Match everything after particular pattern except whitespace till next letter

Question

I have a string like this:

'Medicine Treatment:     Not applicable'

Now, I want to match the according treatment but without the white spaces infront of the treatment. So I just want to match: "Not applicable" and not " Not applicable"

Im pretty sure it must be something like this:

(?<=Medicine Treatment:)[^\s].*

or (?<=Medicine Treatment:)\\S.*

... (?<=Medicine Treatment:).* returns only what I do not want: " Not applicable"

Answer 1

You can do this without using regex but by combining the split() and strip() functions :

s = 'Medicine Treatment:     Not applicable'
s.split('Medicine Treatment:')[1].strip()

Output :

'Not applicable'

Answer 2

如果你有一个字符串s ， s.strip()将是字符串的副本，两边没有空格， s.lstrip()只会从左边删除空格。

Answer 3

Python re does not support patterns matching strings of unknown length (though it is possible with PyPi regex module ). So, you can't use re.search(r'(?<=Medicine Treatment:\\s*).*', txt) (it would work with PyPi regex.search ).

You may use a capturing group instead of a lookbehind:

import re
s = 'Medicine Treatment:     Not applicable'
m = re.search(r'Medicine Treatment:\s*(.*)', s)
if m:
    print(m.group(1)) # => Not applicable

See the Python demo

Details

• Medicine Treatment: - a literal string (serves as left-hand context of the required match)
• \\s* - consumes zero or more whitespace chars
• (.*) - captures into Group 1 any 0 or more chars other than line break chars as many as possble.