问题描述
我一直在研究在给定坐标的特定邻近范围内选择最接近点的性能.
I was looking into performance of selecting closest points within certain proximity to given coordinate.
选项是使用两个decimal(8,6)-lat，长列或单个geography列并进行处理.
Options are to ether use two decimal(8,6) - lat, long columns or single geography column and work with that.
我只对哪个更快感兴趣?
推荐答案
TL; DR 地理要快10倍.
好，所以我已经设置了测试:
Ok so I have set up test:
表中的一个与id,lat,long (int, decimal(8,6),decimal(8,6))耦合，另一个与id,coord (int, geography)耦合.
Couple of tables one with id,lat,long (int, decimal(8,6),decimal(8,6)) other with id,coord (int, geography).
然后插入47k的随机数据.
Then insert 47k of random data.
对于索引第一张表，我在lat，long上使用非聚集升序索引，填充因子为95. 第二个GRIDS =(LEVEL_1 = LOW,LEVEL_2 = MEDIUM,LEVEL_3 = LOW,LEVEL_4 = LOW的填充因子为95.
For indexing first table I used nonclustered Ascending index on lat,long with fill factor of 95. for second one GRIDS =(LEVEL_1 = LOW,LEVEL_2 = MEDIUM,LEVEL_3 = LOW,LEVEL_4 = LOW with fill factor 95.
CREATE TABLE dbo.Temp
(
Id int NOT NULL IDENTITY (1, 1),
Coord geography NOT NULL
) ON [PRIMARY]
TEXTIMAGE_ON [PRIMARY]
GO
ALTER TABLE dbo.Temp ADD CONSTRAINT
PK_Temp PRIMARY KEY CLUSTERED
(
Id
) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
declare @i int =0
declare @lat decimal(8,6) =0.0
declare @long decimal(8,6) =0.0
while (@i < 47000)
begin
set @lat =(select (0.9 -Rand()*1.8)*100)
set @long =(select (0.9 -Rand()*1.8)*100)
insert into Temp
select geography::Point(@lat, @long,4326)
set @i =@i+1
end
go
CREATE SPATIAL INDEX [SpatialIndex_1] ON [dbo].Temp
(
[coord]
)USING GEOGRAPHY_GRID
WITH (GRIDS =(LEVEL_1 = LOW,LEVEL_2 = MEDIUM,LEVEL_3 = LOW,LEVEL_4 = LOW),
CELLS_PER_OBJECT = 16, PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS = OFF, ALLOW_PAGE_LOCKS = ON, FILLFACTOR = 95) ON [PRIMARY]
GO
CREATE TABLE [dbo].[Temp2](
[Id] [int] IDENTITY(1,1) NOT NULL,
[Lat] [decimal](8, 6) NOT NULL,
[Long] [decimal](8, 6) NOT NULL,
CONSTRAINT [PK_Temp2] PRIMARY KEY CLUSTERED
(
[Id] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
declare @i int = 0
declare @lat decimal(8,6) = 0
declare @long decimal(8,6) = 0
while (@i < 47000)
begin
set @lat = (select (0.9 - (RAND()*1.8))*100)
set @long = (select (0.9 - (RAND()*1.8))*100)
insert into Temp2
select @lat , @long
set @i = @i +1
end
go
CREATE NONCLUSTERED INDEX [Coord_IX] ON [dbo].[Temp2]
(
[Lat] ASC,
[Long] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON, FILLFACTOR = 95) ON [PRIMARY]
GO
然后我进行了几次测试:
Then I ran couple of tests:
第一个是Lat，Long.
declare @lat decimal(8,6) = 0.0,
@lon decimal(8,6) = 0.0,
@i int = 0,
@start datetime = getdate()
while(@i < 100)
begin
set @lat = (select (0.9 - Rand()*1.8)*100)
set @lon = (select (0.9 - (RAND()*1.8))*100.0)
DECLARE @lat_s FLOAT = SIN(@lat * PI() / 180),
@lat_c FLOAT = COS(@lat * PI() / 180)
SELECT DISTINCT top 1000 @lat, @lon, *
FROM (
SELECT
lat,
long,
((ACOS(@lat_s * SIN(lat * PI() / 180) + @lat_c * COS(lat * PI() / 180) * COS((@lon - long) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS dist
FROM dbo.Temp2
) t
ORDER BY dist
set @i= @i+1
end
print CONVERT(varchar,(getdate()-@start),108)
go
第二个是针对地理的.
DECLARE @g geography;
declare @point nvarchar(50) =''
declare @i int =0,
@lat decimal(8,6) =0.0,
@long decimal(8,6) =0.0,
@start datetime = getdate()
while (@i < 100)
begin
set @lat =(select (0.9 -Rand()*1.8)*100)
set @long =(select (0.9 -Rand()*1.8)*100)
set @point = (select 'POINT('+CONVERT(varchar(10), @lat)+ ' ' +CONVERT(varchar(10), @long)+')')
SET @g = geography::STGeomFromText(@point, 4326);
SELECT TOP 1000
@lat,
@long,
@g.STDistance(st.[coord]) AS [DistanceFromPoint (in meters)]
, st.[coord]
, st.id
FROM Temp st
ORDER BY @g.STDistance(st.[coord]) ASC
set @i =@i+1
end
print CONVERT(varchar,(getdate()-@start),108)
go
结果:
- 长，长-00:00:10
- 地理-00:02:21
对于那些想知道为什么地理表现如此差的人这是执行计划-请注意，由于行大小为4047字节(十进制为25字节)，因此它不使用空间索引，并且需要花费一定的时间进行排序.尝试来强制索引导致运行时错误
For those who are wondering why geography has such poor performancehere's execution plan - notice that it does not use spatial index, and takes ages to sort since row size is 4047 Bytes (against 25 Bytes in decimal). Trying to force index results in runtime error
PS 我也为平面做了一个，但是与球形的差别很小，只有0.5s(在9.5-10.0秒后回来，这似乎稍快一点)仍然可以满足所有要求在一个地方的脚本中:
P.S I also did one for flat surface but difference from spherical one is very small ~0.5s (comes back in 9.5-10.0 seconds which does seem to be slightly faster) still to have it all in one place here's script:
print 'flat'
declare @lat decimal(8,6) = 0.0,
@lon decimal(8,6) = 0.0,
@i int = 0,
@start datetime = getdate()
while(@i < 100)
begin
set @lat = (select (0.9 - Rand()*1.8)*100)
set @lon = (select (0.9 - (RAND()*1.8))*100.0)
SELECT DISTINCT top 1000 @lat, @lon, *
FROM (
SELECT
lat,
long,
sqrt(power((@lat - lat),2) + (power((@lon - long),2))) AS dist
FROM dbo.Temp2
) t
ORDER BY dist
set @i= @i+1
end
print CONVERT(varchar,(getdate()-@start),108)
go
更新:
切换到SQL 2014并强制使用具有1000万条记录的索引后:
After switching to SQL 2014 and forcing to use the index with 10M records:
- Lat，Long采取了00:00:22.935
- 单位花了00:00:22.988
- 地理占据了00:00:02.427
使用的地理脚本:
DECLARE @g geography;
declare @point nvarchar(50) =''
declare @i int =0,
@lat decimal(8,6) =0.0,
@long decimal(8,6) =0.0,
@start datetime = getdate()
set @lat =(select (0.9 -Rand()*1.8)*100)
set @long =(select (0.9 -Rand()*1.8)*100)
set @point = (select 'POINT('+CONVERT(varchar(10), @lat)+ ' '
+CONVERT(varchar(10), @long)+')')
SET @g = geography::STGeomFromText(@point, 4326);
SELECT TOP 1000
@lat,
@long,
@g.STDistance(st.[coord]) AS [DistanceFromPoint (in meters)]
, st.[coord]
, st.id
FROM Temp st with(index([SpatialIndex_1]))
WHERE @g.STDistance(st.[coord]) IS NOT NULL
ORDER BY @g.STDistance(st.[coord]) asc
这篇关于SQL几何VS十进制(8,6)纬度，长期性能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！