date_range default calendar day return

[zxymark221]

Code Executed:

#0.000016 second less than 24 hours, acrossing 2 calendar day.
print pandas.date_range(start='2016-03-22 12:19:49.053501', end='2016-03-23 12:19:49.053485')

Output of different versions.

• Prior to 0.17
  
  Prior to 0.17 versions, I tested 0.16 and 0.13, they both return two days.
  
  DatetimeIndex(['2016-03-22 12:19:49.053501', '2016-03-23 12:19:49.053501'], dtype='datetime64[ns]', freq='D', tz=None)
• V0.17 & 0.18
  
  Returns one day only
  
  DatetimeIndex(['2016-03-22 12:19:49.053501'], dtype='datetime64[ns]', freq='D')

On the documentation, the freq is set to be calendar daily by default, the test above covers two calendar days, does it make more sense to return two days like the earlier versions do?

[jreback]

This was changed here, because of a bug.

Not really sure what you are trying to do.

[zxymark221]

@jreback Thanks for your reply.

What I am trying to say is: I am expecting the code execute above should return two days(both 2016-03-22 and 2016-03-23) even though the start and end time I specified above is less than 24 hrs. This is because the freq setting in date_range is by default calendar daily. From my perspective, calendar daily is different from 24 hours, otherwise, why should we call it calendar daily.

[jreback]

@zxymark221 I don't think so. you asked for 1 Day from the start effectively as stop time < start time + 1 day:

In [3]: start = Timestamp('2016-03-22 12:19:49.053501')

In [4]: stop = Timestamp('2016-03-23 12:19:49.053485')

In [6]: start+pd.offsets.Day()
Out[6]: Timestamp('2016-03-23 12:19:49.053501')

In [8]: stop-start+pd.offsets.Day()
Out[8]: Timedelta('1 days 23:59:59.999984')

In [9]: start+pd.offsets.Day() > stop
Out[9]: True

[zxymark221]

@jreback Thank you. Understood. By default, freq is 24 hours.

I've been using 0.16 version for a long time, and the idea of calendar daily here made me think it will return all the dates the range has been through.