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DOC: add docstring for Index.get_duplicates #20223

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36 changes: 36 additions & 0 deletions pandas/core/indexes/base.py
Original file line number Diff line number Diff line change
Expand Up @@ -1710,6 +1710,42 @@ def _invalid_indexer(self, form, key):
kind=type(key)))

def get_duplicates(self):
"""
Extract duplicated index elements.

This function returns a sorted list of index elements which appear more
than once in the index.

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

See Also
--------
:meth:`Index.duplicated` : Return boolean array denoting duplicates.
:meth:`Index.drop_duplicates` : Return Index with duplicates removed.
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In the See Also section, it is not needed to make an explicit link with the :meth:`..` syntax, you can just use Index.duplicated


Examples
--------
>>> pd.Index([1, 2, 3, 4]).get_duplicates()
[]
>>> pd.Index([1, 2, 2, 3, 3, 3, 4]).get_duplicates()
[2, 3]
>>> pd.Index([1, 2, 3, 2, 3, 4, 3]).get_duplicates()
[2, 3]
>>> pd.Index(['a', 'b', 'b', 'c', 'c', 'c', 'd']).get_duplicates()
['b', 'c']
>>> dates = pd.to_datetime(['2018-01-01', '2018-01-02',
... '2018-01-03', '2018-01-03'],
... format='%Y-%m-%d')
>>> pd.Index(pd.to_datetime(dates, format='%Y-%m-%d')).get_duplicates()
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you don't need to repeat the to_datetime here.

Can you also add a sentence here before the example explaining that for DatetimeIndex, it does not return a list but a DatetetimeIndex ?

DatetimeIndex(['2018-01-03'], dtype='datetime64[ns]', freq=None)

Notes
-----
Returns empty list in case all index elements are unique.
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I would put this explanation above the actual example where you show this

"""
from collections import defaultdict
counter = defaultdict(lambda: 0)
for k in self.values:
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