API: allow step!=1 slice with IntervalIndex

doc/source/whatsnew/v1.1.0.rst

@@ -145,7 +145,7 @@ Strings
 
 Interval
 ^^^^^^^^
-
+- Bug in :class:`IntervalIndex` slicing that prevented slicing with ``step > 1`` (:issue:`31658`)
 -
 -
 

pandas/core/indexes/interval.py

@@ -884,11 +884,6 @@ def get_indexer_for(self, target: AnyArrayLike, **kwargs) -> np.ndarray:
             return self.get_indexer_non_unique(target)[0]
         return self.get_indexer(target, **kwargs)
 
-    def _convert_slice_indexer(self, key: slice, kind=None):
-        if not (key.step is None or key.step == 1):
-            raise ValueError("cannot support not-default step in a slice")
-        return super()._convert_slice_indexer(key, kind)
-
     @Appender(Index.where.__doc__)
     def where(self, cond, other=None):
         if other is None:

pandas/tests/indexing/interval/test_interval_new.py

@@ -128,6 +128,7 @@ def test_loc_with_slices(self):
         with pytest.raises(NotImplementedError, match=msg):
             s[Interval(3, 4, closed="left") :]
 
+        # FIXME: dont leave commented-out
         # TODO with non-existing intervals ?
         # s.loc[Interval(-1, 0):Interval(2, 3)]
 
@@ -143,9 +144,16 @@ def test_loc_with_slices(self):
         tm.assert_series_equal(expected, s[:2.5])
         tm.assert_series_equal(expected, s[0.1:2.5])
 
+    def test_slice_step_ne1(self):
         # slice of scalar with step != 1
-        with pytest.raises(ValueError):
-            s[0:4:2]
+        s = self.s
+        expected = s.iloc[0:4:2]
+
+        result = s[0:4:2]
+        tm.assert_series_equal(result, expected)
+
+        result2 = s[0:4][::2]
+        tm.assert_series_equal(result2, expected)
 
     def test_loc_with_overlap(self):