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i have the following link:
https://webcache.googleusercontent.com/search?q=cache:jAc7OJyyQboJ:https://cooking.nytimes.com/learn-to-cook+&cd=5&hl=en&ct=clnk
I have multiple links in a dataset. Each link is of same pattern. I want to get a specific part of the link, for the above link i would be the bold part of the link above. I want text starting from 2nd http to before first + sign.
I don't know how to do so using regex. I am working in python. Kindly help me out.
If each link has the same pattern you do not need regex. You can use string.find() and string cutting
link = "https://webcache.googleusercontent.com/search?q=cache:jAc7OJyyQboJ:https://cooking.nytimes.com/learn-to-cook+&cd=5&hl=en&ct=clnk"
# This finds the second occurrence of "https://" and returns the position
second_https = link.find("https://", link.find("https://")+1)
# Index of the end of the link
end_of_link = link.find("+")
new_link = link[second_https:end_of_link]
print(new_link)
This will return "https://cooking.nytimes.com/learn-to-cook" and will work if the link follows the same pattern as described (it is the second https:// in the link and ends with + sign)
I'd go with urlparse (Python 2) or urlparse (Python 3) and a little bit of regex:
import re
from urlparse import urlparse
url_example = "https://webcache.googleusercontent.com/search?q=cache:jAc7OJyyQboJ:https://cooking.nytimes.com/learn-to-cook+&cd=5&hl=en&ct=clnk"
parsed = urlparse(url_example)
result = re.findall('https?.*', parsed.query)[0].split('+')[0]
print(result)
Output:
https://cooking.nytimes.com/learn-to-cook
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I need to replace part of an an html string via a Django custom template tag that I am customising.
The string in it's raw format is as follows:
arg = "<a class='login-password-forget' tabindex='1' href='{% url 'account:pwdreset'%}'>Forgot your password?</a>"
I need to replace the {% url 'account:pwdreset'%} part with a url string using re.search().
The code that I written is clumsy and I would appreciate help with finding a better way of achieving the same.
url_string = re.search("{.*?}", arg)
url_string_inner = re.search("'(.+?)'", url_string.group())
add_html = SafeText(''.join([arg.split('{')[0], reverse(url_string_inner.group(1)), arg.split('}')[1]]))
!!UPDATE!!
The solution that I ran with is as follows:
url_string = re.search("{.*?}", arg)
url_string_inner = re.search("'(.+?)'", url_string.group())
add_html = SafeText(''.join([arg.split('{')[0], reverse(url_string_inner.group(1)), arg.split('}')[1]]))
Thank you Fourth Bird for your help.
If you only want to replace the the part with 'account:pwdreset' you could use re.sub with a capture group and use that group in the replacement between single quotes
'{%\s*url '([^']*)'%}
Regex demo | Python demo
import re
pattern = r"'{%\s*url '([^']*)'%}"
s = "<a class='login-password-forget' tabindex='1' href='{% url 'account:pwdreset'%}>Forgot your password?</a>"
print(re.sub(pattern, r"'\1'", s))
Output
<a class='login-password-forget' tabindex='1' href=account:pwdreset>Forgot your password?</a>
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I am very new to Python and I am trying to create a list out of string in python.
Input = "<html><body><ul style="padding-left: 5pt"><i>(See attached file: File1.pdf)</i><i>(See attached file: File2.ppt)</i><i>(See attached file: File3.docx)</i></ul></body></html>"
Desired Output = [File1.pdf, File2.ppt, File3.docx]
What is the most efficient and pythonic way to achieve this? Any help will be very much appreciated.
Thanks
You can use beatifulsoup, which has HTML parsing utils.
>>> from bs4 import BeautifulSoup
>>> html = """<html><body><ul style="padding-left: 5pt"><i>(See attached file: File1.pdf)</i><i>(See attached file: File2.ppt)</i><i>(See attached file: File3.docx)</i></ul></body></html>"""
>>> soup = BeautifulSoup(html, parser='html')
>>> files_list = [i.text.split('file: ')[1].replace(')', '') for i in soup.find_all('i')]
>>> print(files_list)
['File1.pdf', 'File2.ppt', 'File3.docx']
There might be a nice way to do this using a HTML parser like shree.pat18 suggested but here is a quick and dirty way using string.split()
Output = [s.split(")")[0] for s in Input.split("file: ")[1:]]
By first splitting on "file: " we get list of strings, the first one contains the first part of the original string so we don't care about that one. The others start with the filenames that we want and the first character we don't care about is ")". So split on ")" and take the first part.
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Please, recommend regex expression to extract the last two words from an URL like:
INPUT OUTPUT
'www.abcd.google.com' --> 'google.com'
'www.xyz.stackoverflow.com' --> 'stackoverflow.com'
use this regex with 'negative lookahead' feature:
import re
for url in ['www.abcd.google.com','www.xyz.stackoverflow.com']:
print (re.search (r'\w*\.(?!\w*\.)\w*', url)[0])
google.com
stackoverflow.com
Here is the example at Regexr:
Use split with . as delimetr:
url_strings = 'www.xyz.stackoverflow.com'
s = '.'.join(url_strings.split('.')[-2:])
# stackoverflow.com
print(s)
If input validation is required:
url_strings = 'www.xyz.stackoverflow.com'
def return_last_words(url_string, last_words_count=2):
splitted = url_string.split('.')
if last_words_count < len(splitted):
return '.'.join(splitted[-last_words_count:])
return url_string
print(return_last_words(url_strings))
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I want to remove links in the format Reddit uses
comment = "Hello this is my [website](https://www.google.com)"
no_links = RemoveLinks(comment)
# no_links == "Hello this is my website"
I found a similar question about the same thing, but I don't know how to translate it to python.
I am not that familiar with regex so I would appreciate it if you explained what's happening.
You could do the following:
import re
pattern = re.compile('\[(.*?)\]\(.*?\)')
comment = "Hello this is my [website](https://www.google.com)"
print(pattern.sub(r'\1', comment))
The line:
pattern = re.compile('\[(.*?)\]\(.*?\)')
creates a regex pattern that will search for anything surrounded by square brackets, followed by anything surrounded by parenthesis, the '?' indicates that they should match as little text as possible (non-greedy).
The function sub(r'\1', comment) replaces a match by the first capturing group in this case the text inside the brackets.
For more information about regex I suggest you read this.
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How to open a webpage and search for a word in python?
This is a little simplified:
>>> import urllib
>>> import re
>>> page = urllib.urlopen("http://google.com").read()
# => via regular expression
>>> re.findall("Shopping", page)
['Shopping']
# => via string.find, returns the position ...
>>> page.find("Shopping")
2716
First, get the page (e.g. via urllib.urlopen). Second use a regular expression to find portions of the text, you are interested in. Or use string.find.
you can use urllib2
import urllib2
webp=urllib2.urlopen("the_page").read()
webp.find("the_word")
hope that helps :D
How to open a webpage?
I think the most convinient way is:
from urllib2 import urlopen
page = urlopen('http://www.example.com').read()
How to search for a word?
I guess you are going to search for some pattern in the page next, so here we go:
import re
pattern = re.compile('^some regex$')
match = pattern.search(page)