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Author: jreback

doc/source/advanced.rst

@@ -845,6 +845,39 @@ Of course if you need integer based selection, then use ``iloc``
 
    dfir.iloc[0:5]
 
+.. _indexing.intervallindex:
+
+IntervalIndex
+~~~~~~~~~~~~~
+
+.. versionadded:: 0.20.0
+
+.. warning::
+
+   These indexing behaviors are provisional and may change in a future version of pandas.
+
+.. ipython:: python
+
+   df = pd.DataFrame({'A': [1, 2, 3, 4]},
+                      index=pd.IntervalIndex.from_breaks([0, 1, 2, 3, 4]))
+   df
+
+Label based indexing via ``.loc`` along the edges of an interval works as you would expect,
+selecting that particular interval.
+
+.. ipython:: python
+
+   df.loc[2]
+   df.loc[[2, 3]]
+
+If you select a lable *contained* within an interval, this will also select the interval.
+
+.. ipython:: python
+
+   df.loc[2.5]
+   df.loc[[2.5, 3.5]]
+
+
 Miscellaneous indexing FAQ
 --------------------------
 

doc/source/whatsnew/v0.20.0.txt

@@ -318,7 +318,8 @@ IntervalIndex
 ^^^^^^^^^^^^^
 
 pandas has gain an ``IntervalIndex`` with its own dtype, ``interval`` as well as the ``Interval`` scalar type. These allow first-class support for interval
-notation, specifically as return type for ``pd.cut`` and ``pd.qcut``. (:issue:`7640`, :issue:`8625`)
+notation, specifically as return type for ``pd.cut`` and ``pd.qcut``. The ``IntervalIndex`` allows some unique indexing, see the
+:ref:`docs <indexing.intervallindex>`. (:issue:`7640`, :issue:`8625`)
 
 **Previous behavior**: