I am looking to extract only chars from the given string. but my query is doing exactly opposite
s= "A man, a plan, a canal: Panama"
newS = ''.join(re.findall("[^a-zA-Z]*", s))
print(newS) // my o/p: , , :
expected o/p string is:
"A man a plan a canal Panama"
Your regular expression is inverting the match - that's what the caret symbol (^) does inside square brackets (negated character class). You first need to remove that.
Next, you should be matching a sequence of one or more characters (+) rather than zero or more characters (*) -- using * will match the empty string, which you don't want in this case.
Finally your join should join with a space to get the intended output, rather than an empty string -- which won't retain the spaces between the words.
newS = ' '.join(re.findall(r'[a-zA-Z]+', s))
Though not essential in this case, its advised to use raw strings for regular expressions (r). More in this post.
Full working code:
import re
s = 'A man, a plan, a canal: Panama'
newS = ' '.join(re.findall(r'[a-zA-Z]+', s))
print(newS)
