9/8 每日一題(MYSQL 找出各組的兩次開始與結束經過平均時間)

 Table: Activity

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| machine_id     | int     |
| process_id     | int     |
| activity_type  | enum    |
| timestamp      | float   |
+----------------+---------+
The table shows the user activities for a factory website.
(machine_id, process_id, activity_type) is the primary key (combination of columns with unique values) of this table.
machine_id is the ID of a machine.
process_id is the ID of a process running on the machine with ID machine_id.
activity_type is an ENUM (category) of type ('start', 'end').
timestamp is a float representing the current time in seconds.
'start' means the machine starts the process at the given timestamp and 'end' means the machine ends the process at the given timestamp.
The 'start' timestamp will always be before the 'end' timestamp for every (machine_id, process_id) pair.

 

There is a factory website that has several machines each running the same number of processes. Write a solution to find the average time each machine takes to complete a process.

The time to complete a process is the 'end' timestamp minus the 'start' timestamp. The average time is calculated by the total time to complete every process on the machine divided by the number of processes that were run.

The resulting table should have the machine_id along with the average time as processing_time, which should be rounded to 3 decimal places.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Activity table:
+------------+------------+---------------+-----------+
| machine_id | process_id | activity_type | timestamp |
+------------+------------+---------------+-----------+
| 0          | 0          | start         | 0.712     |
| 0          | 0          | end           | 1.520     |
| 0          | 1          | start         | 3.140     |
| 0          | 1          | end           | 4.120     |
| 1          | 0          | start         | 0.550     |
| 1          | 0          | end           | 1.550     |
| 1          | 1          | start         | 0.430     |
| 1          | 1          | end           | 1.420     |
| 2          | 0          | start         | 4.100     |
| 2          | 0          | end           | 4.512     |
| 2          | 1          | start         | 2.500     |
| 2          | 1          | end           | 5.000     |
+------------+------------+---------------+-----------+
Output: 
+------------+-----------------+
| machine_id | processing_time |
+------------+-----------------+
| 0          | 0.894           |
| 1          | 0.995           |
| 2          | 1.456           |
+------------+-----------------+
Explanation: 
There are 3 machines running 2 processes each.
Machine 0's average time is ((1.520 - 0.712) + (4.120 - 3.140)) / 2 = 0.894
Machine 1's average time is ((1.550 - 0.550) + (1.420 - 0.430)) / 2 = 0.995
Machine 2's average time is ((4.512 - 4.100) + (5.000 - 2.500)) / 2 = 1.456

# Write your MySQL query statement below

SELECT A.machine_id ,

ROUND(SUM((select timestamp from Activity WHERE activity_type ='end'

 and machine_id=A.machine_id

 and process_id=A.process_id)

-(select timestamp from Activity WHERE activity_type ='start'

 and machine_id=A.machine_id

 and process_id=A.process_id))/COUNT(machine_id),3) AS processing_time

 

 

FROM Activity AS A

GROUP BY A.machine_id

-----------------------------------

# Write your MySQL query statement below
select a.machine_id, round(avg(
  b.timestamp - a.timestamp
),3) as processing_time from activity a
join activity b 

on a.machine_id = b.machine_id and a.activity_type='start' and b.activity_type='end'
group by a.machine_id

-------------------------------------

# Write your MySQL query statement below
select  
        s.machine_id, 
        round(avg(p.timestamp-s.timestamp), 3) as processing_time 
from
Activity as s join Activity as p
on s.machine_id = p.machine_id and s.process_id = p.process_id
where s.activity_type = 'start' and p.activity_type = 'end'
group by s.machine_id;