Practice / window functions
A growth team tracks daily sign-ups and wants the day-over-day change.
daily
Your task: return day, signups, and a delta — the change in signups versus the previous day, ordered by day. Keep the first day, whose delta is null (there is no previous day to compare against).
Assign your answer DataFrame to result. Row order doesn't matter.
dailyinput DataFrame| column | type |
|---|---|
| day | string |
| signups | long |
| day | signups |
|---|---|
| 2026-06-01 | 10 |
| 2026-06-02 | 14 |
| 2026-06-03 | 9 |
daysignupsdelta· 3 rowsw = Window.orderBy("day"), then delta = F.col("signups") - F.lag("signups", 1).over(w).
The first row has no previous day, so lag is null and the subtraction is null —
keep that row.
This problem builds on Window Functions (~9 min). Pop it open in a new tab if you want a quick recap.