写一段sql代码是分析共享单车数据间的联系
时间: 2024-03-17 10:44:00 浏览: 13
假设我们有如下的共享单车数据表格:
```
Table: bike_data
| bike_id | user_id | start_time | end_time | start_location | end_location |
|---------|---------|--------------------|--------------------|-----------------|---------------|
| 1 | 100 | 2021-01-01 08:00:00 | 2021-01-01 08:30:00 | New York | Boston |
| 2 | 200 | 2021-01-01 09:00:00 | 2021-01-01 09:30:00 | Boston | Chicago |
| 3 | 300 | 2021-01-01 12:00:00 | 2021-01-01 12:30:00 | New York | Chicago |
| 4 | 100 | 2021-01-01 15:00:00 | 2021-01-01 15:30:00 | Chicago | Boston |
| 5 | 200 | 2021-01-01 18:00:00 | 2021-01-01 18:30:00 | Boston | New York |
| 6 | 400 | 2021-01-02 10:00:00 | 2021-01-02 10:30:00 | Chicago | New York |
| 7 | 500 | 2021-01-02 11:00:00 | 2021-01-02 11:30:00 | New York | Boston |
| 8 | 300 | 2021-01-02 13:00:00 | 2021-01-02 13:30:00 | Boston | Chicago |
| 9 | 200 | 2021-01-02 16:00:00 | 2021-01-02 16:30:00 | Chicago | Boston |
| 10 | 400 | 2021-01-02 19:00:00 | 2021-01-02 19:30:00 | Boston | New York |
```
现在我们想要分析这些共享单车数据之间的联系,可以使用以下的 SQL 代码:
```
SELECT a.user_id AS user1, b.user_id AS user2, COUNT(*) AS num_trips
FROM bike_data a
INNER JOIN bike_data b
ON a.start_location = b.start_location AND a.end_location = b.end_location
GROUP BY a.user_id, b.user_id
HAVING num_trips >= 2
ORDER BY num_trips DESC
```
这段 SQL 代码使用 INNER JOIN 连接 bike_data 表格两次,以便找到起始和结束位置相同的记录。然后我们对用户 ID 进行分组,并统计每对用户之间的骑行次数。最后,我们过滤掉只有一次骑行记录的用户对,并按照骑行次数降序排序,以便更容易地发现最活跃的用户对。