Canonical URL compare in Python?

Are there any tools to do a URL compare in Python?

For example, if I have http://google.com and google.com/ I'd like to know that they are likely to be the same site.

If I were to construct a rule manually, I might Uppercase it, then strip off the http:// portion, and drop anything after the last alpha-numeric character.. But I can see failures of this, as I'm sure you can as wel开发者_StackOverflowl.

Is there a library that does this? How would you do it?

This off the top of my head:

def canonical_url(u):
    u = u.lower()
    if u.startswith("http://"):
        u = u[7:]
    if u.startswith("www."):
        u = u[4:]
    if u.endswith("/"):
        u = u[:-1]
    return u

def same_urls(u1, u2):
    return canonical_url(u1) == canonical_url(u2)

Obviously, there's lots of room for more fiddling with this. Regexes might be better than startswith and endswith, but you get the idea.

There is quite a bit to creating a canonical url apparently. The url-normalize library is best that I have tested.

Depending on the source of your urls you may wish to clean them of other standard parameters such as UTM codes. w3lib.url.url_query_cleaner is useful for this.

Combining this with Ned Batchelder's answer could look something like:

Code:

from w3lib.url import url_query_cleaner
from url_normalize import url_normalize

urls = ['google.com',
'google.com/',
'http://google.com/',
'http://google.com',
'http://google.com?',
'http://google.com/?',
'http://google.com//',
'http://google.com?utm_source=Google']


def canonical_url(u):
    u = url_normalize(u)
    u = url_query_cleaner(u,parameterlist = ['utm_source','utm_medium','utm_campaign','utm_term','utm_content'],remove=True)

    if u.startswith("http://"):
        u = u[7:]
    if u.startswith("https://"):
        u = u[8:]
    if u.startswith("www."):
        u = u[4:]
    if u.endswith("/"):
        u = u[:-1]
    return u

list(map(canonical_url,urls))

Result:

['google.com',
 'google.com',
 'google.com',
 'google.com',
 'google.com',
 'google.com',
 'google.com',
 'google.com']

You could look up the names using dns and see if they point to the same ip. Some minor string processing may be required to remove confusing chars.

from socket import gethostbyname_ex

urls = ['http://google.com','google.com/','www.google.com/','news.google.com']

data = []
for orginalName in urls:
    print 'url:',orginalName
    name = orginalName.strip()
    name = name.replace( 'http://','')
    name = name.replace( 'http:','')
    if name.find('/') > 0:
        name = name[:name.find('/')]
    if name.find('\\') > 0:
        name = name[:name.find('\\')]
    print 'dns lookup:', name
    if name:
        try:
            result = gethostbyname_ex(name)
        except:
            continue # Unable to resolve
        for ip in result[2]:
            print 'ip:', ip
            data.append( (ip, orginalName) )

print data

result:

url: http://google.com
dns lookup: google.com
ip: 66.102.11.104
url: google.com/
dns lookup: google.com
ip: 66.102.11.104
url: www.google.com/
dns lookup: www.google.com
ip: 66.102.11.104
url: news.google.com
dns lookup: news.google.com
ip: 66.102.11.104
[('66.102.11.104', 'http://google.com'), ('66.102.11.104', 'google.com/'), ('66.102.11.104', 'www.google.com/'), ('66.102.11.104', 'news.google.com')]

It's not 'fuzzy', it just find the 'distance' between two strings:

http://pypi.python.org/pypi/python-Levenshtein/

I would remove all portions which are semantically meaningful to URL parsing (protocol, slashes, etc.), normalize to lowercase, then perform a levenstein distance, then from there decide how many difference is an acceptable threshold.

Just an idea.