I need some help for a problem that i am struggling to solve.
Example table:
ID |Identifier1 | Identifier2 --------------------------------- 1 | a | c 2 | b | f 3 | a | g 4 | c | h 5 | b | j 6 | d | f 7 | e | k 8 | i | 9 | l | h
I want to group identifiers that are related with each other between two columns and assign a unique group id.
Desired Output:
Identifier | Gr_ID | Gr.Members --------------------------------------------------- a | 1 | (a,c,g,h,l) b | 2 | (b,d,f,j) c | 1 | (a,c,g,h,l) d | 2 | (b,d,f,j) e | 3 | (e,k) f | 2 | (b,d,f,j) g | 1 | (a,c,g,h,l) h | 1 | (a,c,g,h,l) j | 2 | (b,d,f,j) k | 3 | (e,k) l | 1 | (a,c,g,h,l) i | 4 | (i)
Note:the column Gr.Members is not necessary, mostly is used for a clearer view.
So the definition for a group is: A row belongs to a group if it shares at least one identifier with at least one row of this group
But the group id has to be assigned to each identifier(selected by the union of the two columns) not to the row.
Any help on how to build a query to give the desired output?
Thank you.
Update: Below are some extra sample sets with their expected output.
Given table:
Identifier1 | Identifier2
----------------------------
a | f
a | g
a | NULL
b | c
b | a
b | h
b | j
b | NULL
b | NULL
b | g
c | k
c | b
d | l
d | f
d | g
d | m
d | a
d | NULL
d | a
e | c
e | b
e | NULL
Expected output: all the records should belong to the same group with group ID = 1.
Given Table:
Identifier1 | Identifier2 -------------------------- a | a b | b c | a c | b c | c
Expected output: The records should be in the same group with group ID = 1.
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Answer
Here is a variant that doesn’t use cursor, but uses a single recursive query.
Essentially, it treats the data as edges in a graph and traverses recursively all edges of the graph, stopping when the loop is detected. Then it puts all found loops in groups and gives each group a number.
See the detailed explanations of how it works below. I recommend you to run the query CTE-by-CTE and examine each intermediate result to understand what it does.
Sample 1
DECLARE @T TABLE (ID int, Ident1 char(1), Ident2 char(1)); INSERT INTO @T (ID, Ident1, Ident2) VALUES (1, 'a', 'a'), (2, 'b', 'b'), (3, 'c', 'a'), (4, 'c', 'b'), (5, 'c', 'c');
Sample 2
I added one more row with z value to have multiple rows with unpaired values.
DECLARE @T TABLE (ID int, Ident1 char(1), Ident2 char(1)); INSERT INTO @T (ID, Ident1, Ident2) VALUES (1, 'a', 'a'), (1, 'a', 'c'), (2, 'b', 'f'), (3, 'a', 'g'), (4, 'c', 'h'), (5, 'b', 'j'), (6, 'd', 'f'), (7, 'e', 'k'), (8, 'i', NULL), (88, 'z', 'z'), (9, 'l', 'h');
Sample 3
DECLARE @T TABLE (ID int, Ident1 char(1), Ident2 char(1)); INSERT INTO @T (ID, Ident1, Ident2) VALUES (1, 'a', 'f'), (2, 'a', 'g'), (3, 'a', NULL), (4, 'b', 'c'), (5, 'b', 'a'), (6, 'b', 'h'), (7, 'b', 'j'), (8, 'b', NULL), (9, 'b', NULL), (10, 'b', 'g'), (11, 'c', 'k'), (12, 'c', 'b'), (13, 'd', 'l'), (14, 'd', 'f'), (15, 'd', 'g'), (16, 'd', 'm'), (17, 'd', 'a'), (18, 'd', NULL), (19, 'd', 'a'), (20, 'e', 'c'), (21, 'e', 'b'), (22, 'e', NULL);
Query
WITH
CTE_Idents
AS
(
SELECT Ident1 AS Ident
FROM @T
UNION
SELECT Ident2 AS Ident
FROM @T
)
,CTE_Pairs
AS
(
SELECT Ident1, Ident2
FROM @T
WHERE Ident1 <> Ident2
UNION
SELECT Ident2 AS Ident1, Ident1 AS Ident2
FROM @T
WHERE Ident1 <> Ident2
)
,CTE_Recursive
AS
(
SELECT
CAST(CTE_Idents.Ident AS varchar(8000)) AS AnchorIdent
, Ident1
, Ident2
, CAST(',' + Ident1 + ',' + Ident2 + ',' AS varchar(8000)) AS IdentPath
, 1 AS Lvl
FROM
CTE_Pairs
INNER JOIN CTE_Idents ON CTE_Idents.Ident = CTE_Pairs.Ident1
UNION ALL
SELECT
CTE_Recursive.AnchorIdent
, CTE_Pairs.Ident1
, CTE_Pairs.Ident2
, CAST(CTE_Recursive.IdentPath + CTE_Pairs.Ident2 + ',' AS varchar(8000)) AS IdentPath
, CTE_Recursive.Lvl + 1 AS Lvl
FROM
CTE_Pairs
INNER JOIN CTE_Recursive ON CTE_Recursive.Ident2 = CTE_Pairs.Ident1
WHERE
CTE_Recursive.IdentPath NOT LIKE CAST('%,' + CTE_Pairs.Ident2 + ',%' AS varchar(8000))
)
,CTE_RecursionResult
AS
(
SELECT AnchorIdent, Ident1, Ident2
FROM CTE_Recursive
)
,CTE_CleanResult
AS
(
SELECT AnchorIdent, Ident1 AS Ident
FROM CTE_RecursionResult
UNION
SELECT AnchorIdent, Ident2 AS Ident
FROM CTE_RecursionResult
)
SELECT
CTE_Idents.Ident
,CASE WHEN CA_Data.XML_Value IS NULL
THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END AS GroupMembers
,DENSE_RANK() OVER(ORDER BY
CASE WHEN CA_Data.XML_Value IS NULL
THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END
) AS GroupID
FROM
CTE_Idents
CROSS APPLY
(
SELECT CTE_CleanResult.Ident+','
FROM CTE_CleanResult
WHERE CTE_CleanResult.AnchorIdent = CTE_Idents.Ident
ORDER BY CTE_CleanResult.Ident FOR XML PATH(''), TYPE
) AS CA_XML(XML_Value)
CROSS APPLY
(
SELECT CA_XML.XML_Value.value('.', 'NVARCHAR(MAX)')
) AS CA_Data(XML_Value)
WHERE
CTE_Idents.Ident IS NOT NULL
ORDER BY Ident;
Result 1
+-------+--------------+---------+ | Ident | GroupMembers | GroupID | +-------+--------------+---------+ | a | a,b,c, | 1 | | b | a,b,c, | 1 | | c | a,b,c, | 1 | +-------+--------------+---------+
Result 2
+-------+--------------+---------+ | Ident | GroupMembers | GroupID | +-------+--------------+---------+ | a | a,c,g,h,l, | 1 | | b | b,d,f,j, | 2 | | c | a,c,g,h,l, | 1 | | d | b,d,f,j, | 2 | | e | e,k, | 3 | | f | b,d,f,j, | 2 | | g | a,c,g,h,l, | 1 | | h | a,c,g,h,l, | 1 | | i | i | 4 | | j | b,d,f,j, | 2 | | k | e,k, | 3 | | l | a,c,g,h,l, | 1 | | z | z | 5 | +-------+--------------+---------+
Result 3
+-------+--------------------------+---------+ | Ident | GroupMembers | GroupID | +-------+--------------------------+---------+ | a | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | b | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | c | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | d | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | e | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | f | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | g | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | h | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | j | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | k | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | l | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | | m | a,b,c,d,e,f,g,h,j,k,l,m, | 1 | +-------+--------------------------+---------+
How it works
I’ll use the second set of sample data for this explanation.
CTE_Idents
CTE_Idents gives the list of all Identifiers that appear in both Ident1 and Ident2 columns.
Since they can appear in any order we UNION both columns together. UNION also removes any duplicates.
+-------+ | Ident | +-------+ | NULL | | a | | b | | c | | d | | e | | f | | g | | h | | i | | j | | k | | l | | z | +-------+
CTE_Pairs
CTE_Pairs gives the list of all edges of the graph in both directions. Again, UNION is used to remove any duplicates.
+--------+--------+ | Ident1 | Ident2 | +--------+--------+ | a | c | | a | g | | b | f | | b | j | | c | a | | c | h | | d | f | | e | k | | f | b | | f | d | | g | a | | h | c | | h | l | | j | b | | k | e | | l | h | +--------+--------+
CTE_Recursive
CTE_Recursive is the main part of the query that recursively traverses the graph starting from each unique Identifier.
These starting rows are produced by the first part of UNION ALL.
The second part of UNION ALL recursively joins to itself linking Ident2 to Ident1.
Since we pre-made CTE_Pairs with all edges written in both directions, we can always link only Ident2 to Ident1 and we’ll get all paths in the graph.
At the same time the query builds IdentPath – a string of comma-delimited Identifiers that have been traversed so far.
It is used in the WHERE filter:
CTE_Recursive.IdentPath NOT LIKE CAST('%,' + CTE_Pairs.Ident2 + ',%' AS varchar(8000))
As soon as we come across the Identifier that had been included in the Path before, the recursion stops as the list of connected nodes is exhausted.
AnchorIdent is the starting Identifier for the recursion, it will be used later to group results.
Lvl is not really used, I included it for better understanding of what is going on.
+-------------+--------+--------+-------------+-----+ | AnchorIdent | Ident1 | Ident2 | IdentPath | Lvl | +-------------+--------+--------+-------------+-----+ | a | a | c | ,a,c, | 1 | | a | a | g | ,a,g, | 1 | | b | b | f | ,b,f, | 1 | | b | b | j | ,b,j, | 1 | | c | c | a | ,c,a, | 1 | | c | c | h | ,c,h, | 1 | | d | d | f | ,d,f, | 1 | | e | e | k | ,e,k, | 1 | | f | f | b | ,f,b, | 1 | | f | f | d | ,f,d, | 1 | | g | g | a | ,g,a, | 1 | | h | h | c | ,h,c, | 1 | | h | h | l | ,h,l, | 1 | | j | j | b | ,j,b, | 1 | | k | k | e | ,k,e, | 1 | | l | l | h | ,l,h, | 1 | | l | h | c | ,l,h,c, | 2 | | l | c | a | ,l,h,c,a, | 3 | | l | a | g | ,l,h,c,a,g, | 4 | | j | b | f | ,j,b,f, | 2 | | j | f | d | ,j,b,f,d, | 3 | | h | c | a | ,h,c,a, | 2 | | h | a | g | ,h,c,a,g, | 3 | | g | a | c | ,g,a,c, | 2 | | g | c | h | ,g,a,c,h, | 3 | | g | h | l | ,g,a,c,h,l, | 4 | | f | b | j | ,f,b,j, | 2 | | d | f | b | ,d,f,b, | 2 | | d | b | j | ,d,f,b,j, | 3 | | c | h | l | ,c,h,l, | 2 | | c | a | g | ,c,a,g, | 2 | | b | f | d | ,b,f,d, | 2 | | a | c | h | ,a,c,h, | 2 | | a | h | l | ,a,c,h,l, | 3 | +-------------+--------+--------+-------------+-----+
CTE_CleanResult
CTE_CleanResult leaves only relevant parts from CTE_Recursive and again merges both Ident1 and Ident2 using UNION.
+-------------+-------+ | AnchorIdent | Ident | +-------------+-------+ | a | a | | a | c | | a | g | | a | h | | a | l | | b | b | | b | d | | b | f | | b | j | | c | a | | c | c | | c | g | | c | h | | c | l | | d | b | | d | d | | d | f | | d | j | | e | e | | e | k | | f | b | | f | d | | f | f | | f | j | | g | a | | g | c | | g | g | | g | h | | g | l | | h | a | | h | c | | h | g | | h | h | | h | l | | j | b | | j | d | | j | f | | j | j | | k | e | | k | k | | l | a | | l | c | | l | g | | l | h | | l | l | +-------------+-------+
Final SELECT
Now we need to build a string of comma-separated Ident values for each AnchorIdent.
CROSS APPLY with FOR XML does it.
DENSE_RANK() calculates the GroupID numbers for each AnchorIdent.