I am trying to filter my Polars DataFrame for dates that are nearest to a given date. For example: import polars import datetime data = { "date": ["2025-01-01", "2025-01-01", "2025-01-01", "2026-01-01"], "value": [1, 2, 3, 4], } df = polars.DataFrame(data).with_columns([polars.col("date").cast(polars.Date)])  shape: (4, 2) ┌────────────┬───────┐ │ date ┆ value │ │ --- ┆ --- │ │ date ┆ i64 │ ╞════════════╪═══════╡ │ 2025-01-01 ┆ 1 │ │ 2025-01-01 ┆ 2 │ │ 2025-01-01 ┆ 3 │ │ 2026-01-01 ┆ 4 │ └────────────┴───────┘  Given a date, say: date = datetime.date(2024, 12, 31)  I want to filter the DataFrame for rows where the date column only includes records that are closest to my required date. I know that I can do the following: result = df.with_columns( diff=(polars.col("date") - date).abs() ).filter( polars.col("diff") == polars.min("diff") )  shape: (3, 3) ┌────────────┬───────┬──────────────┐ │ date ┆ value ┆ diff │ │ --- ┆ --- ┆ --- │ │ date ┆ i64 ┆ duration[ms] │ ╞════════════╪═══════╪══════════════╡ │ 2025-01-01 ┆ 1 ┆ 1d │ │ 2025-01-01 ┆ 2 ┆ 1d │ │ 2025-01-01 ┆ 3 ┆ 1d │ └────────────┴───────┴──────────────┘  Is there a more succinct way to achieve this (without creating a new column, for example)?
You don't need to add the temporary column, just filter directly: df.filter((m:=(pl.col('date')-date).abs()).min() == m)  Or, without the walrus operator: diff = (pl.col('date')-date).abs() df.filter(diff.min() == diff)  Output: ┌────────────┬───────┐ │ date ┆ value │ │ --- ┆ --- │ │ date ┆ i64 │ ╞════════════╪═══════╡ │ 2025-01-01 ┆ 1 │ │ 2025-01-01 ┆ 2 │ │ 2025-01-01 ┆ 3 │ └────────────┴───────┘ 