pub fn lexicographical_partition_ranges(
columns: &[SortColumn],
) -> Result<impl Iterator<Item = Range<usize>> + '_, ArrowError>
👎Deprecated: Use partition
Expand description
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.