README_EN.md

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2854. Rolling Average Steps 🔒

中文文档

Description

Table: Steps

+-------------+------+ 
| Column Name | Type | 
+-------------+------+ 
| user_id     | int  | 
| steps_count | int  |
| steps_date  | date |
+-------------+------+
(user_id, steps_date) is the primary key for this table.
Each row of this table contains user_id, steps_count, and steps_date.

Write a solution to calculate 3-day rolling averages of steps for each user.

We calculate the n-day rolling average this way:

• For each day, we calculate the average of n consecutive days of step counts ending on that day if available, otherwise, n-day rolling average is not defined for it.

Output the user_id, steps_date, and rolling average. Round the rolling average to two decimal places.

Return the result table ordered by user_id, steps_date in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Steps table:
+---------+-------------+------------+
| user_id | steps_count | steps_date |
+---------+-------------+------------+
| 1       | 687         | 2021-09-02 |
| 1       | 395         | 2021-09-04 |
| 1       | 499         | 2021-09-05 |
| 1       | 712         | 2021-09-06 |
| 1       | 576         | 2021-09-07 |
| 2       | 153         | 2021-09-06 |
| 2       | 171         | 2021-09-07 |
| 2       | 530         | 2021-09-08 |
| 3       | 945         | 2021-09-04 |
| 3       | 120         | 2021-09-07 |
| 3       | 557         | 2021-09-08 |
| 3       | 840         | 2021-09-09 |
| 3       | 627         | 2021-09-10 |
| 5       | 382         | 2021-09-05 |
| 6       | 480         | 2021-09-01 |
| 6       | 191         | 2021-09-02 |
| 6       | 303         | 2021-09-05 |
+---------+-------------+------------+
Output: 
+---------+------------+-----------------+
| user_id | steps_date | rolling_average | 
+---------+------------+-----------------+
| 1       | 2021-09-06 | 535.33          | 
| 1       | 2021-09-07 | 595.67          | 
| 2       | 2021-09-08 | 284.67          |
| 3       | 2021-09-09 | 505.67          |
| 3       | 2021-09-10 | 674.67          |    
+---------+------------+-----------------+
Explanation: 
- For user id 1, the step counts for the three consecutive days up to 2021-09-06 are available. Consequently, the rolling average for this particular date is computed as (395 + 499 + 712) / 3 = 535.33.
- For user id 1, the step counts for the three consecutive days up to 2021-09-07 are available. Consequently, the rolling average for this particular date is computed as (499 + 712 + 576) / 3 = 595.67.
- For user id 2, the step counts for the three consecutive days up to 2021-09-08 are available. Consequently, the rolling average for this particular date is computed as (153 + 171 + 530) / 3 = 284.67.
- For user id 3, the step counts for the three consecutive days up to 2021-09-09 are available. Consequently, the rolling average for this particular date is computed as (120 + 557 + 840) / 3 = 505.67.
- For user id 3, the step counts for the three consecutive days up to 2021-09-10 are available. Consequently, the rolling average for this particular date is computed as (557 + 840 + 627) / 3 = 674.67.
- For user id 4 and 5, the calculation of the rolling average is not viable as there is insufficient data for the consecutive three days. Output table ordered by user_id and steps_date in ascending order.

Solutions

Solution 1: Window Functions

We can use the window function LAG() OVER() to calculate the difference in days between the current date and the date before the last date for each user. If the difference is $2$, it means that there are continuous data for $3$ days between these two dates. We can use the window function AVG() OVER() to calculate the average of these $3$ data.

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            user_id,
            steps_date,
            ROUND(
                AVG(steps_count) OVER (
                    PARTITION BY user_id
                    ORDER BY steps_date
                    ROWS 2 PRECEDING
                ),
                2
            ) AS rolling_average,
            DATEDIFF(
                steps_date,
                LAG(steps_date, 2) OVER (
                    PARTITION BY user_id
                    ORDER BY steps_date
                )
            ) = 2 AS st
        FROM Steps
    )
SELECT
    user_id,
    steps_date,
    rolling_average
FROM T
WHERE st = 1
ORDER BY 1, 2;