Sort Versions in Python

Question

I'm trying to get it so that 1.7.0 comes after 1.7.0.rc0 but before 1.8.0, as it should if you were sorting versions. I thought the whole point of LooseVersion was that it handled the sorting and comparison of this kind of thing correctly.

>>> from distutils.version import LooseVersion
>>> versions = ["1.7.0", "1.7.0.rc0", "1.8.0"]
>>> lv = [LooseVersion(v) for v in versions]
>>> sorted(lv, reverse=True)
[LooseVersion ('1.8.0'), LooseVersion ('1.7.0.rc0'), LooseVersion ('1.7.0')]

Answer 1

>>> from distutils.version import LooseVersion
>>> versions = ["1.7.0", "1.7.0rc0", "1.11.0"]
>>> sorted(versions, key=LooseVersion)
['1.7.0', '1.7.0rc0', '1.11.0']

from the docs

Version numbering for anarchists and software realists. Implements the standard interface for version number classes as described above. A version number consists of a series of numbers, separated by either periods or strings of letters. When comparing version numbers, the numeric components will be compared numerically, and the alphabetic components lexically.
...
In fact, there is no such thing as an invalid version number under this scheme; the rules for comparison are simple and predictable, but may not always give the results you want (for some definition of "want").

so you see there is no smarts about treating "rc" specially

You can see how the version number is broken down like this

>>> LooseVersion('1.7.0rc0').version
[1, 7, 0, 'rc', 0]