I have a string like this '6\' 3" ( 190 cm )' and I would like to extract '190 cm' only using regular expressions. I can't find the appropriate pattern to look for.
I have tried
string = '6\' 3" ( 190 cm )'
pattern = re.compile(r'[^\\( 0-9+ \\)]')
pattern.findall(a)
but it returns ["'", '"', 'c', 'm']
Thanks for helping!
print re.findall(r'[0-9]+ cm',string)[0]
where string is:
'6\' 3" ( 190 cm )'
too many unrequired and harmful symbols in your expression.
Using surrounding [] made findall match individual characters, which explains the output you're getting.
This needs a full rethink: escape the parentheses, use \d+ to match one or more digits, and explicit cm and spaces.
create a group to match only digits+unit, use search to find the group and display it.
import re
string = '6\' 3" ( 190 cm )'
pattern = re.compile(r'\( (\d+ cm) \)')
>>> pattern.search(string).group(1)
'190 cm'
With regular expressions:
import re
s = '6\' 3" ( 190 cm )'
desired_output = re.search(r'\((.*?)\)',s).group(1).lstrip()
print(desired_output)
>>> 190 cm
Without regular expressions:
s = '6\' 3" ( 190 cm )'
desired_output = s[s.find("(")+1:s.find(")")].lstrip()
print(desired_output)
>>> 190 cm
You could use a capturing group which will be returned by findall:
\(\s*([0-9]+\s*[a-z]+)\s*\)
That will match:
\(\s* match ( and 0+ times a whitespace char( Capturing group
[0-9]+\s*[a-z]+ Match 1+ a digit, 0+ times a whitespace char and 1+ times a-z (or use cm instead of [a-z]+ if you want to match that literally)) Close capturing group\s*\) Match 0+ times a whitespace charFor example:
import re
string = '6\' 3" ( 190 cm )'
pattern = re.compile(r"\(\s*([0-9]+\s*[a-z]+)\s*\)")
print(pattern.findall(string))
(?<=\()[^)]+(?=\))or use a capturing group\(\s*([0-9]+\s*[a-z]+)\s*\)[0-9]+or \d+