Practice / aggregations
Corporate at The Daily Grind wants to spotlight its busiest stores — but "busy" means real money, not a high sticker price on one lucky order.
You have the orders DataFrame:
Your task: for each store, compute the average order price as avg_price, but return only stores whose total revenue is greater than 15.00. Return store and avg_price.
Assign your answer DataFrame to result. Row order doesn't matter, and avg_price is graded with a small float tolerance — no need to round.
ordersinput DataFrame| column | type |
|---|---|
| order_id | long |
| store | string |
| price | double |
| order_id | store | price |
|---|---|---|
| 1 | Downtown | 5 |
| 2 | Downtown | 6 |
| 3 | Downtown | 5.5 |
| 4 | Airport | 4 |
| 5 | Airport | 3 |
| 6 | Pier | 7 |
| 7 | Pier | 9 |
| 8 | Harbor | 15 |
storeavg_price· 2 rowsAggregate two things at once: F.avg("price").alias("avg_price") and
F.sum("price").alias("total"). Then filter on the total, and finally
select just store and avg_price. HAVING is just a filter after the agg.
This problem builds on Aggregations & groupBy (~7 min). Pop it open in a new tab if you want a quick recap.