获取平均日或周值

Question

I have statistical data like this:

time        val1
1424166578  51
1424166877  55
1424167178  57
1424167477  57  

time is a unix timestamp. There is one record every 5 minutes excluding nights and sundays. This continues over several weeks.

Now I want to get these values for an average day and an average week. The result should include values for every 5 minutes like normal but for average past days or weeks.

The result should look like this:

time    val1
0       43.423
300     46.635
600     51.887
...

So time could be a timestamp with relative time since day or week start. Perhaps it is better to use DATETIME ... not sure.

If I use GROUP BY FROM_UNIXTIME(time, '%Y%m%d') for example I get one value for the whole day. But I want all average values for all days.

Answer 1

You seem to be interested in grouping dates by five minute intervals instead of dates. This is fairly straightforward:

SELECT
    HOUR(FROM_UNIXTIME(time)) AS HH,
    (MINUTE(FROM_UNIXTIME(time)) DIV 5) * 5 AS MM,
    AVG(val1) AS VAL
FROM your_table
WHERE time > UNIX_TIMESTAMP(CURRENT_TIMESTAMP - INTERVAL 7 DAY)
GROUP BY HH, MM

The following result will explain how date is clamped:

time        FROM_UNIXTIME(time)  HH  MM
1424166578  2015-02-17 14:49:38  14  45
1424166877  2015-02-17 14:54:37  14  50
1424167178  2015-02-17 14:59:38  14  55
1424167477  2015-02-17 15:04:37  15  00

Answer 2

I would approach this as:

select date(from_unixtime(time)) as day, avg(val)
from table t
group by date(from_unixtime(time))
order by day;

Although you can use the format argument, I think of that more for converting the value to a string than to a date/time.