视图需要相互依赖的逻辑:没有模型可能吗? [英] View Requires Interdependence Logic: Possible without MODEL?
问题描述
我正在尝试编写一些 Oracle 11g SQL，但遇到了一些先有鸡还是先有蛋的问题.我正在寻找类似电子表格的行为.我找到了一个确实使用 Oracle 的 MODEL 子句的解决方案，但性能并不好.所以我想知道非MODEL"解决方案在技术上是否可行.
I'm trying to write some Oracle 11g SQL, but I'm running into a bit of a chicken-and-egg problem. I'm looking for spreadsheet-like behavior. I've found a solution that does use Oracle's MODEL clause, but the performance isn't great. So I'm wondering if a "non-MODEL" solution is even technically feasible.
这是一个演示我正在尝试做的事情的玩具示例.鉴于此表:
Here's a toy example that demonstrates what I'm trying to do. Given this table:
CREATE TABLE t (id NUMBER PRIMARY KEY, n NUMBER);
INSERT INTO t (id, n) VALUES (2, 0);
INSERT INTO t (id, n) VALUES (3, 1);
INSERT INTO t (id, n) VALUES (5, 1);
INSERT INTO t (id, n) VALUES (7, 2);
INSERT INTO t (id, n) VALUES (11, 3);
INSERT INTO t (id, n) VALUES (13, 5);
INSERT INTO t (id, n) VALUES (17, 8);
INSERT INTO t (id, n) VALUES (19, 13);
我想计算两个额外的派生列，将它们称为 X 和 Y.
I want to compute two additional, derived columns, call them X and Y.
以下是如何计算 X 和 Y 的规则:
Here are the rules for how X and Y are to be computed:
X:对于第一行，由 ID 的最小值定义，将 X 设置为 N.对于所有后续行，X 的值应比前一个 Y 的值小 1，按 ID 排序.
X: For the very first row, as defined by the minimum value of ID, set
XtoN. For all subsequent rows, the value ofXshould be one less than the value of the previousY, as sorted byID.
你:两次 N 加上 X.
接下来的几个步骤展示了如果我要手动执行此操作，我将如何填写所需的视图.首先，给定数据的前几行:
These next few steps show how I'd fill out my desired view if I were to do this by hand. First, the first few rows of the given data:
ID N X Y
--- --- --- ---
2 0
3 1
5 1
7 2
....
由于我们在第一行，X 应该设置为 N 或 0.Y 应该是 2 * N + X，或者 0.
Since we're in the first row, X should be set to N, or 0. Y should be 2 * N + X, or 0.
ID N X Y
--- --- --- ---
2 0 0 0
3 1
5 1
7 2
....
现在，由于我们不再位于第一行，从现在开始，X 应该总是比前一行的 Y 小一个.在第二行，这意味着 X = (previous Y) - 1 = 0 - 1 = <代码>-1.第二行的 Y 将是 2 * N + X，或 2 * (1) + (-1) = 1.
Now, since we're not in the first row any longer, X should always be one less than the previous row's Y from here on out. Here in the second row, that means X = (previous Y) - 1 = 0 - 1 = -1. And the second row's Y will be 2 * N + X, or 2 * (1) + (-1) = 1.
ID N X Y
--- --- --- ---
2 0 0 0
3 1 -1 1
5 1
7 2
....
如果你继续数学，这里是你想要的结果:
If you continue with the math, here's the desired outcome:
ID N X Y
--- --- --- ---
2 0 0 0
3 1 -1 1
5 1 0 2
7 2 1 5
11 3 4 10
13 5 9 19
17 8 18 34
19 13 33 59
给定如何计算 X 和 Y 的规则，是否可以在不必求助于 MODEL 子句的情况下获得此结果?
Given the rules for how X and Y are computed, is it possible to get this outcome without having to resort to the MODEL clause?
我不是在寻找基于此特定示例的数学简化；这只是我想出的一个玩具示例，它展示了我在实际问题中面临的那种相互依赖.
I'm not looking for a mathematical simplification based on this particular example; this is just a toy example I came up with that demonstrates the kind of interdependence I'm facing in my actual problem.
P.S.:这是一个 MODEL 示例，我可以拼凑起来生成此输出；也许有可能进行修改以提高性能?
P.S.: Here's a MODEL example I was able to cobble together that does generate this output; maybe there are modifications possible to improve performance?
SQL> WITH u AS (
2 SELECT ROW_NUMBER() OVER (ORDER BY t.id) r
3 , t.id
4 , t.n
5 FROM t
6 )
7 SELECT r
8 , id
9 , n
10 , x
11 , y
12 FROM u
13 MODEL
14 DIMENSION BY (r)
15 MEASURES (id
16 , n
17 , CAST(NULL AS NUMBER) x
18 , CAST(NULL AS NUMBER) y) RULES AUTOMATIC ORDER
19 ( x[1] = n[cv()]
20 , y[r] = 2 * n[cv()] + x[cv()]
21 , x[r > 1] ORDER BY r = y[cv() - 1] - 1
22 )
23 ;
R ID N X Y
---------- ---------- ---------- ---------- ----------
1 2 0 0 0
2 3 1 -1 1
3 5 1 0 2
4 7 2 1 5
5 11 3 4 10
6 13 5 9 19
7 17 8 18 34
8 19 13 33 59
8 rows selected.
SQL>
谢谢.
推荐答案
您可以使用 递归子查询分解(也称为递归 CTE):
You could use recursive subquery factoring (also known as a recursive CTE):
with tmp as (
select t.*,
row_number() over (order by t.id) as rn
from t
),
r (id, n, x, y, rn) as (
select id, n, 0, 0, rn
from tmp
where rn = 1
union all
select tmp.id, tmp.n, r.y - 1, (tmp.n * 2) + r.y - 1, tmp.rn
from r
join tmp on tmp.rn = r.rn + 1
)
select id, n, x, y
from r
order by rn;
ID N X Y
---------- ---------- ---------- ----------
2 0 0 0
3 1 -1 1
5 1 0 2
7 2 1 5
11 3 4 10
13 5 9 19
17 8 18 34
19 13 33 59
它基本上是通过您的手动步骤.锚成员是您的第一个手动步骤，将第一行的 x 和 y 都设置为零.然后递归成员执行您指定的计算.(在计算该行的 y 时，您不能引用新计算的 x 值，因此您必须将其重复为 (tmp.n * 2)+ ry - 1).rn 只是按照 ID 保持行的顺序，同时更容易找到下一行 - 所以你可以寻找 rn + 1 而不是找到下一个最高的 ID直接取值.
It's basically walking through your manual steps. The anchor member is your first manual step, setting x and y both to zero for the first row. The recursive member then does the calculation you specified. (You can't refer to the new-calculated x value when calculating that row's y, so you have to repeat that as (tmp.n * 2) + r.y - 1). The rn is just to keep the rows orders by ID while making it easier to find the next row - so you can look for rn + 1 instead of find the next highest ID value directly.
您的示例数据没有显着的性能差异，但是添加了一千行后，模型子句大约需要 5 秒，递归 CTE 大约需要 1 秒；另一千行模型需要约 20 秒，CTE 需要约 3 秒；另一千行模型需要约 40 秒，CTE 需要约 6 秒；再有一千行(总共 4,008 行)模型用了大约 75 秒，CTE 用了大约 10 秒.(我厌倦了等待具有更多行的模型版本；五分钟后用 10,000 杀死了它).我真的不能说这对您的真实数据的表现如何，但在此基础上，它可能值得一试.
There's no significant performance difference with your sample data, but with a thousand rows added the model clause takes about 5 seconds and the recursive CTE takes about 1 second; with another thousand rows model takes ~20 seconds and the CTE takes ~3 seconds; with another thousand rows model took ~40 seconds and CTE took ~6 seconds; and with another thousand rows (so 4,008 in total) model took ~75 seconds and CTE took ~10 seconds. (I got bored waiting for the model version with more rows than that; killed it after a five minutes with 10,000). I can't really say how this will perform with your real data, but on that basis, it's probably worth trying.
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