Python正则表达式 - 匹配和开始（）

Question

Let's say I need to find the word "water" in a string. This word cannot be part of another word and it can't be preceded by punctuation (so i'm assuming it can only be preceded by a " " or it's the beginning of the string). I need to return the index of the word's first char "w". So I'm trying this code :

import re
s = re.search(r"(\A| )\bwater\b", "Need water") 
return s.start() # This returns the index of the char " " :(

Is it possible to ignore the part of the pattern so that s.start() always returns the index of the char "w"? Or am I thinking this wrong?

Answer 1

You can use

(?<!\S)\bwater\b

See the regex demo

Explanation:

• (?<!\\S) - a negative lookbehind failing a match if there is a non-whitespace character right before a whole word water
• \\bwater\\b - a whole word water .

Here is a Python demo :

import re
s = re.search(r"(?<!\S)\bwater\b", "Need water") 
if s:
    print(s.start())

Answer 2

You don't need to have that "beginning of a string or space" check. You've already applied the word boundaries check:

>>> s = re.search(r"\bwater\b", "Need water")
>>> s.start()
5
>>> s = re.search(r"\bwater\b", "water is needed")
>>> s.start()
0

Answer 3

You don't even need regex. Just match the space and the word, that will get you the character that the space is at, but you want the first letter so add 1

bigString = "I drink water"

if " " not in bigString:
    print(bigString.find("water"))
else:
    print(bigString.find(" water")+1)