BUG fixes tuple agg 18079

Author: bobhaffner

doc/source/whatsnew/v0.21.1.txt

@@ -62,6 +62,7 @@ Bug Fixes
 - Bug in ``pd.Series.rolling.skew()`` and ``rolling.kurt()`` with all equal values has floating issue (:issue:`18044`)
 - Bug in ``pd.DataFrameGroupBy.count()`` when counting over a datetimelike column (:issue:`13393`)
 - Bug in ``pd.concat`` when empty and non-empty DataFrames or Series are concatenated (:issue:`18178` :issue:`18187`)
+- Bug in :class:`NDFrameGroupBy` fixes ValueError: no results error when grouping by a single column and aggregating with a tuple (:issue:`18079`)  
 
 Conversion
 ^^^^^^^^^^

pandas/core/groupby.py

@@ -3022,7 +3022,9 @@ def aggregate(self, func_or_funcs, *args, **kwargs):
         if isinstance(func_or_funcs, compat.string_types):
             return getattr(self, func_or_funcs)(*args, **kwargs)
 
-        if hasattr(func_or_funcs, '__iter__'):
+        if isinstance(func_or_funcs, collections.Iterable):
+            # Catch instances of lists / tuples
+            # but not the class list / tuple itself.
             ret = self._aggregate_multiple_funcs(func_or_funcs,
                                                  (_level or 0) + 1)
         else:

pandas/tests/groupby/test_groupby.py

@@ -2718,6 +2718,22 @@ def test_empty_dataframe_groupby(self):
 
         assert_frame_equal(result, expected)
 
+    def test_tuple_aggregation(self):
+        # Issue #18079
+        df = pd.DataFrame({'A': [1, 1, 1, 3, 3, 3],
+                           'B': [1, 1, 1, 4, 4, 4], 'C': [1, 1, 1, 3, 4, 4]})
+
+        result = df.groupby(['A', 'B']).aggregate(tuple)
+        expected = pd.DataFrame({'C': {(1, 1): (1, 1, 1), (3, 4): (3, 4, 4)}})
+        expected.index.names = ['A', 'B']
+        assert_frame_equal(result, expected)
+
+        result = df.groupby('A')['C'].aggregate(tuple)
+        expected = pd.Series([(1, 1, 1), (3, 4, 4)], index=[1, 3])
+        expected.name = 'C'
+        expected.index.name = 'A'
+        assert_series_equal(result, expected)
+
 
 def _check_groupby(df, result, keys, field, f=lambda x: x.sum()):
     tups = lmap(tuple, df[keys].values)