BUG: Fixed latex output for multi-indexed dataframes - GH9778

Author: yred

doc/source/whatsnew/v0.16.1.txt

@@ -157,3 +157,5 @@ Bug Fixes
 - Bug where dividing a dataframe containing values of type ``Decimal`` by another ``Decimal`` would raise. (:issue:`9787`)
 - Bug where using DataFrames asfreq would remove the name of the index. (:issue:`9885`)
 - Changed caching in ``AbstractHolidayCalendar`` to be at the instance level rather than at the class level as the latter can result in unexpected behaviour. (:issue:`9552`)
+
+- Fixed latex output for multi-indexed dataframes (:issue:`9778`)

pandas/core/format.py

@@ -613,8 +613,12 @@ def get_col_type(dtype):
             name = any(self.frame.columns.names)
             for i, lev in enumerate(self.frame.index.levels):
                 lev2 = lev.format(name=name)
-                width = len(lev2[0])
-                lev3 = [' ' * width] * clevels + lev2
+                blank = ' ' * len(lev2[0])
+                lev3 = [blank] * clevels
+                for level_idx, group in itertools.groupby(
+                        self.frame.index.labels[i]):
+                    count = len(list(group))
+                    lev3.extend([lev2[level_idx]] + [blank] * (count - 1))
                 strcols.insert(i, lev3)
 
         if column_format is None:

pandas/tests/test_format.py

@@ -2194,6 +2194,28 @@ def test_to_latex_multiindex(self):
 x & y &  a \\
 \bottomrule
 \end{tabular}
+"""
+        self.assertEqual(result, expected)
+
+        df = DataFrame.from_dict({
+            ('c1', 0): pd.Series(dict((x, x) for x in range(4))),
+            ('c1', 1): pd.Series(dict((x, x + 4) for x in range(4))),
+            ('c2', 0): pd.Series(dict((x, x) for x in range(4))),
+            ('c2', 1): pd.Series(dict((x, x + 4) for x in range(4))),
+            ('c3', 0): pd.Series(dict((x, x) for x in range(4))),
+        }).T
+        result = df.to_latex()
+        expected = r"""\begin{tabular}{llrrrr}
+\toprule
+   &   &  0 &  1 &  2 &  3 \\
+\midrule
+c1 & 0 &  0 &  1 &  2 &  3 \\
+   & 1 &  4 &  5 &  6 &  7 \\
+c2 & 0 &  0 &  1 &  2 &  3 \\
+   & 1 &  4 &  5 &  6 &  7 \\
+c3 & 0 &  0 &  1 &  2 &  3 \\
+\bottomrule
+\end{tabular}
 """
         self.assertEqual(result, expected)