Fix GroupBy nth Handling with Observed=False (#26419)

* Added test coverage for observed=False with ops

* Fixed issue with observed=False and nth

Author: WillAyd

doc/source/whatsnew/v0.25.1.rst

@@ -121,6 +121,7 @@ Groupby/resample/rolling
 ^^^^^^^^^^^^^^^^^^^^^^^^
 
 - Bug in :meth:`pandas.core.groupby.DataFrameGroupBy.transform` where applying a timezone conversion lambda function would drop timezone information (:issue:`27496`)
+- Bug in :meth:`pandas.core.groupby.GroupBy.nth` where ``observed=False`` was being ignored for Categorical groupers (:issue:`26385`)
 - Bug in windowing over read-only arrays (:issue:`27766`)
 - Fixed segfault in `pandas.core.groupby.DataFrameGroupBy.quantile` when an invalid quantile was passed (:issue:`27470`)
 -

pandas/core/groupby/groupby.py

@@ -1773,7 +1773,11 @@ def nth(self, n: Union[int, List[int]], dropna: Optional[str] = None) -> DataFra
             if not self.as_index:
                 return out
 
-            out.index = self.grouper.result_index[ids[mask]]
+            result_index = self.grouper.result_index
+            out.index = result_index[ids[mask]]
+
+            if not self.observed and isinstance(result_index, CategoricalIndex):
+                out = out.reindex(result_index)
 
             return out.sort_index() if self.sort else out
 

pandas/tests/groupby/test_categorical.py

@@ -434,6 +434,21 @@ def test_observed_groups_with_nan(observed):
     tm.assert_dict_equal(result, expected)
 
 
+def test_observed_nth():
+    # GH 26385
+    cat = pd.Categorical(["a", np.nan, np.nan], categories=["a", "b", "c"])
+    ser = pd.Series([1, 2, 3])
+    df = pd.DataFrame({"cat": cat, "ser": ser})
+
+    result = df.groupby("cat", observed=False)["ser"].nth(0)
+
+    index = pd.Categorical(["a", "b", "c"], categories=["a", "b", "c"])
+    expected = pd.Series([1, np.nan, np.nan], index=index, name="ser")
+    expected.index.name = "cat"
+
+    tm.assert_series_equal(result, expected)
+
+
 def test_dataframe_categorical_with_nan(observed):
     # GH 21151
     s1 = Categorical([np.nan, "a", np.nan, "a"], categories=["a", "b", "c"])