Edit Distance

Mar 18, 2008

Hi,
please, it is possible to know the edit distance used in the fuzzy lookup/grouping.
On this forum I read fuzzy lookup use 4-gram with fix size.
Does exist any document explaining how fuzzy lookup calculate the similarity? In other word, what kind of edit distance, algorithm is used by fuzzy lookup/grouping?
I hope I was enough clear with my poor english.
Thanks All

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Levenshtein Edit Distance Algorithm

Jun 24, 2005

See here www.merriampark.com/ld.htm for information about the algorithm. This page has a link (http://www.merriampark.com/ldtsql.htm) to a T-SQL implementation by Joseph Gama: unfortunately, that function doesn't work. There is a debugged version in the also-referenced package of TSQL functions (http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=502&lngWId=5), but this still has the fundamental problem that it only works on pairs of strings up to 49 characters.


CREATE FUNCTION edit_distance(@s1 nvarchar(3999), @s2 nvarchar(3999))
RETURNS int
AS
BEGIN
DECLARE @s1_len int, @s2_len int, @i int, @j int, @s1_char nchar, @c int, @c_temp int,
@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

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Jan 21, 2004

Hi
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Help will be appreciated

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Mar 1, 2007

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Any comments??

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Mar 28, 2007

I'm trying to run a dyncamic query that returns all records within a specific distance of a certain point. The longitude and latitude of each record is stored in the database. The query is constructed from two dynamic variables $StartLatitude and $StartLongitude with represent the starting point.

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FROM HPN_Painters
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It runs fine until I add the 'HAVING (Distance <= 150)' clause, in which I recieve the error: Invalid column name 'Distance' It seems that Distance cannot be referenced in the HAVING clause.

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Mar 20, 2007

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Mar 27, 2007

I am new to data mining so please excuse my ignorance. Lets assume

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- identified 3 clusters ( a, b, c)

- each record consists of 15 columns

- collecting new records( 15 variables) real time

what i would like to do is plot these new records programmatically as i collect them realtime. I assume this new record will belong to one of these three clusters. I believe we can find the cluster this new record belongs to by ' SELECT Cluster()....' and distance from the center of the cluster by ClusterDistance(). To plot this on a 2-dimentional space i need (x, y).

ClusterDistance() could be Y but what will be X.



thanks.

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Distance Between Two Points Lat/long

Jan 12, 2008

I have a user defined function, I want to determine the distance between the 2 points. I have it working but i'm having a problem getting to print.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++




Code Snippetcreate function dbo.Distance( @lat1 float , @long1 float , @lat2 float , @long2 float)
returns float

as

begin

declare @DegToRad as float
declare @Ans as float
declare @Miles as float

set @DegToRad = 57.29577951
set @Ans = 0
set @Miles = 0

if @lat1 is null or @lat1 = 0 or @long1 is null or @long1 = 0 or @lat2 is
null or @lat2 = 0 or @long2 is null or @long2 = 0

begin

return ( @Miles )

end

set @Ans = SIN(@lat1 / @DegToRad) * SIN(@lat2 / @DegToRad) + COS(@lat1 / @DegToRad ) * COS( @lat2 / @DegToRad ) * COS(ABS(@long2 - @long1 )/@DegToRad)

set @Miles = 3959 * ATAN(SQRT(1 - SQUARE(@Ans)) / @Ans)

set @Miles = CEILING(@Miles)

return ( @Miles )

end

DECLARE @RC float
EXEC Distance '39.943762', '-78.122265', '32.334709', '-96.633546'
PRINT @RC /* in miles */

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Oct 15, 2007

Great Circle distance calculation
Is there any stored procedure or application that implements Great Circle distance calculation 
 

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Oct 15, 2015

DECLARE @Latitude NUMERIC(9, 6), @Longitude NUMERIC(9, 6)

DECLARE @MyLatitude NUMERIC(9, 6), @MyLongitude NUMERIC(9, 6)
Set @Latitude = 42.329596;
Set @Longitude = -83.709286;
Set @MyLatitude = 42.430883;
Set @MyLongitude = -82.923642;

Question: How do we calculate the distance in miles between the 2 points.

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Nov 22, 2013

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Table: enquiry

Primary key: enquiry_number

Co-ordinates data fields: enquiry.enquiry_easting and enquiry.enquiry_northing.

I will need to self-search on the same table to find possible enquiries within 10m distance.

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Jul 20, 2005

I am trying to use the haversine function to find the distance betweentwo points on a sphere, specifically two zip codes in my database. I'mneither horribly familiar with SQL syntax nor math equations :), so Iwas hoping I could get some help. Below is what I'm using and it is,as best as I can figure, the correct formula. It is not however,giving me correct results. Some are close, others don't seem right atall. Any ideas?SET @lat1 = RADIANS(@lat1)SET @log1 = RADIANS(@log1)SET @lat2 = RADIANS(@lat2)SET @log2 = RADIANS(@log2)SET @Dlat = ABS(@lat2 - @lat1)SET @Dlog = ABS(@log2 - @log1)SET @R = 3956 /*Approximate radius of earth in miles*/SET @A = SQUARE(SIN(@Dlat/2)) + COS(@lat1) * COS(@lat2) *SQUARE(SIN(@Dlog/2))SET @C = 2 * ATN2(SQRT(@A), SQRT(1 - @A))/*SET @C = 2 * ASIN(min(SQRT(@A))) Alternative calculation*/SET @distance = @R * @Cthnx,cjrsumner

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Oct 15, 2007


Great Circle distance calculation
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Feb 1, 2007

hi everyone:

the report show two tables two matrixs

how can i control the distance between them

I want to set the same distance between the table and matrix

or (table and table )





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Jan 2, 2008

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Jun 14, 2006

I need to be able to take the latitude and logitude of two locations and compare then to determine the number of miles between each point. It doesn't need to account for elevation, but assumes a flat plane with lat and long.

Does anyone have any algorithms in T-SQL to do this?

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T-SQL (SS2K8) :: How To GROUP BY With Shortest Distance By Account Number

Mar 11, 2014

Given the following example;

declare @CustIfno table (AccountNumber int, StoreID int, Distance decimal(14,10))
insert into @CustIfno values ('1','44','2.145223'),('1','45','4.567834'),
('1','46','8.4325654'),('2','44','7.8754345'),('2','45','1.54654323'),
('2','46','11.5436543'), ('3','44','9.145223'),('3','45','8.567834'),
('3','46','17.4325654'),('4','44','7.8754345'),('4','45','1.54654323'),
('4','46','11.5436543')

How can I show the shortest Distance by AccountID and StoreID. Results would look like this;

AccountNumberStoreID Distance
1 44 2.1452230000
2 45 1.5465432300
3 45 8.5678340000
4 45 1.5465432300

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Great Circle Distance Function - Haversine Formula

Mar 28, 2007

This function computes the great circle distance in Kilometers using the Haversine formula distance calculation.

If you want it in miles, change the average radius of Earth to miles in the function.

create function dbo.F_GREAT_CIRCLE_DISTANCE
(
@Latitude1 float,
@Longitude1 float,
@Latitude2 float,
@Longitude2 float
)
returns float
as
/*
fUNCTION: F_GREAT_CIRCLE_DISTANCE

Computes the Great Circle distance in kilometers
between two points on the Earth using the
Haversine formula distance calculation.

Input Parameters:
@Longitude1 - Longitude in degrees of point 1
@Latitude1 - Latitude in degrees of point 1
@Longitude2 - Longitude in degrees of point 2
@Latitude2 - Latitude in degrees of point 2

*/
begin
declare @radius float

declare @lon1 float
declare @lon2 float
declare @lat1 float
declare @lat2 float

declare @a float
declare @distance float

-- Sets average radius of Earth in Kilometers
set @radius = 6371.0E

-- Convert degrees to radians
set @lon1 = radians( @Longitude1 )
set @lon2 = radians( @Longitude2 )
set @lat1 = radians( @Latitude1 )
set @lat2 = radians( @Latitude2 )

set @a = sqrt(square(sin((@lat2-@lat1)/2.0E)) +
(cos(@lat1) * cos(@lat2) * square(sin((@lon2-@lon1)/2.0E))) )

set @distance =
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return @distance

end


Edit: corrected spelling


CODO ERGO SUM

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Apr 28, 2008

Hi All,
Does anyone have a Stored Procedure that works perfectly to retrieve all zipcodes within a specified zipcode and distance radius - a zipcode and radius is passed and the Store Procedure result shows all zipcodes that falls within that range.

Thanks in advance

Ade

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Apr 29, 2015

I have the two following locations.

They're both towns in Australia , State of Victoria

Fitzroy,-37.798701, 144.978687
Footscray,-37.799736, 144.899734

After running geography::Point(Latitude, Longitude , 4326) on the latitude and longitude provided for each location, my Geography column for each row is populated with the following:

Fitzroy, 0xE6100000010C292499D53BE642C0A7406667511F6240
Footscray, 0xE6100000010C89B7CEBF5DE642C02D23F59ECA1C6240

In my SQL Query, I have the following which works out the distance between both towns. Geo being my Geography column

DECLARE @s geography = 0xE6100000010C292499D53BE642C0A7406667511F6240 -- Fitzroy
DECLARE @t geography = 0xE6100000010C89B7CEBF5DE642C02D23F59ECA1C6240 -- Footscray
SELECT @s.STDistance(@t)

The result I get is

6954.44911927616

I then looked at formatting this as in Australia we go by KM so after some searching I found two solutions one for Miles and the other KM

So I changed Select statement to look like this

select @s.STDistance(@t)/1000 -- format to KM

My result is then

6.95444911927616

When I go to google maps and do a direction request between the locations provided above it says 10.2km (depending on traffic)

Now I'm new to this spatial data within SQL, why would I get a different result from google maps?

Also I would like to round this number so its easier to use within my where statement so I'm using Ceiling as shown here:

SELECT CEILING(@s.STDistance(@t)/1000)

Is ceiling the correct way to go?

Reason I need to round this is because we are allowing the end user to search by radius so if they pass in 50km I will then say

Where CEILING(@s.STDistance(@t)/1000) < 50

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Sample DDL: 2 tables
create table dim_lead
(
date_created datetime,
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[Code] .....

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Edski writes "Help please,
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.........................
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