code will throw an AttributeError, which isn't very helpful to the caller. I'd recommend raising an explicit exception with a more helpful message.

  4. Your last case, direct_link_regex is never reached with valid input since it is handled by image_regex.


  1. Concatenating the regex to handle each case is somewhat messy. It would be better to have a single regex which handles all cases.

  2. Regular expressions are usually expressed using raw strings, that is, strings with an r prefix. This helps with escaping characters correctly. In this case I'm guessing you just got lucky that it worked as you expected.

  3. Including a docstring is always a good idea, and you can even embed tests using doctest.

How I would implement this function:

def imgur_url_parser(url):
    Parses an imgur url into components.

    >>> imgur_url_parser("http://imgur.com/a/cjh4E") == {"type": "album", "id": "cjh4E"}
    >>> imgur_url_parser("http://imgur.com/gallery/59npG") == {"type": "gallery", "id": "59npG"}
    >>> imgur_url_parser("http://i.imgur.com/altd8Ld.png") == {"type": "image", "id": "altd8Ld"}
    >>> imgur_url_parser("http://imgur.com/OzZUNMM") == {"type": "image", "id": "OzZUNMM"}
    match = re.match(r"^https?://(?:www\.|i\.)?imgur\.com/([\w.]+)/?(\w*)$", url)
    if not match:
        raise ValueError('The string "{}" is not a valid imgur link'.format(url))
    # Empty when this is an image link
    if not match.group(2):
        # Remove image extension, if it exists
        image_id = re.sub(r"(\.\w+)?$", "", match.group(1))
        return { "id": image_id, "type": "image" }
    url_type = match.group(1) == "a" and "album" or "gallery"
    return { "id": match.group(2), "type": url_type}

if __name__ == "__main__":
    import doctest
  • 1
    \$\begingroup\$ While slightly longer and a different order, "album" if match.group(1) == "a" else "gallery" is the built-in Python way to do ternary expressions. \$\endgroup\$
    – Graipher
    Commented Sep 25, 2018 at 14:00

The other answers are great, but I'll address something more fundamental: Regex is not the right tool for URL parsing. Python has terrific built-in modules. Take advantage of them. urlparse is great!

acceptable_hostname_regex = re.compile("(i.(stack.)?)?imgur.com")

def parse_imgur_url(url):
    parsed = urlparse(url)

    if not acceptable_hostname_regex.match(parsed.hostname):
        raise ValueError(f"The string {url} is not a valid imgur link")

    path_components = [c for c in parsed.path.split("/") if c]

    if len(path_components) == 1:
        image_name = path_components[0]
        id = image_name[:image_name.index(".")]
        type = "image"

    elif len(path_components) == 2:
        type = path_components[0]
        id = path_components[1]

        type_overwrite_table = { "a": "album" }

        type = type_overwrite_table.get(type, type)

        raise ValueError(f"The imgur link {url} has too many path components.")

    return { 'id': id, 'type': type }

Another issue: your gallery_regex, album_regex, etc. local variables don't actually store regex objects (of type re.Pattern) like their name would imply, rather, they're storing re.Match objects.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.