Series.nlargest(self, n=5, keep='first')
[source]
Return the largest n
elements.
Parameters: |
|
---|---|
Returns: |
|
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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Licensed under the 3-clause BSD License.
https://pandas.pydata.org/pandas-docs/version/0.25.0/reference/api/pandas.Series.nlargest.html