import re
pattern=r'\d+'
pattern1= r'\d+\s'
place = re.search((pattern | pattern1),s)
I need this regex to work for s='2m' as well as s='2 m' so that I can find m and separate it also it should for in case of s='2 mins' or s= '2mins'
can someone please help me with this. Also PFA Thanks
Just use a single pattern with an alternation:
\b\d+\s*(?:m|mins)\b
Sample script:
inp = 'I saw it 2 mins ago'
if (re.search(r'\b\d+\s*(?:m|mins)\b', inp)):
print("MATCH")
Use the following pattern that does not use alternation but rather optional sub-expressions:
\b\d+ ?m(?:ins)?\b
\b Asserts position at a word boundary.\d+ matches 1 or more digits. ? Optional space. If you want to allow multiple spaces, then use *.m Matches 'm'.(?:ins)? Matches optional 'ins' in a non-capturing group.\b Asserts position at a word boundary.Use:
import re
m = re.search(r'\b\d+ ?m(?:ins)?\b', 'See you in 2 mins or sooner')
if m:
print(m[0])
Prints:
2 mins
re.search("|".join([pattern, pattern1]), s)?