How to remove any URL within a string in Python
PythonRegexUrlReplacePython 2.7Python Problem Overview
I want to remove all URLs inside a string (replace them with "") I searched around but couldn't really find what I want.
Example:
text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6
http://url.com/bla3/blah3/
I want the result to be:
text1
text2
text3
text4
text5
text6
Python Solutions
Solution 1 - Python
the shortest way
re.sub(r'http\S+', '', stringliteral)
Solution 2 - Python
Python script:
import re
text = re.sub(r'^https?:\/\/.*[\r\n]*', '', text, flags=re.MULTILINE)
Output:
text1
text2
text3
text4
text5
text6
Test this code here.
Solution 3 - Python
This worked for me:
import re
thestring = "text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"
URLless_string = re.sub(r'\w+:\/{2}[\d\w-]+(\.[\d\w-]+)*(?:(?:\/[^\s/]*))*', '', thestring)
print URLless_string
Result:
text1
text2
text3
text4
text5
text6
Solution 4 - Python
Removal of HTTP links/URLs mixed up in any text:
import re
re.sub(r'''(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))''', " ", text)
Solution 5 - Python
This solution caters for http, https and the other normal url type special characters :
import re
def remove_urls (vTEXT):
vTEXT = re.sub(r'(https|http)?:\/\/(\w|\.|\/|\?|\=|\&|\%)*\b', '', vTEXT, flags=re.MULTILINE)
return(vTEXT)
print( remove_urls("this is a test https://sdfs.sdfsdf.com/sdfsdf/sdfsdf/sd/sdfsdfs?bob=%20tree&jef=man lets see this too https://sdfsdf.fdf.com/sdf/f end"))
Solution 6 - Python
I wasn't able to find any that handled my particular situation, which was removing urls in the middle of tweets that also have whitespaces in the middle of urls so I made my own:
(https?:\/\/)(\s)*(www\.)?(\s)*((\w|\s)+\.)*([\w\-\s]+\/)*([\w\-]+)((\?)?[\w\s]*=\s*[\w\%&]*)*
here's an explanation:
(https?:\/\/) matches http:// or https://
(\s)* optional whitespaces
(www\.)? optionally matches www.
(\s)* optionally matches whitespaces
((\w|\s)+\.)* matches 0 or more of one or more word characters followed by a period
([\w\-\s]+\/)* matches 0 or more of one or more words(or a dash or a space) followed by ''
([\w\-]+) any remaining path at the end of the url followed by an optional ending
((\?)?[\w\s]*=\s*[\w\%&]*)* matches ending query params (even with white spaces,etc)
test this out here:https://regex101.com/r/NmVGOo/8
Solution 7 - Python
What you really want to do is to remove any string that starts with either http:// or https:// plus any combination of non white space characters. Here is how I would solve it. My solution is very similar to that of @tolgayilmaz
#Define the text from which you want to replace the url with "".
text ='''The link to this post is https://stackoverflow.com/questions/11331982/how-to-remove-any-url-within-a-string-in-python'''
import re
#Either use:
re.sub('http://\S+|https://\S+', '', text)
#OR
re.sub('http[s]?://\S+', '', text)
And the result of running either code above is
>>> 'The link to this post is '
I prefer the second one because it is more readable.
Solution 8 - Python
In order to remove any URL within a string in Python, you can use this RegEx function :
import re
def remove_URL(text):
"""Remove URLs from a text string"""
return re.sub(r"http\S+", "", text)
Solution 9 - Python
I know this has already been answered and its stupid late but I think this should be here. This is a regex that matches any kind of url.
[^ ]+\.[^ ]+
It can be used like
re.sub('[^ ]+\.[^ ]+','',sentence)
Solution 10 - Python
You could also look at it from the other way around...
from urlparse import urlparse
[el for el in ['text1', 'FTP://somewhere.com', 'text2', 'http://blah.com:8080/foo/bar#header'] if not urlparse(el).scheme]
Solution 11 - Python
The following regular expression in Python works well for detecting URL(s) in the text:
source_text = '''
text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6 '''
import re
url_reg = r'[a-z]*[:.]+\S+'
result = re.sub(url_reg, '', source_text)
print(result)
Output:
text1
text2
text3
text4
text5
text6
Solution 12 - Python
import re
s = '''
text1
text2
http://url.com/bla1/blah1/
text3
text4
http://url.com/bla2/blah2/
text5
text6
http://url.com/bla3/blah3/'''
g = re.findall(r'(text\d+)',s)
print ('list',g)
for i in g:
print (i)
Out
list ['text1', 'text2', 'text3', 'text4', 'text5', 'text6']
text1
text2
text3
text4
text5
text6
Solution 13 - Python
I think the most general URL regex pattern is this one:
URL_PATTERN = r'[A-Za-z0-9]+://[A-Za-z0-9%-_]+(/[A-Za-z0-9%-_])*(#|\\?)[A-Za-z0-9%-_&=]*'
There is a small module that does what do you want:
pip install mysmallutils
from mysutils.text import remove_urls
remove_urls(text)
Solution 14 - Python
A simple .* with a positive look behind should do the job.
text="text1\ntext2\nhttp://url.com/bla1/blah1/\ntext3\ntext4\nhttp://url.com/bla2/blah2/\ntext5\ntext6"
req=re.sub(r'http.*?(?=\s)', " ", text)
print(req)