Python正则表达式查找所有单个字母字符

Question

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 B stack peanut c <b>"
words = re.finditer('\S+', s)
has_alpha = re.compile(??????).search
for word in words:
    if has_alpha(word.group()):
        print (word.start())

Desired output:

9
24

Answer 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()

Answer 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:
    if len(word.group()) == 1 and has_alpha(word.group()):
        print (word.start())
>>> 
9
24

Answer 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".