Add convenience function to get year from Day

[mchav]

Given a Day you can get the year, month, and day as a tuple by writing fromGregorian day. Getting the year requires pattern matching on a 3-tuple or using lens' to get the first field. I think it's a common enough operation that it might be worth defining a convenience function called gregorianYear or year in the Gregorian package.

[AshleyYakeley]

You can just use the YearMonthDay pattern.

\(YearMonthDay y _ _) -> y

[mchav]

Yeah that's what I was referring to by tuple matching.

This might be too niche a request but figured it'd be good to ask anyway. I'm working on a dataframe library and I'm porting some datetime related examples from Python. For people starting out with Haskell and I suggest the approach you mention in the tutorial but I thought this might be the sort of thing that makes that would make the library easier to use so could be an upstream change.

So the above example would make sense to write as:

df & D.as "birth_year" D.apply "birthdate" gregorianYear

What do you think?

[mchav]

I also just realized your approach is different from the tuple approach. Sorry. Should have read that more closely. I guess that makes my concern less pressing but the gist of it still holds, I think.