Standalone numbers Regex?

纵饮孤独 提交于 2019-12-02 18:06:35

问题


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?


回答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)



回答2:


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+)



回答3:


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.)




回答4:


try this

(?<!\S)\d+(?!\S)

this will only match integers




回答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.




回答6:


>>>r = re.match("\d+", "23423 in 3.4")
>>>r.group(0)
'23423'


来源:https://stackoverflow.com/questions/11277029/standalone-numbers-regex

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