Function arrow::compute::kernels::partition::lexicographical_partition_ranges

pub fn lexicographical_partition_ranges(
    columns: &[SortColumn],
) -> Result<impl Iterator<Item = Range<usize>>, ArrowError>

👎Deprecated: Use partition

Use partition instead. Given a list of already sorted columns, find partition ranges that would partition lexicographically equal values across columns.

The returned vec would be of size k where k is cardinality of the sorted values; Consecutive values will be connected: (a, b) and (b, c), where start = 0 and end = n for the first and last range.