concat([sparse]) should be a Series / DataFrame

[TomAugspurger]

I messed up the logic somewhere along the way.

In [11]: df = pd.DataFrame({"A": pd.SparseArray([0, 1])})

In [12]: type(pd.concat([df, df]))
Out[12]: pandas.core.sparse.frame.SparseDataFrame

That should be a DataFrame with sparse values.

When should concat return a SparseSeries/Frame?

a.) When any of the inputs are SparseSeries / SparseDataFrame
b.) When all of the inputs are SparseSeries/Frame.

Option b is probably least disruptive. Implementing it isn't a pain.

[WillAyd]

What's the downside to option a? I admittedly don't use these objects a lot but to cast a vote in there that seems more intuitive to me

[TomAugspurger]

For 0.25.0, I want to ensure that pandas only returns a SparseDataFrame if you're already using them. I hoped we did that for 0.24.0, but I missed this case apparently. Option a is a (slightly) larger API-breaking change.

…

On Tue, Mar 12, 2019 at 9:51 PM William Ayd ***@***.***> wrote: What's the downside to option a? I admittedly don't use these objects a lot but to cast a vote in there that seems more intuitive to me — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub <#25702 (comment)>, or mute the thread <https://github.com/notifications/unsubscribe-auth/ABQHIgQSCvTNXliqdmNCdsjk39dZuGrAks5vWGfCgaJpZM4bsOnJ> .

[TomAugspurger]

Note that this only affects DataFrame. concat(List[Series[Sparse]]) has the desired behavior of returning a Series with sparse values.

In [4]: type(pd.concat([pd.Series(pd.SparseArray([0, 1]))] * 2))
Out[4]: pandas.core.series.Series