Expression to remove URL links from Twitter tweet

The following regex will capture two matched groups: the first includes everything in the tweet until the url and the second will catch everything that will come after the URL (empty in the example you posted above):

import re
str = 'This is a tweet with a url: http://t.co/0DlGChTBIx'
clean_tweet = re.match('(.*?)http.*?\s?(.*?)', str)
if clean_tweet: 
    print clean_tweet.group(1)
    print clean_tweet.group(2) # will print everything after the URL 

you can use:

text = 'Amazing save #FACup #zeebox https://stackoverflow.com/tiUya56M Ok'
text = re.sub(r'https?:\/\/\S*', '', text, flags=re.MULTILINE)

# output: 'Amazing save #FACup #zeebox  Ok'

• r The solution is to use Python’s raw string notation for regular expression patterns; backslashes are not handled in any special way in a string literal prefixed with 'r'
• ? Causes the resulting RE to match 0 or 1 repetitions of the preceding RE. https? will match either ‘http’ or ‘https’.
• https?:\/\/ will match any "http://" and "https://" in string
• \S Returns a match where the string DOES NOT contain a white space character
• * Zero or more occurrences

You could try the below re.sub function to remove URL link from your string,

>>> str = 'This is a tweet with a url: http://t.co/0DlGChTBIx'
>>> m = re.sub(r':.*$', ":", str)
>>> m
'This is a tweet with a url:'

It removes everything after first : symbol and : in the replacement string would add : at the last.

This would prints all the characters which are just before to the : symbol,

>>> m = re.search(r'^.*?:', str).group()
>>> m
'This is a tweet with a url:'

Do this:

result = re.sub(r"http\S+", "", subject)

• http matches literal characters
• \S+ matches all non-whitespace characters (the end of the url)
• we replace with the empty string