2

当 db 时间片大于想要的时间片时，如何从数据中查询时间片。最终结果将用于绘制堆积条形图。

示例数据：

START_TS (int)| END_TS (int) | DATA (int) | GROUP
-----------------------------------
0       | 179      | 2000  | G1
180     | 499      | 1000  | G2
500     | 699      | 1000  | G1
845 ...

需要使用时间片作为 100 个“单位”的输出。输出中不需要 End_ts，但有助于理解计算。

START_TS |  END_TS  | DATA (equation = amount in that time slice) | GROUP
-------------------------------------------------------
0       |    99   | (2000 / 180) * 100 =  1111 | G1
100     |   199   | (2000 / 180) *  80 =   889 | G1
100     |   199   | (1000 / 320) *  20 =    63 | G2
200     |   299   | (1000 / 320) * 100 =   313 | G2 
300     |   399   | (1000 / 320) * 100 =   313 | G2
400     |   499   | (1000 / 320) * 100 =   313 | G2

从中获得时间序列是这样的。

SELECT (startts/100)*100, ...
FROM TABLE
    FULL JOIN
        ( SELECT startts from generate_series(0,700,100) startts ) s1
    USING (startts)
GROUP BY  startts/100

所以它会是这样的（没有分组）

 STARTTS | ENDTS | DATA | GROUP
   0     | 179      | 2000   | G1
   100   |      
   180   | 499      | 1000   | G2
   200   |
   300   |
   400   | 
   500   | 699      | 1000   | G1
   600   |
   700

但是如何将 DATA 拆分为两个或多个生成的行（时间片行），以便在时间片中计算。

** 这基本上有效，但在大数据集上没有真正的功能。行如 1-100M 行。

这是执行此操作的查询+更多聚合不重叠时间片的值

SELECT (start_ts/100)*100 as start_ts, sum(part) as data, cgroup
FROM (
SELECT *, ( data * (overlap_end-overlap_start + 1 ) / ( end_ts - tts + 1 ) ) as part
FROM 
    (
    SELECT (case when s1.start_ts > t.start_ts then s1.start_ts else t.start_ts end) as overlap_start,
        (case when s1.start_ts+100 < t.end_ts then s1.start_ts+100-1 else t.end_ts end) as overlap_end,
        t.start_ts as tts, s1.start_ts as start_ts, t.end_ts, cgroup, data
    FROM (SELECT start_ts from generate_series(0,800,100) start_ts ) s1 
        LEFT OUTER JOIN test t on t.start_ts < s1.start_ts+100 and t.end_ts >= s1.start_ts
    ) t
) t2
GROUP BY start_ts/100, cgroup
4

2 回答 2

1

您需要将不同的时隙分成由序列定义的箱。以下查询通过修改连接条件并计算两者之间的重叠来完成此操作：

SELECT (startts/100)*100, ...
from (select (case when s1.starts > t.start_ts then s1.starts else t.start_t2 end) as overlap_start,
             (case when s1.starts+100 < t.end_ts then s1.starts+100-1 else t.end_ts end) as overlap_end,
             ts.*
      FROM (SELECT startts from generate_series(0,700,100) startts ) s1 left outer join
           TABLE t
           on t.startts < s1.starts+100 and
              t.end_ts >= s1.starts
     ) t
于 2013-01-07T16:59:11.843 回答
0

SQL 小提琴。为了清楚起见，它显示了每个步骤的所有计算列。

with data_avg as (
    select start_ts, end_ts, "data" * 1.0 / ((end_ts + 1) - start_ts) data_avg
    from test
), gs as (
    select start_ts, start_ts + 99 end_ts
    from generate_series(
        (select min(start_ts) from test),
        (select max(end_ts) from test),
        100
    ) gs(start_ts)
)
select 
    t_start, t_end,
    gs_start, gs_end,
    cgroup,
    s."start", s."end",
    da.start_ts da_start, da.end_ts da_end
    ,round((s."end" - s."start" + 1) * da.data_avg) "data"
from (
    select
        t.start_ts t_start, t.end_ts t_end,
        gs.start_ts gs_start, gs.end_ts gs_end,
        cgroup,
        greatest(t.start_ts, gs.start_ts) "start", least(t.end_ts, gs.end_ts) "end"
    from
        test t
        inner join
        gs on
            gs.start_ts between t.start_ts and t.end_ts
            or
            gs.end_ts between t.start_ts and t.end_ts
    ) s
    inner join
    data_avg da on
        da.start_ts between t_start and t_end
        and
        da.end_ts between t_start and t_end
order by s."start"

结果：

 t_start | t_end | gs_start | gs_end | cgroup | start | end | da_start | da_end | data 
---------+-------+----------+--------+--------+-------+-----+----------+--------+------
       0 |   179 |        0 |     99 | G1     |     0 |  99 |        0 |    179 | 1111
       0 |   179 |      100 |    199 | G1     |   100 | 179 |        0 |    179 |  889
     180 |   499 |      100 |    199 | G2     |   180 | 199 |      180 |    499 |   63
     180 |   499 |      200 |    299 | G2     |   200 | 299 |      180 |    499 |  313
     180 |   499 |      300 |    399 | G2     |   300 | 399 |      180 |    499 |  313
     180 |   499 |      400 |    499 | G2     |   400 | 499 |      180 |    499 |  313
     500 |   699 |      500 |    599 | G1     |   500 | 599 |      500 |    699 |  500
     500 |   699 |      600 |    699 | G1     |   600 | 699 |      500 |    699 |  500
于 2013-01-08T13:38:30.103 回答