Read Primitives — MAX_BY with Ties

Read Primitives benchmark · all Read Primitives queries

Find the supplier with the highest supply cost for each part

Representative SQL

Rendered for DataFusion with default parameters.

WITH ranked AS (
  SELECT
    p.p_partkey,
    p.p_name,
    s.s_name,
    ps.ps_supplycost,
    ROW_NUMBER() OVER (PARTITION BY p.p_partkey, p.p_name ORDER BY ps.ps_supplycost DESC, s.s_name) AS rn
  FROM partsupp AS ps
  JOIN part AS p
    ON ps.ps_partkey = p.p_partkey
  JOIN supplier AS s
    ON ps.ps_suppkey = s.s_suppkey
)
SELECT
  p_partkey,
  p_name,
  s_name AS supplier_name,
  ps_supplycost AS max_supply_cost
FROM ranked
WHERE
  rn = 1
ORDER BY
  max_supply_cost DESC,
  p_partkey,
  p_name
LIMIT 100

Representative DataFrame

Expression family (Polars, PySpark, DataFusion), default parameters.

def max_by_with_ties_expression_impl(ctx: DataFrameContext) -> Any:
    return _supply_cost_extreme_expr(ctx, descending=True, value_alias="max_supply_cost", limit=100)

Get the exact query for your run

The blocks above are representative. To get the precise statement a specific platform runs at a specific scale, substitute your own values:

from benchbox import ReadPrimitives

bench = ReadPrimitives(scale_factor=0.01)            # your scale factor
queries = bench.get_queries(dialect="datafusion")   # your platform's dialect; drop dialect= for native SQL
print(queries['max_by_with_ties'])

For the DataFrame form:

from benchbox.core.query_catalog import get_dataframe_render

print(get_dataframe_render('read_primitives', 'max_by_with_ties', family="expression").source)  # or family="pandas"

Source

Query text: benchbox/core/read_primitives/queries.py