pattern = (1|2|3|4|5|6|7|8|9|10|11|12)
str = '11'
This only matches '1', not '11'. How to match the full '11'? I changed it to:
pattern = (?:1|2|3|4|5|6|7|8|9|10|11|12)
It is the same.
I am testing here first:
Asked
Active
Viewed 213 times
-1
marlon
- 6,029
- 8
- 42
- 76
-
1Could you please post the code which you are using for performing these tests? – Ganesh Tata Oct 15 '20 at 05:06
-
Put the longer inputs in the front - `(10|11|12|1|2|3)` instead of `(1|2|3|10|11|12)`. The earlier an option is in the chain of 'or's, the more precedence it takes, because the check will short-circuit. – Green Cloak Guy Oct 15 '20 at 05:08
-
In fact, your alternation should now match both 1 and 11. Please include the relevant Python code here. – Tim Biegeleisen Oct 15 '20 at 05:08
-
It matches `1`, so it stops looking, since a regex won't give you a match for all the ways in which it matches, just the first way in which it matches. What are you trying to achieve? – Grismar Oct 15 '20 at 05:10
-
Use boundary to match whole string like `\b11\b` – Ambrish Pathak Oct 15 '20 at 05:15
-
1@AmbrishPathak, yes, I can use '^' and '$' to mark the boundary. – marlon Oct 15 '20 at 05:17
-
1Btw, just a note that regex doesn't really have an order of precedent, each construct performs an atomic action and from left to right. – Oct 15 '20 at 16:41
3 Answers
1
It is matching 1 instead of 11 because you have 1 before 11 in your alternation. If you use re.findall then it will match 1 twice for input string 11.
However to match numbers from 1 to 12 you can avoid alternation and use:
\b[1-9]|1[0-2]?\b
It is safer to use word boundary to avoid matching within word digits.
anubhava
- 761,203
- 64
- 569
- 643
0
Regex always matches left before right.
On an alternation you'd put the longest first.
However, factoring should take precedense.
(1|2|3|4|5|6|7|8|9|10|11|12)
then it turns into
1[012]?|[2-9]
https://regex101.com/r/qmlKr0/1
I purposely didn't add boundary parts as everybody has their own preference.
