Calling shift on a DatetimeIndex of length 0 returns an Index instead of a DatetimeIndex

[ssanderson]

Minimal repro:

In [5]: pandas.__version__
Out[5]: '0.16.0-151-gcef3c85'

In [7]: DatetimeIndex([]).shift(3, 'H')
Out[7]: Index([], dtype='object')

[ssanderson]

(This happens because DatetimeIndex delegates to Index.shift, which then calls the Index.new with an empty list, and it can't infer that the output should be a DatetimeIndex.)

[ssanderson]

This is fixed by #9904.

[sinhrks]

This has been already fixed in current master (maybe by #11211?). Adding tests then close.

[jreback]

@sinhrks looks good.

In [1]: DatetimeIndex([]).shift(3, 'H')
Out[1]: DatetimeIndex([], dtype='datetime64[ns]', freq=None)

I since we have 0 len we don't infer the freq, but in this case we know the freq (as we are shifting by it), or is that too much carry-over?

[sinhrks]

DatetimeIndex.freq is common frequencies between elem, and it can be different from shifting freq. Thus it sounds making an exception in length=0 case. Thoughts?

s = pd.DatetimeIndex(['2011-01-01', '2011-01-02'], freq='D')
s.shift(1, 'H')
# DatetimeIndex(['2011-01-01 01:00:00', '2011-01-02 01:00:00'], dtype='datetime64[ns]', freq=None)

[jreback]

cc @ssanderson
cc @llllllllll

[jreback]

@sinhrks no, I was more thinking that If I shift an index, then I want to have it set the freq (as opposed to infer it which won't work if its < 3 elements). not really strong here opinion here, except that this was from a user action.

[ssanderson]

@jreback I don't think that shifting an index by a specific frequency should be taken to imply that the index's data has that frequency. If I have an hourly-frequency index, and I shift it forward by 5 minutes, it's still hourly frequency. That doesn't change if my index is empty.

[ssanderson]

#13026 looks good to me btw. Thanks @sinhrks.

[jreback]

@ssanderson hmm, that is right. ok then!