unable to replcae a pattern using method re.sub()

[Jeevananda]

my question here
I am unable to replace a pattern using the method re.sub()

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import re   #initialize string   x='<?xml version="1.0" encoding="UTF-8"?><rule stage="Source">'   #extracting a part of string   b=re.findall('(<.+\s.+\?>)',x) ##result will be b=['<?xml version="1.0" encoding="UTF-8"?>']   #now I want to remove the extract part so I am using   print(re.sub(b[0],'',x))

# expected result is <rule stage="Source"> how ever I am getting <?xml version="1.0" encoding="UTF-8"?#><rule stage="Source"> please help

[DeaD_EyE]

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text = '<?xml version="1.0" encoding="UTF-8"?><rule stage="Source">' regex = r'(<.+\s.+\?>)' re.sub(regex, 'Replacement for regex', text)

Output:
'Replacement for regex<rule stage="Source">'

re.sub replaces the pattern, which is your regex, with the replacement (2nd argument) of the string (3rd argument).

You can address a group match inside the replacement with \1 - \n

1	re.sub(regex, r'\1Foo', text)

Output:
'<?xml version="1.0" encoding="UTF-8"?>Foo<rule stage="Source">'

Are you sure, that you want to use regex for this task?
Regex it's not a good choice for parsing html/xml.

You can use instead xml from the stdlib,
which is for my opinion not easier, but it's safer.

Programmer 1: We have a problem.
Programmer 2: Let’s use RegEx!
Programmer 1: Now we have two problems.

[nilamo]

re.sub() takes a regex as it's first parameter.  You're passing it a random string that you got from a different regex.  If you skip the findall() step, you'll have what you're looking for.

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>>> import re >>> text = '<?xml version="1.0" encoding="UTF-8"?><rule stage="Source">' >>> re.findall('(<.+\s.+\?>)', text) ['<?xml version="1.0" encoding="UTF-8"?>'] >>> re.sub('(<.+\s.+\?>)', '', text) '<rule stage="Source">'

edit: I was ninja'd :(

[snippsat]

You can use instead xml from the stdlib, which is for my opinion not easier, but it's safer.

That's an option,but i never use parser from stdlib when BeautifulSoup and lxml are better and easier to use.

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>>> from bs4 import BeautifulSoup >>> text = '<?xml version="1.0" encoding="UTF-8"?><rule stage="Source">' >>> soup = BeautifulSoup(text, 'lxml') >>> s = soup.find('rule') >>> s <rule stage="Source"></rule> >>> s.attrs {'stage': 'Source'}

As mention regex is wrong tool for this this,
sure for 1 line it's not problem to use regex.
But HTML/XML are usually a lot lines and many way a regex can break.

This funny answer  which have been posted many times is worth reading.

[oneclick]

I tried to use same method but I'm getting different error.

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import re   desc = "fsafdsdf [23]" reg = r'\[[0-9]?[0-9]\]' re.sub(re, "hi", desc)

Error:
Traceback (most recent call last):
  File "E:\PYTHON\123.py", line 6, in <module>
    re.sub(re, "hi", desc)
  File "C:\Program Files (x86)\Python37\lib\re.py", line 192, in sub
    return _compile(pattern, flags).sub(repl, string, count)
  File "C:\Program Files (x86)\Python37\lib\re.py", line 285, in _compile
    raise TypeError("first argument must be string or compiled pattern")
TypeError: first argument must be string or compiled pattern

could you tell me what is going wrong

my question here
I am unable to replace a pattern using the method re.sub()

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import re   #initialize string   x='<?xml version="1.0" encoding="UTF-8"?><rule stage="Source">'   #extracting a part of string   b=re.findall('(<.+\s.+\?>)',x) ##result will be b=['<?xml version="1.0" encoding="UTF-8"?>']   #now I want to remove the extract part so I am using   print(re.sub(b[0],'',x))

# expected result is <rule stage="Source"> how ever I am getting <?xml version="1.0" encoding="UTF-8"?#><rule stage="Source"> please help