I currently use this regex:
(d+)
the problem that i can get 2 strings:
“2112343 and alot of 4.99”
OR
“4.99 and alot of 2112343 “
I get this from both:
[2112343, 4, 99]
I need to get only the 2112343…
How can i achieve this?
Answers:
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Method 1
Using lookaround, you can restrict your capturing to only digits which are not surrounded by other digits or decimal points:
(?<![0-9.])(d+)(?![0-9.])
Alternatively, if you want to only match stand-alone numbers (e.g. if you don’t want to match the 123 in abc123def):
(?<!S)d+(?!S)
Method 2
try this
(?<!S)d+(?!S)
this will only match integers
Method 3
If I understand you right, you want to match those numbers with a point inside, too, but dont want to have these in the resulting collection.
I would approach this via 2 steps, first select all numbers, also those with a dot:
(d+(?:.d+)*)
then filter out everything that is not purely numbers, and use your first regex and apply it to each item of the resulting collection from the first step:
(d+)
Method 4
As I posted in my comment:
(?:^| )(d+)(?:$| )
It will match all “words” that are entirely composed of digits(a word being a string of non-space characters surrounded by space characters and or the beginning/end of the string.)
Method 5
Try this
(?<![0-9.])d+(?![0-9.])
It usees the pattern
(?<!prefix)position(?!suffix)
where (?<!prefix)position means: Match position not following prefix.
and position(?!suffix) means: Match position not preceeding suffix.
finally [0-9.] means: Any digit or the decimal point.
Method 6
>>>r = re.match("d+", "23423 in 3.4")
>>>r.group(0)
'23423'
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0