BUG: TimedeltaIndex raising ValueError when slice indexing (pandas-dev#16637)

closes pandas-dev#16637

Author: Jean-Mathieu Deschenes

doc/source/whatsnew/v0.20.2.txt

@@ -51,6 +51,7 @@ Bug Fixes
 - Bug in :func:`unique` on an array of tuples (:issue:`16519`)
 - Bug in :func:`cut` when ``labels`` are set, resulting in incorrect label ordering (:issue:`16459`)
 - Fixed a compatibility issue with IPython 6.0's tab completion showing deprecation warnings on ``Categoricals`` (:issue:`16409`)
+- Bug in ``TimedeltaIndex`` raising a ``ValueError`` when slice indexing with ``loc`` (:issue:`16637`)
 
 Conversion
 ^^^^^^^^^^

pandas/core/indexes/timedeltas.py

@@ -680,8 +680,7 @@ def get_loc(self, key, method=None, tolerance=None):
         -------
         loc : int
         """
-
-        if is_bool_indexer(key):
+        if is_bool_indexer(key) or is_timedelta64_dtype(key):
             raise TypeError
 
         if isnull(key):

pandas/tests/indexing/test_timedelta.py

@@ -18,3 +18,20 @@ def test_boolean_indexing(self):
                                     index=pd.to_timedelta(range(10), unit='s'),
                                     columns=['x'])
             tm.assert_frame_equal(expected, result)
+
+    def test_slice_indexing(self):
+        # GH 16637
+        df = pd.DataFrame({'x': range(10)})
+        df.index = pd.to_timedelta(range(10), unit='s')
+
+        conditions = [df.index[0], df.index[4:8], df.index[[3, 5]]]
+        expected_data = [[20, 1, 2, 3, 4, 5, 6, 7, 8, 9],
+                         [0, 1, 2, 3, 20, 20, 20, 20, 8, 9],
+                         [0, 1, 2, 20, 4, 20, 6, 7, 8, 9]]
+        for cond, data in zip(conditions, expected_data):
+            result = df.copy()
+            result.loc[cond, 'x'] = 20
+            expected = pd.DataFrame(data,
+                                    index=pd.to_timedelta(range(10), unit='s'),
+                                    columns=['x'])
+            tm.assert_frame_equal(expected, result)