python: dot separated versions comparizon

Question

I`m writing a script, where i need to compare versions of binnaries , so for example i need to get 10.2.3 greater than 10.1.30 . If i remove dots, i will get 1023 and 10130, which twists my comparison upside down.

10.2.3 > 10.1.30  ==  1023 !> 10130

Till now, the only solution i came out is :

1. do split(".") on version number 
2. for each elemet of list i got, check  if len() is eq 1 , add one zero from  the left.
3. glue all  elements together and get  int (10.1.30 will became 100130).
4. process second version number same way and compare both as regular ints.

This way :

10.2.3 > 10.1.30  ==  100203 > 100130

Is this the only way to compare versions or there are others?

score 1 · Accepted Answer · answered Nov 07 '13 at 15:49

1

You can borrow the code that distutils and its successors use to compare version numbers

from distutils.version import StrictVersion # Or LooseVersion, if you prefer

if StrictVersion('10.2.3') > StrictVersion('10.2'):
    print "10.2.3 is newer"

answered Nov 07 '13 at 15:49

chepner

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Thanks, and what to do if i need to compare N uniq version numbers ? Does distutils.version has something for this ? – Danylo Gurianov Nov 07 '13 at 15:59
How do you define comparison for more than 2 items? – chepner Nov 07 '13 at 16:02
i got it ) . stupid question. – Danylo Gurianov Nov 07 '13 at 16:24

score 0 · Answer 2 · answered Nov 07 '13 at 15:52

0

You can try:

list(map(int, '10.2.3'.split('.'))) > list(map(int, '10.1.30'.split('.')))
Out[216]: True

answered Nov 07 '13 at 15:52

Ankur Ankan

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Thanks a lot Aknur, will check this as well. – Danylo Gurianov Nov 07 '13 at 16:26

python: dot separated versions comparizon

2 Answers2