DOC: default value for pd.DateOffset() is 24 hours instead of 1 calendar day

[GYHHAHA]

>>>pd.__version__
1.1.1
>>>ts = pd.Timestamp('2011-3-27 00:00:00', tz='Europe/Helsinki')
>>>ts + pd.DateOffset(days=1)
Timestamp('2011-03-28 00:00:00+0300', tz='Europe/Helsinki')
>>>ts + pd.DateOffset()
Timestamp('2011-03-28 01:00:00+0300', tz='Europe/Helsinki')
>>>ts + pd.DateOffset(hours=24)
Timestamp('2011-03-28 01:00:00+0300', tz='Europe/Helsinki')

The default action for pd.DateOffset seems to be equal to pd.DateOffset(hours=24) instead of pd.DateOffset(days=1).
But the user guide says DateOffset object defaults to 1 calendar day.

[arw2019]

What would you expect to get instead?

[GYHHAHA]

I want to get Timestamp('2011-03-28 00:00:00+0300', tz='Europe/Helsinki') when use ts + pd.DateOffset().

[arw2019]

Hmmm so I'm not sure whether this is the expected behavior from DateOffset

For your specific use case would round be a solution?

In [7]: ts.round('D')                                                                                  
Out[7]: Timestamp('2011-03-28 00:00:00+0300', tz='Europe/Helsinki')

[GYHHAHA]

Oh, I know how to convert it to this form.
But just that is not matching the doc description.

For example, a Timedelta day will always increment datetimes by 24 hours, while a DateOffset day will increment datetimes to the same time the next day whether a day represents 23, 24 or 25 hours due to daylight savings time.

Thus maybe it's more clear to point out the default value is 24 hours not a calendar day here.

[arw2019]

Ok, makes sense! Are you interested in opening a PR to add this clarification to the doc?

[GYHHAHA]

glad to

[arw2019]

Fantastic. Here's the contributions guide with the instructions