If the string is I am Fine it is giving me output as I.
import re
string='hello I am Fine'
print(re.search(r'[A-Z]?',string).group())
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If the sting is "I am Fine" it is giving me output as 'I'.Prateek Sharma– Prateek Sharma2016-12-28 05:10:39 +00:00Commented Dec 28, 2016 at 5:10
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Welcome to SO Prateek. How I interpret the question is your regex is not matching all the Uppercase letter in your string and just fetching the first match that is 'I'. And please do mark whichever answer you find helpful as 'Accepted'.Dhaval Simaria– Dhaval Simaria2016-12-28 06:36:26 +00:00Commented Dec 28, 2016 at 6:36
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2 Answers
You can use the findall method.
From Python docs, section 7.2.5.6,
findall() matches all occurrences of a pattern, not just the first one as search() does.
In your case,
>>> re.findall(r'[A-Z]',"hello I am Fine")
['I', 'F']
Comments
The ? specifies that the preceding character or class may or may not exist. When re.search starts searching the string, it does not find that class at the beginning of the string... and that is an acceptable match because of the ?. It is simply returning the empty string.
>>> re.search(r'[A-Z]?', 'hello I am Fine').group()
''
If you want it to find the first capital letter, don't use a ?:
>>> re.search(r'[A-Z]', 'hello I am Fine').group()
'I'
