BUG: Fix MultiIndex DataFrame to_csv() segfault

Author: Matias Heikkilä

doc/source/whatsnew/v0.25.0.rst

@@ -336,7 +336,7 @@ Indexing
 - Improved exception message when calling :meth:`DataFrame.iloc` with a list of non-numeric objects (:issue:`25753`).
 - Bug in :meth:`DataFrame.loc` and :meth:`Series.loc` where ``KeyError`` was not raised for a ``MultiIndex`` when the key was less than or equal to the number of levels in the :class:`MultiIndex` (:issue:`14885`).
 - Bug in which :meth:`DataFrame.append` produced an erroneous warning indicating that a ``KeyError`` will be thrown in the future when the data to be appended contains new columns (:issue:`22252`).
--
+- Bug in which :meth:`DataFrame.to_csv` caused a segfault for a reindexed data frame, when the indices were single-level :class:`MultiIndex` (:issue:`26303`).
 
 
 Missing

pandas/core/indexes/multi.py

@@ -946,7 +946,8 @@ def _format_native_types(self, na_rep='nan', **kwargs):
             new_codes.append(level_codes)
 
         if len(new_levels) == 1:
-            return Index(new_levels[0])._format_native_types()
+            # a single-level multi-index
+            return Index(new_levels[0].take(new_codes[0]))._format_native_types()
         else:
             # reconstruct the multi-index
             mi = MultiIndex(levels=new_levels, codes=new_codes,

pandas/tests/frame/test_to_csv.py

@@ -1220,6 +1220,18 @@ def test_multi_index_header(self):
         expected = tm.convert_rows_list_to_csv_str(expected_rows)
         assert result == expected
 
+    def test_to_csv_single_level_multi_index(self):
+        # see gh-26303
+        index = pd.Index([(1,), (2,), (3,)])
+        df = pd.DataFrame([[1, 2, 3]], columns=index)
+        with ensure_clean() as path:
+            df = df.reindex(columns=[(1,), (3,)])
+            df.to_csv(path, line_terminator='\n')
+            expected = b",1,3\n0,1,3\n"
+
+            with open(path, mode='rb') as f:
+                assert f.read() == expected
+
     def test_gz_lineend(self):
         # GH 25311
         df = pd.DataFrame({'a': [1, 2]})