计算雪花中的警报洪水

Question

我正在尝试在雪花中进行警报洪水计算。我使用雪花窗口函数创建了以下数据集。因此，如果该值大于或等于 3，则警报泛洪将开始，对于下一个 0 值，它将结束。所以在下面的例子中，警报洪水开始于“9:51”，结束于“9:54”，持续了3分钟。下一次洪水开始于“9:57”，结束于“10:02”，即for 5 minutes.FYI, value at 9:59 is 3, but as a flood is already started,我们不必考虑它。下一次洪水是在10:03但没有0值，所以我们必须考虑边缘值 10:06。 所以洪水的总时间是 3+5+4= 12 分钟。

   DateTime    Value
3/10/2020 9:50  1
3/10/2020 9:51  3
3/10/2020 9:52  1
3/10/2020 9:53  2
3/10/2020 9:54  0
3/10/2020 9:55  0
3/10/2020 9:56  1
3/10/2020 9:57  3
3/10/2020 9:58  2
3/10/2020 9:59  3
3/10/2020 10:00 2
3/10/2020 10:01 2
3/10/2020 10:02 0
3/10/2020 10:03 3
3/10/2020 10:04 1
3/10/2020 10:05 1
3/10/2020 10:06 1

所以，简而言之，我期待低于输出

我在 SQL 下尝试过，但它没有给我正确的输出，它在第二次洪水时间失败（因为在下一个 0 之前再次出现值 3）

select t.*,
       (case when value >= 3
             then datediff(minute,
                           datetime,
                           min(case when value = 0 then datetime end) over (order by datetime desc)
                          )
        end) as diff_minutes
from t;

Answer 1

javascript udf 版本：

select d, v, iff(3<=v and 1=row_number() over (partition by N order by d),
    count(*) over (partition by N), null) trig_duration
from t, lateral flood_count(t.v::float) 
order by d;

其中 flood_count() 定义为：

create or replace function flood_count(V float) 
returns table (N float)
language javascript AS
$${

  initialize: function() { 
    this.n = 0 
    this.flood = false
  },

  processRow: function(row, rowWriter) { 
    if (3<=row.V && !this.flood) {
        this.flood = true
        this.n++
    }
    else if (0==row.V) this.flood=false
    rowWriter.writeRow({ N: this.flood ? this.n : null })  
  },

}$$;

假设这个输入：

create or replace table t as
select to_timestamp(d, 'mm/dd/yyyy hh:mi') d, v 
from values
    ('3/10/2020 9:50',  1),
    ('3/10/2020 9:51',  3),
    ('3/10/2020 9:52',  1),
    ('3/10/2020 9:53',  2),
    ('3/10/2020 9:54',  0),
    ('3/10/2020 9:55',  0),
    ('3/10/2020 9:56',  1),
    ('3/10/2020 9:57',  3),
    ('3/10/2020 9:58',  2),
    ('3/10/2020 9:59',  3),
    ('3/10/2020 10:00', 2),
    ('3/10/2020 10:01', 2),
    ('3/10/2020 10:02', 0),
    ('3/10/2020 10:03', 3),
    ('3/10/2020 10:04', 1),
    ('3/10/2020 10:05', 1),
    ('3/10/2020 10:06', 1)
    t(d,v)
;

Answer 2

WITH data as (
  select time::timestamp as time, value from values
    ('2020-03-10 9:50', 1 ),
    ('2020-03-10 9:51', 3 ),
    ('2020-03-10 9:52', 1 ),
    ('2020-03-10 9:53', 2 ),
    ('2020-03-10 9:54', 0 ),
    ('2020-03-10 9:55', 0 ),
    ('2020-03-10 9:56', 1 ),
    ('2020-03-10 9:57', 3 ),
    ('2020-03-10 9:58', 2 ),
    ('2020-03-10 9:59', 3 ),
    ('2020-03-10 10:00', 2 ),
    ('2020-03-10 10:01', 2 ),
    ('2020-03-10 10:02', 0 ),
    ('2020-03-10 10:03', 3 ),
    ('2020-03-10 10:04', 1 ),
    ('2020-03-10 10:05', 1 ),
    ('2020-03-10 10:06', 1 )
     s( time, value)
) 
select 
    a.time
    ,a.value
    ,min(trig_time)over(partition by reset_time_group order by time) as first_trigger_time
    ,iff(a.time=first_trigger_time, datediff('minute', first_trigger_time, reset_time_group), null) as trig_duration
from (
select d.time
   ,d.value 
   ,iff(d.value>=3,d.time,null) as trig_time
   ,iff(d.value=0,d.time,null) as reset_time
   ,max(time)over(order by time ROWS BETWEEN 1 PRECEDING AND UNBOUNDED FOLLOWING) as max_time
   ,coalesce(lead(reset_time)ignore nulls over(order by d.time), max_time) as lead_reset_time
   ,coalesce(reset_time,lead_reset_time) as reset_time_group
from data as d
) as a
order by time;

这给出了您似乎期望/描述的结果..

TIME                     VALUE  FIRST_TRIGGER_TIME         TRIG_DURATION
2020-03-10 09:50:00.000    1        
2020-03-10 09:51:00.000    3    2020-03-10 09:51:00.000    3
2020-03-10 09:52:00.000    1    2020-03-10 09:51:00.000    
2020-03-10 09:53:00.000    2    2020-03-10 09:51:00.000    
2020-03-10 09:54:00.000    0    2020-03-10 09:51:00.000    
2020-03-10 09:55:00.000    0        
2020-03-10 09:56:00.000    1        
2020-03-10 09:57:00.000    3    2020-03-10 09:57:00.000    5
2020-03-10 09:58:00.000    2    2020-03-10 09:57:00.000    
2020-03-10 09:59:00.000    3    2020-03-10 09:57:00.000    
2020-03-10 10:00:00.000    2    2020-03-10 09:57:00.000    
2020-03-10 10:01:00.000    2    2020-03-10 09:57:00.000    
2020-03-10 10:02:00.000    0    2020-03-10 09:57:00.000    
2020-03-10 10:03:00.000    3    2020-03-10 10:03:00.000    3
2020-03-10 10:04:00.000    1    2020-03-10 10:03:00.000    
2020-03-10 10:05:00.000    1    2020-03-10 10:03:00.000    
2020-03-10 10:06:00.000    1    2020-03-10 10:03:00.000    

所以它的工作原理是我们找到触发时间和重置时间，然后计算出最后一行边缘情况的 max_time。之后我们找到下一个reset_time向前，如果没有就使用max_time，然后选择当前的reset时间或之前的lead_reset_time，对于你在这里做的工作，这一步可以忽略，因为你的数据不能触发和重置同一行。鉴于我们正在对触发行进行数学运算，重置行知道它属于哪个组并不重要。

然后我们进入一个新的选择层，因为我们已经达到了嵌套/相关 SQL 的雪花限制，并在 reset_group 上做一分钟以找到第一个触发时间，然后我们将其与行时间进行比较并做一个日期差异。

附注 date_diff 的数学有点幼稚，'2020-01-01 23:59:59' '2020-01-02 00:00:01' 相隔 2 秒，但那是 1 分钟相隔 1 小时和 1 天，因为该函数将时间戳转换为选定的单位（并截断），然后对这些结果进行区分..

要获得请求中要求的值为 4 的最终批次，请将lead_reset_time 行更改为：

,coalesce(lead(reset_time)ignore nulls over(order by d.time), dateadd('minute', 1, max_time)) as lead_reset_time

将此 max_time 向前移动一分钟，如果您想假设在未来有数据之外，10:06 的现有行状态有效 1 分钟。这不是我会怎么做的......但是你想要的代码......

Answer 3

我对这段代码不是最自豪的，但它确实有效并提供了一个起点。我相信它可以被清理或简化。而且我还没有评估大型表的性能。

我使用的关键见解是，如果您将 date_diff 添加到日期，那么您会发现它们都添加到相同值的情况，这意味着它们都计数到相同的“0”记录。希望这个概念对您有所帮助。

此外，第一个 cte 是在结果结束时获得 4 的一种半骇人听闻的方式。

--Add a fake zero at the end of the table to provide a value for
-- comparing high values that have not been resolved
-- added a flag so this fake value can be removed later
with fakezero as
(
SELECT datetime, value, 1 flag
FROM test

UNION ALL

SELECT dateadd(minute, 1, max(datetime)) datetime, 0 value, 0 flag
FROM test  
)

-- Find date diffs between high values and subsequent low values
,diffs as (
select t.*,
       (case when value >= 3
             then datediff(minute,
                           datetime,
                           min(case when value = 0 then datetime end) over (order by datetime desc)
                          )
        end) as diff_minutes
from fakezero t
)

--Fix cases where two High values are "resolved" by the same low value
--i.e. when adding the date_diff to the datetime results in the same timestamp
-- this means that the prior high value record that still hasn't been "resolved"
select
  datetime
  ,value
  ,case when 
      lag(dateadd(minute, diff_minutes, datetime)) over(partition by value order by datetime)
      = dateadd(minute, diff_minutes, datetime)
    then null 
    else diff_minutes 
  end as diff_minutes
from diffs
where flag = 1
order by datetime;