Version number comparison in Python

Question

I want to write a cmp-like function that compares two version numbers and returns -1, 0, or 1 based on their compared values.

• Return -1 if version A is older than version B
• Return 0 if versions A and B are equivalent
• Return 1 if version A is newer than version B

Each subsection is supposed to be interpreted as a number, therefore 1.10 > 1.1.

Desired function outputs are

mycmp('1.0', '1') == 0 mycmp('1.0.0', '1') == 0 mycmp('1', '1.0.0.1') == -1 mycmp('12.10', '11.0.0.0.0') == 1 ... 

And here is my implementation, open for improvement:

def mycmp(version1, version2):     parts1 = [int(x) for x in version1.split('.')]     parts2 = [int(x) for x in version2.split('.')]      # fill up the shorter version with zeros ...     lendiff = len(parts1) - len(parts2)     if lendiff > 0:         parts2.extend([0] * lendiff)     elif lendiff < 0:         parts1.extend([0] * (-lendiff))      for i, p in enumerate(parts1):         ret = cmp(p, parts2[i])         if ret: return ret     return 0 

I'm using Python 2.4.5 btw. (installed at my working place ...).

Here's a small 'test suite' you can use

assert mycmp('1', '2') == -1 assert mycmp('2', '1') == 1 assert mycmp('1', '1') == 0 assert mycmp('1.0', '1') == 0 assert mycmp('1', '1.000') == 0 assert mycmp('12.01', '12.1') == 0 assert mycmp('13.0.1', '13.00.02') == -1 assert mycmp('1.1.1.1', '1.1.1.1') == 0 assert mycmp('1.1.1.2', '1.1.1.1') == 1 assert mycmp('1.1.3', '1.1.3.000') == 0 assert mycmp('3.1.1.0', '3.1.2.10') == -1 assert mycmp('1.1', '1.10') == -1 

Answer 1

How about using Python's distutils.version.StrictVersion?

>>> from distutils.version import StrictVersion >>> StrictVersion('10.4.10') > StrictVersion('10.4.9') True 

So for your cmp function:

>>> cmp = lambda x, y: StrictVersion(x).__cmp__(y) >>> cmp("10.4.10", "10.4.11") -1 

If you want to compare version numbers that are more complex distutils.version.LooseVersion will be more useful, however be sure to only compare the same types.

>>> from distutils.version import LooseVersion, StrictVersion >>> LooseVersion('1.4c3') > LooseVersion('1.3') True >>> LooseVersion('1.4c3') > StrictVersion('1.3')  # different types False 

LooseVersion isn't the most intelligent tool, and can easily be tricked:

>>> LooseVersion('1.4') > LooseVersion('1.4-rc1') False 

To have success with this breed, you'll need to step outside the standard library and use setuptools's parsing utility parse_version.

>>> from pkg_resources import parse_version >>> parse_version('1.4') > parse_version('1.4-rc2') True 

So depending on your specific use-case, you'll need to decide whether the builtin distutils tools are enough, or if it's warranted to add as a dependency setuptools.

Answer 2

Remove the uninteresting part of the string (trailing zeroes and dots), and then compare the lists of numbers.

import re  def mycmp(version1, version2):     def normalize(v):         return [int(x) for x in re.sub(r'(\.0+)*$','', v).split(".")]     return cmp(normalize(version1), normalize(version2)) 

This is the same approach as Pär Wieslander, but a bit more compact:

Here are some tests, thanks to "How to compare two strings in dot separated version format in Bash?":

assert mycmp("1", "1") == 0 assert mycmp("2.1", "2.2") < 0 assert mycmp("3.0.4.10", "3.0.4.2") > 0 assert mycmp("4.08", "4.08.01") < 0 assert mycmp("3.2.1.9.8144", "3.2") > 0 assert mycmp("3.2", "3.2.1.9.8144") < 0 assert mycmp("1.2", "2.1") < 0 assert mycmp("2.1", "1.2") > 0 assert mycmp("5.6.7", "5.6.7") == 0 assert mycmp("1.01.1", "1.1.1") == 0 assert mycmp("1.1.1", "1.01.1") == 0 assert mycmp("1", "1.0") == 0 assert mycmp("1.0", "1") == 0 assert mycmp("1.0", "1.0.1") < 0 assert mycmp("1.0.1", "1.0") > 0 assert mycmp("1.0.2.0", "1.0.2") == 0