How can I create a SEO friendly dash-delimited url from a string in PYTHON?

Question

Basically, my question is the same as this but in Python (and GAE), not C#.

Requirements:

• Separate each word by a dash and remove all punctuation (taking into account not all words are separated by spaces.)
• Function takes in a max length, and gets all tokens below that max length. Example: ToSeoFriendly("hello world hello world", 14) returns "hello-world"
• All words are converted to lower case.

Answer 1

def ToSeoFriendly(s, maxlen):
    '''Join with dashes, eliminate punction, clip to maxlen, lowercase.

        >>> ToSeoFriendly("The quick. brown4 fox jumped", 14)
        'the-quick-brow'

    '''
    t = '-'.join(s.split())                                # join words with dashes
    u = ''.join([c for c in t if c.isalnum() or c=='-'])   # remove punctation   
    return u[:maxlen].rstrip('-').lower()                  # clip to maxlen

if __name__ == '__main__':
    import doctest
    print doctest.testmod()

Answer 2

The term you're searching for is 'slugify'.

Answer 3

As alternative (and probably more tested version) I'm suggesting you use the (minimal modified) slugify code from Django:

import unicodedata
import re

def slugify(value):
    """
    Normalizes string, converts to lowercase, removes non-alpha characters,
    and converts spaces to hyphens.
    """
    value = unicodedata.normalize('NFKD', value).encode('ascii', 'ignore').decode('ascii')
    value = re.sub('[^\w\s-]', '', value).strip().lower()
    return re.sub('[-\s]+', '-', value)

See: https://github.com/django/django/blob/master/django/utils/text.py#L435