polars.Series.top_k#

Series.top_k(k: int | IntoExprColumn = 5) → Series[source]#

Return the k largest elements.

This has time complexity:

$$\begin{array}{r}O(n+k\\ logn-\frac{k}{2})\end{array}$$

Parameters:

k

Number of elements to return.

See also

bottom_k

Examples

>>> s = pl.Series("a", [2, 5, 1, 4, 3])
>>> s.top_k(3)
shape: (3,)
Series: 'a' [i64]
[
    5
    4
    3
]