计算没有间隙的岛屿的大小

Question

我的数据有 3 列； ID、名称、值 ID 是像行数一样的序号我在数据中显示了 3 个不同的名称，但可能有 10 个以上有 2 个 VALUE

我想以 'local' 方式对 NAME 和 VALUE 进行分组，return 这些 'local' 组的大小。这似乎是 Islands and Gaps 场景的一个版本，但没有重复值的 Gaps and Islands。我已经尝试了各种解决 Islands and Gaps 的方法，但到目前为止，没有真正的进展。

我尝试部分解决方案的失败尝试之一（Itzik Ben-Gan 的工作变体）：

WITH A
AS (SELECT ROW_NUMBER() OVER (ORDER BY id) RN,
           NAME,
           VALUE
    FROM #data
    WHERE NAME LIKE '%Joe%'),
T
AS (
SELECT ROW_NUMBER() OVER (ORDER BY RN) - RN AS Grp,
           RN
    FROM A
    WHERE VALUE = 0)
SELECT MIN(T.RN) AS [From],
       MAX(T.RN) AS [To],
       MAX(T.RN) - MIN(T.RN) AS [Length]
FROM T
GROUP BY Grp
ORDER BY MIN(T.RN);

输出如下：

NAME	VALUE	Count
Joe	0	15
Joe	1	3
Joe	0	5
Joe	1	19

Sue	1	23
Sue	0	17
Sue	1	4

Mary	0	16
Mary	1	10
Mary	0	7
Mary	1	19
Mary	0	3

数据是这样的：

ID	NAME	VALUE
1	Joe	0
2	Joe	0
3	Joe	0
4	Joe	0
5	Joe	0
6	Joe	0
7	Joe	0
8	Joe	0
9	Joe	0
10	Joe	0
11	Sue	1
12	Sue	1
13	Sue	1
14	Sue	1
15	Sue	1
16	Sue	1
17	Sue	1
18	Sue	1
19	Sue	1
20	Sue	1
21	Sue	1
22	Sue	1
23	Sue	1
24	Sue	1
25	Sue	1
26	Sue	1
27	Mary	0
28	Mary	0
29	Mary	0
30	Mary	0
31	Mary	0
32	Mary	0
33	Mary	0
34	Mary	0
35	Joe	0
36	Joe	0
37	Joe	0
38	Joe	0
39	Joe	0
40	Joe	1
41	Joe	1
42	Joe	1
43	Joe	0
44	Joe	0
45	Joe	0
46	Joe	0
47	Joe	0
48	Joe	1
49	Joe	1
50	Mary	0
51	Mary	0
52	Mary	0
53	Mary	0
54	Mary	0
55	Mary	0
56	Mary	0
57	Mary	0
58	Mary	1
59	Mary	1
60	Mary	1
61	Mary	1
62	Mary	1
63	Mary	1
64	Mary	1
65	Mary	1
66	Mary	1
67	Mary	1
68	Mary	0
69	Sue	1
70	Sue	1
71	Sue	1
72	Sue	1
73	Sue	1
74	Sue	1
75	Sue	1
76	Sue	0
77	Sue	0
78	Sue	0
79	Sue	0
80	Sue	0
81	Sue	0
82	Sue	0
83	Sue	0
84	Mary	0
85	Mary	0
86	Mary	0
87	Mary	0
88	Mary	0
89	Mary	0
90	Mary	1
91	Mary	1
92	Mary	1
93	Mary	1
94	Mary	1
95	Mary	1
96	Mary	1
97	Mary	1
98	Mary	1
99	Mary	1
100	Mary	1
101	Sue	0
102	Sue	0
103	Sue	0
104	Sue	0
105	Sue	0
106	Sue	0
107	Sue	0
108	Sue	0
109	Sue	0
110	Sue	1
111	Sue	1
112	Sue	1
113	Sue	1
114	Joe	1
115	Joe	1
116	Joe	1
117	Joe	1
118	Joe	1
119	Joe	1
120	Joe	1
121	Joe	1
122	Joe	1
123	Joe	1
124	Mary	1
125	Mary	1
126	Mary	1
127	Mary	1
128	Mary	1
129	Mary	1
130	Mary	1
131	Mary	1
132	Mary	0
133	Mary	0
134	Mary	0
135	Joe	1
136	Joe	1
137	Joe	1
138	Joe	1
139	Joe	1
140	Joe	1
141	Joe	1

Answer 1

DROP TABLE IF EXISTS #data;
CREATE TABLE #data 
(
    ID INT,
    NAME VARCHAR(50),
    VALUE INT
);
INSERT INTO #data values
(1  ,'Joe', 0    ),
(2  ,'Joe', 0    ),
(3  ,'Joe', 0    ),
(4  ,'Joe', 0    ),
(5  ,'Joe', 0    ),
(6  ,'Joe', 0    ),
(7  ,'Joe', 0    ),
(8  ,'Joe', 0    ),
(9  ,'Joe', 0    ),
(10 ,'Joe', 0    ),
(11 ,'Sue', 1    ),
(12 ,'Sue', 1    ),
(13 ,'Sue', 1    ),
(14 ,'Sue', 1    ),
(15 ,'Sue', 1    ),
(16 ,'Sue', 1    ),
(17 ,'Sue', 1    ),
(18 ,'Sue', 1    ),
(19 ,'Sue', 1    ),
(20 ,'Sue', 1    ),
(21 ,'Sue', 1    ),
(22 ,'Sue', 1    ),
(23 ,'Sue', 1    ),
(24 ,'Sue', 1    ),
(25 ,'Sue', 1    ),
(26 ,'Sue', 1    ),
(27 ,'Mary',    0    ),
(28 ,'Mary',    0    ),
(29 ,'Mary',    0    ),
(30 ,'Mary',    0    ),
(31 ,'Mary',    0    ),
(32 ,'Mary',    0    ),
(33 ,'Mary',    0    ),
(34 ,'Mary',    0    ),
(35 ,'Joe', 0    ),
(36 ,'Joe', 0    ),
(37 ,'Joe', 0    ),
(38 ,'Joe', 0    ),
(39 ,'Joe', 0    ),
(40 ,'Joe', 1    ),
(41 ,'Joe', 1    ),
(42 ,'Joe', 1    ),
(43 ,'Joe', 0    ),
(44 ,'Joe', 0    ),
(45 ,'Joe', 0    ),
(46 ,'Joe', 0    ),
(47 ,'Joe', 0    ),
(48 ,'Joe', 1    ),
(49 ,'Joe', 1    ),
(50 ,'Mary',    0    ),
(51 ,'Mary',    0    ),
(52 ,'Mary',    0    ),
(53 ,'Mary',    0    ),
(54 ,'Mary',    0    ),
(55 ,'Mary',    0    ),
(56 ,'Mary',    0    ),
(57 ,'Mary',    0    ),
(58 ,'Mary',    1    ),
(59 ,'Mary',    1    ),
(60 ,'Mary',    1    ),
(61 ,'Mary',    1    ),
(62 ,'Mary',    1    ),
(63 ,'Mary',    1    ),
(64 ,'Mary',    1    ),
(65 ,'Mary',    1    ),
(66 ,'Mary',    1    ),
(67 ,'Mary',    1    ),
(68 ,'Mary',    0    ),
(69 ,'Sue', 1    ),
(70 ,'Sue', 1    ),
(71 ,'Sue', 1    ),
(72 ,'Sue', 1    ),
(73 ,'Sue', 1    ),
(74 ,'Sue', 1    ),
(75 ,'Sue', 1    ),
(76 ,'Sue', 0    ),
(77 ,'Sue', 0    ),
(78 ,'Sue', 0    ),
(79 ,'Sue', 0    ),
(80 ,'Sue', 0    ),
(81 ,'Sue', 0    ),
(82 ,'Sue', 0    ),
(83 ,'Sue', 0    ),
(84 ,'Mary',    0    ),
(85 ,'Mary',    0    ),
(86 ,'Mary',    0    ),
(87 ,'Mary',    0    ),
(88 ,'Mary',    0    ),
(89 ,'Mary',    0    ),
(90 ,'Mary',    1    ),
(91 ,'Mary',    1    ),
(92 ,'Mary',    1    ),
(93 ,'Mary',    1    ),
(94 ,'Mary',    1    ),
(95 ,'Mary',    1    ),
(96 ,'Mary',    1    ),
(97 ,'Mary',    1    ),
(98 ,'Mary',    1    ),
(99 ,'Mary',    1    ),
(100    ,'Mary',    1),
(101    ,'Sue', 0),
(102    ,'Sue', 0),
(103    ,'Sue', 0),
(104    ,'Sue', 0),
(105    ,'Sue', 0),
(106    ,'Sue', 0),
(107    ,'Sue', 0),
(108    ,'Sue', 0),
(109    ,'Sue', 0),
(110    ,'Sue', 1),
(111    ,'Sue', 1),
(112    ,'Sue', 1),
(113    ,'Sue', 1),
(114    ,'Joe', 1),
(115    ,'Joe', 1),
(116    ,'Joe', 1),
(117    ,'Joe', 1),
(118    ,'Joe', 1),
(119    ,'Joe', 1),
(120    ,'Joe', 1),
(121    ,'Joe', 1),
(122    ,'Joe', 1),
(123    ,'Joe', 1),
(124    ,'Mary',    1),
(125    ,'Mary',    1),
(126    ,'Mary',    1),
(127    ,'Mary',    1),
(128    ,'Mary',    1),
(129    ,'Mary',    1),
(130    ,'Mary',    1),
(131    ,'Mary',    1),
(132    ,'Mary',    0),
(133    ,'Mary',    0),
(134    ,'Mary',    0),
(135    ,'Joe', 1),
(136    ,'Joe', 1),
(137    ,'Joe', 1),
(138    ,'Joe', 1),
(139    ,'Joe', 1),
(140    ,'Joe', 1),
(141    ,'Joe', 1);

-- first find the changes
WITH breaks AS 
(
    SELECT ID, NAME, VALUE, 
        IIF(LAG(VALUE, 1) OVER (PARTITION BY Name ORDER BY ID) != VALUE, 1, 0) AS NewGrp 
    FROM #data
), 
-- then count how many breaks we have passed 
grouped AS 
(
    SELECT ID, NAME, VALUE, SUM(NewGrp) OVER (PARTITION BY Name ORDER BY ID) grp
    FROM breaks
)
-- group by that count
SELECT NAME, VALUE, COUNT(*)
FROM grouped
GROUP BY NAME, VALUE, grp
ORDER BY NAME, MIN(ID)

计算没有间隙的岛屿的大小

Count size of islands with no gaps

sql

sql-server

gaps-and-islands

sql-server-2012