Extract everything before a particular string in python

Question

Let's say I have a string

s = 'ab@cD!.2e.cp'

I want to extract only ab@cD!.2e out of it. I am trying this:

print(re.search(r'^(.*?)\.cp',s).group())

But still getting the output as ab@cD!.2e.cp . Can someone please tell me where I am doing it wrong and what should be the correct regex for this?

Answer 1

You probably meant to add 1 as parameter to group:

import re
s = 'ab@cD!.2e.cp'
re.search(r'^(.*?)\.cp',s).group()      # 'ab@cD!.2e.cp'
re.search(r'^(.*?)\.cp',s).group(0)     # 'ab@cD!.2e.cp'
re.search(r'^(.*?)\.cp',s).group(1)     # 'ab@cD!.2e'

Answer 2

Instead of re.search , use re.findall :

import re
s = 'ab@cD!.2e.cp'
print(re.findall(r'^(.*?)\.cp',s)[0])

Output:

ab@cD!.2e

Answer 3

If it is really just about extracting everything before a certain string - as your title suggests - you don't need a regex at all but a simple split will do:

res = s.split('.cp')[0]

yields

'ab@cD!.2e'

Please be aware that this will return the original string if .cp was not found:

s = 'foo'
s.split('.cp')[0]

will return

'foo'