Regular expression for version number bigger than 1.18.10

Question

I need to verify if the version number of application bigger than 1.18.10. How regular expression should look like in this case?

to understand why you don't want to use a RE, you should read courses on [automata](http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-045j-automata-computability-and-complexity-spring-2011/lecture-notes/MIT6_045JS11_lec03.pdf) and [Regular Expressions/NFA](http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-045j-automata-computability-and-complexity-spring-2011/lecture-notes/MIT6_045JS11_lec04.pdf) to understand what RE really are. — zmo, Jun 11 '13 at 09:36

Magnus Hoff · Accepted Answer · 2013-06-11T09:12:25.657

15

Don't use regular expressions for this. Use split and tuple comparison:

def is_recent(version):
    version_as_ints = (int(x) for x in version.split('.'))
    return tuple(version_as_ints) > (1, 18, 10)

And then check is_recent("1.10.11") or is_recent("1.18.12")

edited Jun 11 '13 at 09:12

answered Jun 11 '13 at 09:06

Magnus Hoff

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Appreciate the feedback Magnus – Preet Kukreti Jun 11 '13 at 09:58

Preet Kukreti · Answer 2 · 2013-06-11T11:00:17.280

Seems like this battery has already been included in Python in distutils.version :

from distutils.version import LooseVersion
LooseVersion("1.18.11") > LooseVersion("1.18.10")
#True

LooseVersion("1.2.11") > LooseVersion("1.18.10")
#False (note that "2">"18" is True)

LooseVersion("1.18.10a") > LooseVersion("1.18.10")
#True

This takes into account splitting and comparing both version number parts as integers, and non-numeric parts (e.g alphabetic extension) seperately and correctly. (If you want the alternate behaviour, (lexicographical comparison), you can directly compare the tuples of strings that result on a version_num.split("."))

Note that there is also a StrictVersion variant that will throw an exception (ValueError) on alphabetic characters in the version string. See also PEP386 which is planning to deprecate both, replacing them with a NormalizedVersion.

score 0 · Answer 3 · answered Jun 11 '13 at 09:01

0

Don't use regular expression for that, but something like:

major, minor, patch = v.split('.')
if int(major) > 1 or (int(major) == 1 and int(minor) >= 18):
    ...

answered Jun 11 '13 at 09:01

Guillaume

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score 0 · Answer 4 · answered Jun 11 '13 at 09:02

Not sure exactly why you need a regex, it's not a particularly good tool for doing complex range checking.

I would just split the string into a three-element array and then check each element, something like:

(p1, p2, p3) = verstr.split(".")

# May want to check they're numeric first.

if int(p1) < 1: return False
if int(p1) == 1:
    if int(p2) < 18: return False
    if int(p2) == 18:
        if int(p3) < 10: return False
return True

Regular expression for version number bigger than 1.18.10

4 Answers4

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