def get_duplicates(self):
warnings.warn("<STR_LIT>"<EOL>"<STR_LIT>"<EOL>"<STR_LIT>",<EOL>FutureWarning, stacklevel=<NUM_LIT:2>)<EOL>return self[self.duplicated()].unique()<EOL>
Extract duplicated index elements.

.. deprecated:: 0.23.0
    Use idx[idx.duplicated()].unique() instead

Returns a sorted list of index elements which appear more than once in
the index.

Returns
-------
array-like
    List of duplicated indexes.

See Also
--------
Index.duplicated : Return boolean array denoting duplicates.
Index.drop_duplicates : Return Index with duplicates removed.

Examples
--------

Works on different Index of types.

>>> pd.Index([1, 2, 2, 3, 3, 3, 4]).get_duplicates()  # doctest: +SKIP
[2, 3]

Note that for a DatetimeIndex, it does not return a list but a new
DatetimeIndex:

>>> dates = pd.to_datetime(['2018-01-01', '2018-01-02', '2018-01-03',
...                         '2018-01-03', '2018-01-04', '2018-01-04'],
...                        format='%Y-%m-%d')
>>> pd.Index(dates).get_duplicates()  # doctest: +SKIP
DatetimeIndex(['2018-01-03', '2018-01-04'],
              dtype='datetime64[ns]', freq=None)

Sorts duplicated elements even when indexes are unordered.

>>> pd.Index([1, 2, 3, 2, 3, 4, 3]).get_duplicates()  # doctest: +SKIP
[2, 3]

Return empty array-like structure when all elements are unique.

>>> pd.Index([1, 2, 3, 4]).get_duplicates()  # doctest: +SKIP
[]
>>> dates = pd.to_datetime(['2018-01-01', '2018-01-02', '2018-01-03'],
...                        format='%Y-%m-%d')
>>> pd.Index(dates).get_duplicates()  # doctest: +SKIP
DatetimeIndex([], dtype='datetime64[ns]', freq=None)