Python Regex Find All Single Alphabetical Characters

I want to find all indexes for each occurrence of single alphabetical characters in a string. I don't want to catch single char html codes. Here is my code: import re s = 'fish oil

Solution 1:

This does it:

r'(?i)\b[a-z]\b'

Breaking it down:

• Case insensitive match
• A word boundary
• A letter
• A word boundary

Your code can be simplified to this:

for match in re.finditer(r'(?i)\b[a-z]\b', s):
   print match.start()

Solution 2:

Using your format (as you wanted) but adding only a simple check.

import re
s = "fish oil B stack peanut c <b>"
words = re.finditer('\S+', s)
has_alpha = re.compile(r'[a-zA-Z]').search
for word in words:
    iflen(word.group()) == 1and has_alpha(word.group()):
        print (word.start())
>>> 924

Solution 3:

In the most general case I'd say:

re.compile(r'(?i)(?<![a-z])[a-z](?![a-z])').search

Using lookarounds to say "a letter not preceded by another letter nor followed by another letter".