Added version sorting on Web and Removed obsolete versions

[Masnan-306]

• closes WEB: Better management of releases in the pandas home page #50885
• Tests added and passed
• All code checks passed.

[datapythonista]

Very nice, thanks @Masnan-306

I added few more comments, but looking great.

[datapythonista]

I think you can use v here instead of release. Since str(v) == release I think the render on the website should stay the same. And you don't need to make releases and sorted_releases lists (or generators) of tuples, but only of Version objects, which makes things easier.

[Masnan-306]

I don't think str(v) == release is true judging from the code below. I think there are fields like assets and tag_name in each release. So I may not understand what you mean here.

[Masnan-306]

@datapythonista Hi, can you give some more clarifications on this?

[datapythonista]

Sorry for the delay. So, here you keep a tuple to deal with versions: (Version(1, 5, 2), '1.5.2'). You use the first element to compare versions, and the second to display it in the template. But the template will parse to string whatever you give it when creating the html. And this is true:

>>> from packaging import version
>>> v152 = version.Version('1.5.2')
>>> str(v152)
'1.5.2'

The second element in the tuple is not needed, since you can give the first one to the template, and when rendered it will become the second. So, you can just use Version(...) instead of the tuple, and simplify the code, which is a bit tricky to follow.

[Masnan-306]

Hi, I am still a little confused. The first element in the tuple is indeed the objectVersion(...), but it is parsed from release[tag_name] but not from release directly. Since each release is a dictionary with many fields including assets, tag_name, published_at etc, I can only think of making them key-values pairs in the list (v, release) and sort the list by the key values v. In other word, the list latest_releases consists of tuple like (Version(1,5,2), {...}) instead of (Version(1, 5, 2), '1.5.2'). So, I'm sorting the dictionaries by the versions and then feed the dictionaries to the next loop.

[frank-monkey]

@datapythonista could you please review this?

[datapythonista]

Sorry for the delay @Masnan-306, and for not being clear in my previous review. This is looking better, but I think we can still simplify your implementation significantly, added couple of ideas.

[datapythonista]

Sorry for the delay. So, here you keep a tuple to deal with versions: (Version(1, 5, 2), '1.5.2'). You use the first element to compare versions, and the second to display it in the template. But the template will parse to string whatever you give it when creating the html. And this is true:

>>> from packaging import version
>>> v152 = version.Version('1.5.2')
>>> str(v152)
'1.5.2'

The second element in the tuple is not needed, since you can give the first one to the template, and when rendered it will become the second. So, you can just use Version(...) instead of the tuple, and simplify the code, which is a bit tricky to follow.

[Masnan-306]

@datapythonista could you please review my reply?
I still think that my implementation of creating (version.parse(release["tag_name"]), release) tuple is necessary, since the first serves as the key and the second is a dictionary object that can not be parsed directly. So do you mean that the first element in the tuple can be avoided by using lambda function in sorted? to save space? Thanks for your effort in advance.

[mroeschke]

Thanks for the pull request, but it appears to have gone stale. If interested in continuing, please merge in the main branch, address any review comments and/or failing tests, and we can reopen.