value is a list (of dates in this case). Wanted to subtract another column (scalar date) from all the entries.

[zbenmo]

Description

PanicException: sub operation not supported for dtype list[date]

[zbenmo]

May be related: https://stackoverflow.com/questions/76817818/how-can-i-perform-operations-between-a-list-and-scalar-column-in-polars

#8006

[alexander-beedie]

You may find list eval suitable if you can get away with subtracting a constant?

from datetime import date
import polars as pl

d = date.today()

df = pl.DataFrame(
  data={"dts": [[d,d],[d,d]]},
).with_columns(
  sub = pl.col("dts").list.eval(pl.element() - date(2024,1,1))
)
# shape: (2, 2)
# ┌──────────────────────────┬────────────────────┐
# │ dts                      ┆ sub                │
# │ ---                      ┆ ---                │
# │ list[date]               ┆ list[duration[ms]] │
# ╞══════════════════════════╪════════════════════╡
# │ [2024-04-17, 2024-04-17] ┆ [107d, 107d]       │
# │ [2024-04-17, 2024-04-17] ┆ [107d, 107d]       │
# └──────────────────────────┴────────────────────┘

If not, the current solutions are all bit more exotic 😅
(See @zbenmo links above).

[zbenmo]

.list.eval is indeed promising, it works at least at the moment with a constant scalar yet not with a scalar from another column.

For my current use case, I'll solve it by having the list column first as long format (exploded), with the needed date duplicated for all rows in another column, and only then group_by and agg.