Regex matching a set of characters except when one of them is last in python

Question

I am trying to extract site names from articles but sometimes the domain names that are at the end of a sentence have an undesired dot character such as in "You can find more information at www.website.gov.us. Last year we had..."

I want to get the domain name and avoid including the last dot character. My current regex in python is:

Regex = r'[www.\w\.]+'

Possible duplicate of [Extracting a URL in Python](https://stackoverflow.com/questions/839994/extracting-a-url-in-python) — Allan, Jun 11 '18 at 08:58

score 3 · Accepted Answer · answered Jun 15 '18 at 00:04

3

thanks everybody for your answers. This regex solved my problem:

Regex = r'https?://(?:ww\w\.)?([a-zA-Z\d-]+(?:\.[a-zA-Z\d-]+)+)'

answered Jun 15 '18 at 00:04

TheYogi

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score 1 · Answer 2 · answered Jun 11 '18 at 01:19

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You could use the following regex: (?:http://|www.)[^\' ]+

That would get the website address with the dot character and then use rstrip('.') to remove it.

"www.website.gov.us.".rstrip('.') => "www.website.gov.us"

answered Jun 11 '18 at 01:19

Ryan

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Easier to just do the whole thing in regex, I think. – Jonas Jun 11 '18 at 01:49
Great solution, I had the same approach at first! However I am looking for a full regex answer. – TheYogi Jun 15 '18 at 00:02

score 1 · Answer 3 · answered Jun 11 '18 at 01:47

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Good news: No need for hacking in python, you can do it all in regex!

r'www(\.[a-z]+)+'

First, you match the 'www' and then look for the repeated pattern of a dot followed by letters. If your url might have uppercase letters in it, change the '[a-z]' to '[a-zA-Z]'.

answered Jun 11 '18 at 01:47

Jonas

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Hey Jonas, this was not the full solution I needed but your comment definitely pointed me in the right direction. Thanks. FYI I posted the solution to my problem below. – TheYogi Jun 15 '18 at 00:03
Glad it helped! Next time, if you need the https etc in your regex, make sure to include examples like that in your question. – Jonas Jun 15 '18 at 02:15

Allan · Answer 4 · 2018-06-11T08:57:18.303

You can use the following monstrous regex:

\b((?:https?://)?(?:(?:www\.)?(?:[\da-z\.-]+)\.(?:[a-z]{2,6})|(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)|(?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])))(?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])?(?:/[\w\.-]*)*/?)\b

Demo regex101

This regex will accept urls in the following format:

INPUT:

add1 http://mit.edu.com abc
add2 https://facebook.jp.com.2. abc
add3 www.google.be. uvw
add4 https://www.google.be. 123
add5 www.website.gov.us test2
Hey bob on www.test.com. 
another test with ipv4 http://192.168.1.1/test.jpg. toto2
website with different port number www.test.com:8080/test.jpg not port 80
www.website.gov.us/login.html
test with ipv4 192.168.1.1/test.jpg.
search at google.co.jp/maps.
test with ipv6 2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg.

OUTPUT:

http://mit.edu.com
https://facebook.jp.com
www.google.be
https://www.google.be
www.website.gov.us
www.test.com
http://192.168.1.1/test.jpg
www.test.com:8080/test.jpg
www.website.gov.us/login.html
192.168.1.1/test.jpg
google.co.jp/maps
2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg

Explanations:

\b is used for word boundary to delimit the URL and the rest of the text
(?:https?://)? to match http:// or https// if present
(?:(?:www\.)?(?:[\da-z\.-]+)\.(?:[a-z]{2,6}) to match standard url (that might start with www. (lets call it STANDARD_URL)
(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?) to match standard Ipv4 (lets call it IPv4)
to match the IPv6 URLs: (?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])) (lets call it IPv6)
to match the port part (lets call it PORT) if present: (?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])
to match the (?:/[\w\.-]*)*/?) target object part of the url (html file, jpg,...) (lets call it RESSOURCE_PATH)

This gives the following regex:

\b((?:https?://)?(?:STANDARD_URL|IPv4|IPv6)(?:PORT)?(?:RESSOURCE_PATH)\b

Sources:

IPv6: Regular expression that matches valid IPv6 addresses

IPv4: https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9780596802837/ch07s16.html

PORT: https://stackoverflow.com/a/12968117/8794221

Other sources: https://code.tutsplus.com/tutorials/8-regular-expressions-you-should-know--net-6149

$ more url.py

import re

inputString = """add1 http://mit.edu.com abc
add2 https://facebook.jp.com.2. abc
add3 www.google.be. uvw
add4 https://www.google.be. 123
add5 www.website.gov.us test2
Hey bob on www.test.com. 
another test with ipv4 http://192.168.1.1/test.jpg. toto2
website with different port number www.test.com:8080/test.jpg not port 80
www.website.gov.us/login.html
test with ipv4 (192.168.1.1/test.jpg).
search at google.co.jp/maps.
test with ipv6 2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg."""

regex=ur"\b((?:https?://)?(?:(?:www\.)?(?:[\da-z\.-]+)\.(?:[a-z]{2,6})|(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)|(?:(?:[0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,7}:|(?:[0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|(?:[0-9a-fA-F]{1,4}:){1,5}(?::[0-9a-fA-F]{1,4}){1,2}|(?:[0-9a-fA-F]{1,4}:){1,4}(?::[0-9a-fA-F]{1,4}){1,3}|(?:[0-9a-fA-F]{1,4}:){1,3}(?::[0-9a-fA-F]{1,4}){1,4}|(?:[0-9a-fA-F]{1,4}:){1,2}(?::[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:(?:(?::[0-9a-fA-F]{1,4}){1,6})|:(?:(?::[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(?::[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(?:ffff(?::0{1,4}){0,1}:){0,1}(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])|(?:[0-9a-fA-F]{1,4}:){1,4}:(?:(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])\.){3,3}(?:25[0-5]|(?:2[0-4]|1{0,1}[0-9]){0,1}[0-9])))(?::[0-9]{1,4}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])?(?:/[\w\.-]*)*/?)\b"

matches = re.findall(regex, inputString)
print(matches)

OUTPUT:

$ python url.py 
['http://mit.edu.com', 'https://facebook.jp.com', 'www.google.be', 'https://www.google.be', 'www.website.gov.us', 'www.test.com', 'http://192.168.1.1/test.jpg', 'www.test.com:8080/test.jpg', 'www.website.gov.us/login.html', '192.168.1.1/test.jpg', 'google.co.jp/maps', '2001:0db8:0000:85a3:0000:0000:ac1f:8001/test.jpg']

score -1 · Answer 5 · answered Jun 11 '18 at 01:26

-1

Using your current regex, you can just make it like this:

Regex = r'[www.\w\.]+[^\.]'

And it will work. But this is not the best regex to do the job. The suggestion by Ryan is better, and will solve it, and you can add also the [^\.] to it's end and skip the rstrip call.

answered Jun 11 '18 at 01:26

MoustafaS

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Doesn't work for me, matches the "You " in the example string. – Jonas Jun 11 '18 at 01:48

Regex matching a set of characters except when one of them is last in python

5 Answers5