SQL Function - Fuzzy Matching with Levenshtein Distance Algorithm - Return Lowest Value Only - sql

Question

Problem: Need SQL function to return the 'lowest' matching value using the Levenshtein algorithm.
Code:
CREATE FUNCTION ufn_levenshtein(#s1 nvarchar(3999), #s2 nvarchar(3999))
RETURNS int
AS
BEGIN
DECLARE #s1_len int, #s2_len int
DECLARE #i int, #j int, #s1_char nchar, #c int, #c_temp int
DECLARE #cv0 varbinary(8000), #cv1 varbinary(8000)
SELECT
#s1_len = LEN(#s1),
#s2_len = LEN(#s2),
#cv1 = 0x0000,
#j = 1, #i = 1, #c = 0
WHILE #j <= #s2_len
SELECT #cv1 = #cv1 + CAST(#j AS binary(2)), #j = #j + 1
WHILE #i <= #s1_len
BEGIN
SELECT
#s1_char = SUBSTRING(#s1, #i, 1),
#c = #i,
#cv0 = CAST(#i AS binary(2)),
#j = 1
WHILE #j <= #s2_len
BEGIN
SET #c = #c + 1
SET #c_temp = CAST(SUBSTRING(#cv1, #j+#j-1, 2) AS int) +
CASE WHEN #s1_char = SUBSTRING(#s2, #j, 1) THEN 0 ELSE 1 END
IF #c > #c_temp SET #c = #c_temp
SET #c_temp = CAST(SUBSTRING(#cv1, #j+#j+1, 2) AS int)+1
IF #c > #c_temp SET #c = #c_temp
SELECT #cv0 = #cv0 + CAST(#c AS binary(2)), #j = #j + 1
END
SELECT #cv1 = #cv0, #i = #i + 1
END
RETURN #c
END
IF OBJECT_ID('tempdb..#ExistingCustomers') IS NOT NULL
DROP TABLE #ExistingCustomers;
CREATE TABLE #ExistingCustomers
(
Customer VARCHAR(255),
ID INT
)
INSERT #ExistingCustomers SELECT 'Ed''s Barbershop', 1002
INSERT #ExistingCustomers SELECT 'GroceryTown', 1003
INSERT #ExistingCustomers SELECT 'Candy Place', 1004
INSERT #ExistingCustomers SELECT 'Handy Man', 1005
IF OBJECT_ID('tempdb..#POTENTIALCUSTOMERS') IS NOT NULL
DROP TABLE #POTENTIALCUSTOMERS;
CREATE TABLE #POTENTIALCUSTOMERS(Customer VARCHAR(255));
INSERT #POTENTIALCUSTOMERS SELECT 'Eds Barbershop'
INSERT #POTENTIALCUSTOMERS SELECT 'Grocery Town'
INSERT #POTENTIALCUSTOMERS SELECT 'Candy Place'
INSERT #POTENTIALCUSTOMERS SELECT 'Handee Man'
INSERT #POTENTIALCUSTOMERS SELECT 'The Apple Farm'
INSERT #POTENTIALCUSTOMERS SELECT 'Ride-a-Long Bikes'
SELECT A.Customer,
b.ID,
b.Customer as cust,
dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) as ValueLev
FROM #POTENTIALCUSTOMERS a
LEFT JOIN #ExistingCustomers b ON dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) < 15;
This returns:
What I would like to return:
Explanation: The results are the 'lowest' values from the Levenshtein algorithm. There are two rows where the Levenshtein scores are the same The Apple Farm and Ride-a-Long Bikes, in which case any of the values is fine, just as long as it is one value.
References:
SQL Fuzzy Join - MSSQL
http://www.kodyaz.com/articles/fuzzy-string-matching-using-levenshtein-distance-sql-server.aspx

Answer 1

You can use CTE to get the result you want if you partition by the potential customer and use the ValueLev to order the results:
;WITH CTE AS
(
SELECT RANK() OVER (PARTITION BY a.Customer ORDER BY dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) ASC) AS RowNbr,
A.Customer,
b.ID,
b.Customer as cust,
dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) as ValueLev
FROM #POTENTIALCUSTOMERS a
LEFT JOIN #ExistingCustomers b ON dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) < 15
)
SELECT Customer,
MIN(ID) AS ID,
MIN(cust) AS cust,
ValueLev
FROM CTE
WHERE CTE.RowNbr = 1
GROUP BY Customer, ValueLev
As you don't mind which result is returned in the case of duplicate ValueLev, use GROUP BY and MIN to scale the results down to one per potential customer.
Output:
Customer ID cust ValueLev
Candy Place 1004 Candy Place 0
Grocery Town 1003 GroceryTown 0
Eds Barbershop 1002 Ed's Barbershop 1
Handee Man 1005 Handy Man 2
The Apple Farm 1004 Candy Place 9
Ride-a-Long Bikes 1003 Candy Place 14