正则表达式跳过某些特定字符

Question

I am trying to clean the string such that it does not have any punctuation or number, it must only have az and AZ. For example,given String is:

"coMPuter scien_tist-s are,,,  the  rock__stars of tomorrow_ <cool>  ????"

Required output is :

['computer', 'scientists', 'are', 'the', 'rockstars', 'of', 'tomorrow']

My solution is

re.findall(r"([A-Za-z]+)" ,string)

My output is

['coMPuter', 'scien', 'tist', 's', 'are', 'the', 'rock', 'stars', 'of', 'tomorrow', 'cool']

Answer 1

You don't need to use regular expression:

(Convert the string into lower case if you want all lower-cased words), Split words, then filter out word that starts with alphabet:

>>> s = "coMPuter scien_tist-s are,,,  the  rock__stars of tomorrow_ <cool>  ????"
>>> [filter(str.isalpha, word) for word in s.lower().split() if word[0].isalpha()]
['computer', 'scientists', 'are', 'the', 'rockstars', 'of', 'tomorrow']

In Python 3.x, filter(str.isalpha, word) should be replaced with ''.join(filter(str.isalpha, word)) , because in Python 3.x, filter returns a filter object.

Answer 2

With the recommendation of all of the people who answered I got the correct solution that i really wants , Thanks to every one...

s = "coMPuter scien_tist-s are,,,  the  rock__stars of tomorrow_ <cool>  ????"    
cleaned = re.sub(r'(<.*>|[^a-zA-Z\s]+)', '', s).split()
print cleaned

Answer 3

using re , although I'm not sure this is what you want because you said you didn't want "cool" leftover.

import re

s = "coMPuter scien_tist-s are,,,  the  rock__stars of tomorrow_ <cool>  ????"

REGEX = r'([^a-zA-Z\s]+)'

cleaned = re.sub(REGEX, '', s).split()
# ['coMPuter', 'scientists', 'are', 'the', 'rockstars', 'of', 'tomorrow', 'cool']

EDIT

WORD_REGEX = re.compile(r'(?!<?\S+>)(?=\w)(\S+)')
CLEAN_REGEX = re.compile(r'([^a-zA-Z])')

def cleaned(match_obj):
    return re.sub(CLEAN_REGEX, '', match_obj.group(1)).lower()

[cleaned(x) for x in re.finditer(WORD_REGEX, s)]
# ['computer', 'scientists', 'are', 'the', 'rockstars', 'of', 'tomorrow']

WORD_REGEX uses a positive lookahead for any word characters and a negative lookahead for <...>. Whatever non-whitespace that makes it past the lookaheads is grouped:

(?!<?\S+>) # negative lookahead
(?=\w) # positive lookahead
(\S+) #group non-whitespace

cleaned takes the match groups and removes any non-word characters with CLEAN_REGEX