polars.Series.top_k

Series.top_k(*, k: int = 5, descending: bool = False) → Series

Return the k largest elements.

If ‘descending=True` the smallest elements will be given.

This has time complexity:

\[\begin{split}O(n + k \\log{}n - \frac{k}{2})\end{split}\]

Parameters:

k

Number of elements to return.

descending

Return the smallest elements.