examples/tpch/q13_customer_distribution.py

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"""
TPC-H Problem Statement Query 13:

This query determines the distribution of customers by the number of orders they have made,
including customers who have no record of orders, past or present. It counts and reports how many
customers have no orders, how many have 1, 2, 3, etc. A check is made to ensure that the orders
counted do not fall into one of several special categories of orders. Special categories are
identified in the order comment column by looking for a particular pattern.

The above problem statement text is copyrighted by the Transaction Processing Performance Council
as part of their TPC Benchmark H Specification revision 2.18.0.

Reference SQL (from TPC-H specification, used by the benchmark suite)::

    select
        c_count,
        count(*) as custdist
    from
        (
                select
                        c_custkey,
                        count(o_orderkey)
                from
                        customer left outer join orders on
                                c_custkey = o_custkey
                                and o_comment not like '%special%requests%'
                group by
                        c_custkey
        ) as c_orders (c_custkey, c_count)
    group by
        c_count
    order by
        custdist desc,
        c_count desc;
"""

from datafusion import SessionContext, col, lit
from datafusion import functions as F
from util import get_data_path

WORD_1 = "special"
WORD_2 = "requests"

# Load the dataframes we need

ctx = SessionContext()

df_orders = ctx.read_parquet(get_data_path("orders.parquet")).select(
    "o_custkey", "o_comment"
)
df_customer = ctx.read_parquet(get_data_path("customer.parquet")).select("c_custkey")

# Use a regex to remove special cases
df_orders = df_orders.filter(
    F.regexp_match(col("o_comment"), lit(f"{WORD_1}.?*{WORD_2}")).is_null()
)

# Customers with no orders still participate, so this is a left join. Count the
# orders per customer, then count customers per order-count value.
df = (
    df_customer.join(df_orders, left_on="c_custkey", right_on="o_custkey", how="left")
    .aggregate(["c_custkey"], [F.count(col("o_custkey")).alias("c_count")])
    .aggregate(["c_count"], [F.count_star().alias("custdist")])
    .sort(
        col("custdist").sort(ascending=False),
        col("c_count").sort(ascending=False),
    )
)

df.show()