ENH: Arithmetic with Timestamp-based intervals

[gerritholl]

Is your feature request related to a problem?

It is currently possible to perform arithmetic on an Interval iff it is an Interval of numbers. For Interval iv, iv - c == Interval(iv.left-c, iv.right-c). This does not work for intervals of Timestamps:

In [90]: iv = pd.Interval(3, 4)

In [91]: pd.Interval(iv.left-2, iv.right-2)
Out[91]: Interval(1, 2, closed='right')

In [92]: iv - 2
Out[92]: Interval(1, 2, closed='right')

In [93]: iv = pd.Interval(pd.Timestamp("1900-01-01"), pd.Timestamp("1900-01-02"))

In [94]: pd.Interval(iv.left - pd.Timestamp("1900-01-01"), iv.right - pd.Timestamp("1900-01-0
    ...: 1"))
Out[94]: Interval(Timedelta('0 days 00:00:00'), Timedelta('1 days 00:00:00'), closed='right')

In [95]: iv - pd.Timestamp("1900-01-01")
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-95-b61ec78c0d66> in <module>
----> 1 iv - pd.Timestamp("1900-01-01")

TypeError: unsupported operand type(s) for -: 'pandas._libs.interval.Interval' and 'Timestamp'

Describe the solution you'd like

I would like that the behaviour that iv - b == Interval(iv.left-b, iv.right-b) holds for all types that Interval can hold, including Timestamps and Timedeltas.

API breaking implications

No backward compatibility issues expected.

Describe alternatives you've considered

Users can currently resort to Interval(iv.left-b, iv.right-b). This is more typing and required repeating the expression b.

[mroeschke]

IIRC arithmetic between integers and Timestamp was deprecated + removed some time ago cc @jbrockmendel

In [3]: In [94]: pd.Interval(iv.left - pd.Timestamp("1900-01-01"), iv.right - pd.Timestamp("1900-01-01"))
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-3-1d12c6d4a390> in <module>
----> 1 pd.Interval(iv.left - pd.Timestamp("1900-01-01"), iv.right - pd.Timestamp("1900-01-01"))

TypeError: unsupported operand type(s) for -: 'int' and 'Timestamp'

In [4]: pd.__version__
Out[4]: '1.2.0.dev0+174.g9c5c0f2cf.dirty'

Therefore, I don't think it will be likely to support arithmetic between an Interval with integers and a Timestamp

[gerritholl]

My intervals contain only Timestamp, no integers. I want to subtract a Timestamp from another Timestamp. I hope that isn't deprecated.

[mroeschke]

Ah sorry, I had copied the incorrect example locally.

Yes, that request sounds reasonable. Given your integer example works, TImestamp arithmetic should work as well.

[souris-dev]

Sounds good, it would be great if I could work on this. Do we need further discussion about this feature?

[mroeschke]

Go for it @souris-dev

[souris-dev]

take