sql - 在 SQL Developer/Oracle 10g 中插入缺失值

Question

我有一张包含 stop_id、sched_time 和 act_time 的表格，我希望在实际时间中填写空白有(使用线性插值)基于预定时间(因此保留停止之间的相对时间)。所以我想从这样的事情开始:

  stop_id  |  sched_time  |  act_time  |  actual
------------------------------------------------
  001      |  13:47:00    |  13:45:00  |  TRUE
  002      |  13:50:00    |  null      |  FALSE
  003      |  13:52:00    |  13:53:00  |  TRUE
  004      |  13:59:00    |  null      |  FALSE
  005      |  14:01:00    |  null      |  FALSE
  006      |  14:04:00    |  14:04:00  |  TRUE

像这样:

  stop_id  |  sched_time  |  act_time
-------------------------------------
  001      |  13:47:00    |  13:45:00
  002      |  13:50:00    |  13:49:48
  003      |  13:52:00    |  13:53:00
  004      |  13:59:00    |  13:59:25
  005      |  14:01:00    |  14:01:15
  006      |  14:04:00    |  14:04:00

如果要让插值尊重停止之间的原始时间要求太多，act_time 列上的简单线性插值将是一个很好的起点，因为没有太多停止之间的时间差的可变性。

提前致谢!

注意:第一个 act_time 可以在第一个 sched_time 并且可能有多个连续行没有实际时间。

Answer 1

这是一种“第三好的”解决方案，因为一旦您有了实际时间，它就会跟踪您比计划提前或落后多少，并将其应用于最近的没有实际时间的计划时间:

with q1 as (
  select
    t.stop_id, sched_time, act_time,
    nvl2(act_time, t.sched_time - t.act_time, null) ahead,
    sum (nvl2(act_time, 1, 0)) over
      (partition by 1 order by stop_id) as actual_count
  from schedule t
)
select
  stop_id, sched_time,
  act_time,
  nvl (act_time, sched_time - min (ahead) over
    (partition by actual_count)) as act_time2
from q1

结果与您所追求的不完全匹配，但它可能是您可以构建的东西:

STOP_ID   SCHED_TIME  ACT_TIME  ACT_TIME2
001       13:47       13:45     13:45
002       13:50                 13:48
003       13:52       13:53     13:53
004       13:59                 14:00
005       14:01                 14:02
006       14:04       14:04     14:04

-- 7/24/14 编辑 --

假设您的时间已按照您的建议转换为整数 (30s = 1)，我试了一下。这是一个可怕的解决方案，但我认为它会按照您的建议进行。我不确定它是否比您的程序循环更快。我很好奇是不是。 Oracle 的分析功能很棒，但您可以看到我确实使用了它们来完成我认为您描述的事情:

with q1 as (
  select
    t.stop_id, t.sched_time, t.act_time,
    sum (nvl2(act_time, 1, 0)) over 
        (partition by 1 order by stop_id) as group_id,
    lead (sched_time) over (order by stop_id) as next_sched
  from schedule2 t
), q2 as (
  select
    stop_id, sched_time, act_time, group_id, next_sched,
    next_sched - sched_time as elapsed,
    row_number() over (partition by group_id order by stop_id) as stops,
    min (act_time) over (partition by group_id) as min_time,
    min (sched_time) over (partition by group_id) as min_sched
  from q1
), q3 as (
  select
    stop_id, sched_time, act_time, group_id, stops, min_time,
    min_sched, next_sched,
    sum (elapsed) over (partition by group_id order by stop_id) as elapsed,
    max (stops) over (partition by group_id) as grp_stops,
    lead (min_time, 1) over (order by stop_id) as next_grp_actual,
    lead (min_sched, 1) over (order by stop_id) as next_grp_sched
  from q2
), q4 as (
  select
    stop_id, sched_time, act_time, stops, grp_stops,
    min_time, lag (elapsed, 1, 0) over
      (partition by group_id order by stop_id) as elapsed,
    max (next_grp_sched) over (partition by group_id) - min_sched
        as time_btw_sched,
    max (next_grp_actual) over (partition by group_id) - min_time
        as time_btw_actuals
  from q3
)
select 
  stop_id, sched_time, act_time,
  nvl (act_time, min_time + (elapsed / time_btw_sched) * 
      time_btw_actuals) as act_time2
from q4

这是我从你的样本中得到的结果:

id     sched   actual  actual (calc)
001    1654    1650    1650
002    1660            1659.6
003    1664    1666    1666
004    1678            1678.83333333333
005    1682            1682.5
006    1688    1688    1688

我认为这可以在编程语言包装器中做得更干净(也更有效)。我只精通 C# 和 Perl，但他们都可以做得很好