Return the largest n elements.
Return this many descending sorted values.
When there are duplicate values that cannot all fit in a Series of n elements:
first : return the first n occurrences in order of appearance.
last : return the last n occurrences in reverse order of appearance.
all : keep all occurrences. This can result in a Series of size larger than n.
The n largest values in the Series, sorted in decreasing order.
See also
Series.nsmallestGet the n smallest elements.
Series.sort_valuesSort Series by values.
Series.headReturn the first n rows.
Notes
Faster than .sort_values(ascending=False).head(n) for small n relative to the size of the Series object.
Examples
>>> countries_population = {"Italy": 59000000, "France": 65000000,
... "Malta": 434000, "Maldives": 434000,
... "Brunei": 434000, "Iceland": 337000,
... "Nauru": 11300, "Tuvalu": 11300,
... "Anguilla": 11300, "Montserrat": 5200}
>>> s = pd.Series(countries_population)
>>> s
Italy 59000000
France 65000000
Malta 434000
Maldives 434000
Brunei 434000
Iceland 337000
Nauru 11300
Tuvalu 11300
Anguilla 11300
Montserrat 5200
dtype: int64
The n largest elements where n=5 by default.
>>> s.nlargest()
France 65000000
Italy 59000000
Malta 434000
Maldives 434000
Brunei 434000
dtype: int64
The n largest elements where n=3. Default keep value is ‘first’ so Malta will be kept.
>>> s.nlargest(3)
France 65000000
Italy 59000000
Malta 434000
dtype: int64
The n largest elements where n=3 and keeping the last duplicates. Brunei will be kept since it is the last with value 434000 based on the index order.
>>> s.nlargest(3, keep='last')
France 65000000
Italy 59000000
Brunei 434000
dtype: int64
The n largest elements where n=3 with all duplicates kept. Note that the returned Series has five elements due to the three duplicates.
>>> s.nlargest(3, keep='all')
France 65000000
Italy 59000000
Malta 434000
Maldives 434000
Brunei 434000
dtype: int64
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© 2011–2025, Open source contributors
Licensed under the 3-clause BSD License.
https://pandas.pydata.org/pandas-docs/version/2.3.0/reference/api/pandas.core.groupby.SeriesGroupBy.nlargest.html