list.agg

ColExpr.list.agg(

*,

partition_by: Col | ColName | str | Iterable[Col | ColName | str] | None = None,

arrange: ColExpr | Iterable[ColExpr] | None = None,

filter: ColExpr[Bool] | Iterable[ColExpr[Bool]] | None = None,

) → ColExpr[List]

Collect the elements of each group in a list.