Regex parsing in steps

Question

Let's say I have the following string:

s = "once upon a time, there was once a person"

Without using findall to get all onces in the string:

>>> re.findall(r'\bonce\b', s)
['once', 'once']

Is there a way to use search incrementally, so it only returns the first occurrence and then increments the input string?

while (s):
    x = re.search(r'\bonce\b', s) # return 'once' and increment the string to s[4:]
    yield x

@TimBiegeleisen -- oh, it seems there is a builtin that I hadn't known before: `>>> for i in re.finditer(r'\bonce\b',s): ... print i.group()` — David542, Mar 03 '20 at 01:35
Also, the `Match` object returned by `re.search` contains the position in the string that matched the regex so you can move forward in the string by slicing. — tdelaney, Mar 03 '20 at 01:36
@Barmar why not post an answer, I can accept that. Just show the options and how to use it. — David542, Mar 03 '20 at 01:41

score 0 · Accepted Answer · answered Mar 03 '20 at 01:45

0

Use re.finditer()

for match in re.finditer(r'\bonce\b', s):
    yield match

Or you can just return the iterator rather than writing your own loop.

return re.finditer(r'\bonce\b', s)

answered Mar 03 '20 at 01:45

Barmar

1 Answers1