Series.nlargest(n=5, keep='first')
[source]
Return the largest n
elements.
Parameters: |
n : int Return this many descending sorted values keep : {‘first’, ‘last’}, default ‘first’ Where there are duplicate values: - |
---|---|
Returns: |
top_n : Series The n largest values in the Series, in sorted order |
See also
Faster than .sort_values(ascending=False).head(n)
for small n
relative to the size of the Series
object.
>>> import pandas as pd >>> import numpy as np >>> s = pd.Series(np.random.randn(10**6)) >>> s.nlargest(10) # only sorts up to the N requested 219921 4.644710 82124 4.608745 421689 4.564644 425277 4.447014 718691 4.414137 43154 4.403520 283187 4.313922 595519 4.273635 503969 4.250236 121637 4.240952 dtype: float64
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http://pandas.pydata.org/pandas-docs/version/0.23.4/generated/pandas.Series.nlargest.html