Series.nlargest(self, n=5, keep='first') [source]
Return the largest n elements.
| Parameters: |
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| Returns: |
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See also
Series.nsmallest
n smallest elements.Series.sort_values
Series.head
n rows.Faster than .sort_values(ascending=False).head(n) for small n relative to the size of the Series object.
>>> countries_population = {"Italy": 59000000, "France": 65000000,
... "Malta": 434000, "Maldives": 434000,
... "Brunei": 434000, "Iceland": 337000,
... "Nauru": 11300, "Tuvalu": 11300,
... "Anguilla": 11300, "Monserat": 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
Monserat 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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https://pandas.pydata.org/pandas-docs/version/0.25.0/reference/api/pandas.Series.nlargest.html