Multiple results when using consecutive patterns regex

Question

Test String: "Version 3.1.A"

RegEx: "(\d\.){2}."

Returning: [('1.3.A', '3.')]

Why does this return 2 matches, the second only matches a non-reoccuring (\d.)

Is there a way I can force only the complete match (1.3.A) to return using the {*} operator (not explicit \d.\d..)

Use a non-capturing group, like this `(?:\d\.){2}.`. `re.findall(r"((?:\d\.){2}.)", "Version 3.1.A")` gives `['3.1.A']` — dcg, Jul 18 '19 at 13:39
@dcg, please add your comment as an answer as thats solved it, cheers! — LazyMoggy, Jul 18 '19 at 13:42
I think this is actually a dupe of https://stackoverflow.com/questions/31915018/re-findall-behaves-weird — The fourth bird, Jul 18 '19 at 13:43

score 1 · Accepted Answer · answered Jul 18 '19 at 13:49

1

By using a non-capturing group you can get what you want, like the following:

>>> import re
>>> text = "Version 3.1.A"
>>> re.findall(r"((?:\d\.){2}.)", text)
['3.1.A']

answered Jul 18 '19 at 13:49

dcg

1 Answers1