我需要计算mssql中Value_A
,Value_B
和Value_C
的平均值。
我的问题是我需要的每一个信息都在一行中。
每个价值都有自己的重量:
(sum of values * weight) / (sum weight):
每列都可以为null。如果有值而不是重量,则重量为100,
如果有重量但没有值,则当然不考虑具体值
e.g。
第1栏:
(2*100+1*80)/(100+80)= 2.55 ≈ 2.6
第二栏:
(1*100+2*80)/(100+80)
+------+---------+---------+---------+----------+----------+----------+-----+
| ID | VALUE_A | VALUE_B | VALUE_C | Weight_A | Weight_B | Weight_C | AVG |
+------+---------+---------+---------+----------+----------+----------+-----+
| 1111 | 2 | 1 | null | 100 | 80 | 60 | 2.6 |
+------+---------+---------+---------+----------+----------+----------+-----+
| 2222 | 1 | 2 | null | 100 | 80 | 60 | |
+------+---------+---------+---------+----------+----------+----------+-----+
我得到了这么远,以获得没有权重的AVG值
select ID, VALUE_A, VALUE_B, VALUE_C, Weight_A, Weight_B, Weight_C,
(SELECT AVG(Cast(c as decimal(18,1)))
FROM (VALUES(VALUE_A),
(VALUE_B),
(VALUE_C)) T (c)) AS [Average]
FROM table
第二次尝试是选择值的总和乘以它们的权重,然后将它们除以权重的总和。重量总和缺失。无法弄清楚如何添加它
select *,
(SELECT SUM(Cast(c as decimal(18,1)))
FROM (VALUES(VALUE_A* ISNULL(Weight_A,100)),
(VALUE_B* ISNULL(Weight_B,100)),
(VALUE_C* ISNULL(Weight_C,100))
) T (c)) AS [Average]
FROM table
答案 0 :(得分:2)
这是你在找什么?
SELECT SUM(val * COALESCE(w, 100)) / SUM(w) as weighted_average,
SUM(val * COALESCE(w, 100)) as weighted_sum
FROM table t CROSS APPLY
(VALUES (t.VALUE_A, t.Weight_A),
(t.VALUE_B, t.Weight_B),
(t.VALUE_C, t.Weight_C)
) a(val, w)
WHERE a.val IS NOT NULL;
答案 1 :(得分:0)
这是平均值的计算方法
SELECT *
,CASE
WHEN (W.weight_A + W.Weight_B+ W.Weight_C) = 0
THEN 0
ELSE (ISNULL(VALUE_A, 0 * W.Weight_A)
+ (ISNULL(VALUE_B, 0) * W.Weight_B)
+ (ISNULL(VALUE_C, 0) * W.Weight_C))
/ (W.weight_A + w.Weight_B+ W.Weight_C)
END Average
FROM TABLE t
CROSS APPLY (Select CASE WHEN VALUE_A is null then 0 ELSE ISNULL(Weight_A,100) END [Weight_A]
,CASE WHEN VALUE_B is null then 0 ELSE ISNULL(Weight_B,100) END [Weight_B]
,CASE WHEN VALUE_C is null then 0 ELSE ISNULL(Weight_C,100) END [Weight_C]) W