这是一种“第三好”的解决方案，因为一旦您有了实际时间，它就会跟踪您提前或落后于计划的时间，并将其应用于最近的预定时间，而不需要实际时间：
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--
假设您的时间按照您的建议转换为整数（30s = 1），我玩了一下。这是一个可怕的解决方案，但我认为它符合您的建议。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它们来做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
以下是我从您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简洁（也更有效）。我只精通 C# 和 Perl，但他们中的任何一个都可以做得很好