Date Operations

[ywhcuhk]

IMHO, two date functions from SAS can be very useful and straight forward. I wonder if they can be incorporated in Pandas.

intck()

This function computes the difference between 2 dates/datetime with specified units. For example
Assume: d1=1992-03-20 and d2=1992-06-12. And these are all of date type in SAS.
intck('month', d1, d2) yields 3 (regardless of what day it is). intck('qtr', d1, d2) yields 1. intck('year', d1, d2) yields 0.

intnx()

This function increments a given date by an interval. Assuming d1=1992-03-20, intnx('month', d1, 3, 'end') yields 1992-06-30 (increase month by 3 and return the last day of that month. intnx('month', d1, 3, 'same') yields 1992-06-20 (increase month by 3 and return the same day of that month.

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As you can see, these two functions are very flexible and simple. The whole offset system seems to be a little too cumbersome comparing to these 2.

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See this question too. Especially the solution offered by @hayd

[hayd]

I think the trickiness is that Months/Years are no longer a set length (in nanonseconds), so it's trickier to divide them... when you subtract you get a timedelta but you can't write this in terms of Month/Year. I think this means you have to check each tuple part to get this info, you can't do anything clever.

Hmmmm, I was going to say that intnx is solved with offsets (this looks like a bug; seems like the way it ought to work):

In [11]: d1
Out[11]: Timestamp('1992-03-20 00:00:00')

In [12]: d1 + pd.offsets.MonthOffset(3)  # increments the days?? BUG??
Out[12]: Timestamp('1992-03-23 00:00:00')

In [13]: d1 + pd.offsets.MonthEnd(3)  # does what you want... modulo 1.
Out[13]: Timestamp('1992-05-31 00:00:00')

The names of these SAS functions are truly awful!!!

[jreback]

You want this. @hayd the first argument to MonthOffset is a day, This is a bug see #7707

In [23]: t = Timestamp('1992-03-20 00:00:00')

In [24]: t
Out[24]: Timestamp('1992-03-20 00:00:00')

In [25]: t+pd.offsets.MonthOffset(months=3)
Out[25]: Timestamp('1992-06-20 00:00:00')

In [26]: t+pd.offsets.MonthEnd(3)
Out[26]: Timestamp('1992-05-31 00:00:00')

[jreback]

I suppose the docs could be improved slightly. pandas provides extreme flexibility. http://pandas.pydata.org/pandas-docs/stable/timeseries.html#dateoffset-objects. Though they are pretty complete.