CODEWITHSUNDEEP
pythonregex
RedBaron
RedBaron

Reputation: 4755

Representing version number as regular expression

I need to represent version numbers as regular expressions. The broad definition is

  1. Consist only of numbers
  2. Allow any number of decimal points (but not consecutively)
  3. No limit on maximum number

So 2.3.4.1,2.3,2,9999.9999.9999 are all valid whereas 2..,2.3. is not.

I wrote the following simple regex

'(\d+\.{0,1})+'

Using it in python with re module and searching in '2.6.31' gives

>>> y = re.match(r'(\d+\.{0,1})+$','2.6.31')
>>> y.group(0)
'2.6.31'
>>> y.group(1)
'31'

But if I name the group, then the named group only has 31.

Is my regex representation correct or can it be tuned/improved? It does not currently handle the 2.3. case.

Upvotes: 4

Views: 3982

Answers (2)

jcollado
jcollado

Reputation: 40374

Alternatively, you might want to use pyparsing:

>>> from pyparsing import *
>>> integer = Word(nums)
>>> parser = delimitedList(integer, delim='.') + StringEnd()
>>> list(parser.parseString('1.2.3.4'))
['1', '2', '3', '4']

or lepl:

>>> from lepl import *
>>> with Separator(~Literal('.')):
...    parser = Integer()[1:]
>>> parser.parse('1.2.3.4')
['1', '2', '3', '4']

Upvotes: 2

Greg Hewgill
Greg Hewgill

Reputation: 992985

The notation {0,1} can be shortened to just ?:

r'(\d+\.?)+$'

However, the above will allow a trailing .. Perhaps try:

r'\d+(\.\d+)*$'

Once you have validated the format matches what you expect, the easiest way to get the numbers out is with re.findall():

>>> ver = "1.2.3.4"
>>> re.findall(r'\d+', ver)
['1', '2', '3', '4']

Upvotes: 6

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