查找两个纬度/经度点之间距离的最快方法 [英] Fastest Way to Find Distance Between Two Lat/Long Points

问题描述

我目前在mysql数据库中的位置不足一百万，所有位置都包含经度和纬度信息.

I currently have just under a million locations in a mysql database all with longitude and latitude information.

我正在尝试通过查询查找一个点与许多其他点之间的距离.它的速度不如我希望的那么快，尤其是每秒100次以上的命中率.

I am trying to find the distance between one point and many other points via a query. It's not as fast as I want it to be especially with 100+ hits a second.

是否有更快的查询，或者可能是比mysql更快的系统?我正在使用此查询:

Is there a faster query or possibly a faster system other than mysql for this? I'm using this query:

SELECT 
  name, 
   ( 3959 * acos( cos( radians(42.290763) ) * cos( radians( locations.lat ) ) 
   * cos( radians(locations.lng) - radians(-71.35368)) + sin(radians(42.290763)) 
   * sin( radians(locations.lat)))) AS distance 
FROM locations 
WHERE active = 1 
HAVING distance < 10 
ORDER BY distance;

注意:提供的距离以英里为单位.如果需要公里，请使用6371代替3959.

Note: The provided distance is in Miles. If you need Kilometers, use 6371 instead of 3959.

推荐答案

• 使用MyISAM表中Geometry数据类型的Point值创建点. InnoDB表现在还支持SPATIAL索引.

  • Create your points using Point values of Geometry data types in MyISAM table. As of Mysql 5.7.5, InnoDB tables now also support SPATIAL indices.

    在这些点上创建SPATIAL索引

    使用MBRContains()查找值:

    SELECT  *
    FROM    table
    WHERE   MBRContains(LineFromText(CONCAT(
            '('
            , @lon + 10 / ( 111.1 / cos(RADIANS(@lon)))
            , ' '
            , @lat + 10 / 111.1
            , ','
            , @lon - 10 / ( 111.1 / cos(RADIANS(@lat)))
            , ' '
            , @lat - 10 / 111.1 
            , ')' )
            ,mypoint)
    

  ，或者在MySQL 5.1及更高版本中:

  , or, in MySQL 5.1 and above:

      SELECT  *
      FROM    table
      WHERE   MBRContains
                      (
                      LineString
                              (
                              Point (
                                      @lon + 10 / ( 111.1 / COS(RADIANS(@lat))),
                                      @lat + 10 / 111.1
                                    ),
                              Point (
                                      @lon - 10 / ( 111.1 / COS(RADIANS(@lat))),
                                      @lat - 10 / 111.1
                                    ) 
                              ),
                      mypoint
                      )
  

  这将大约在(@lat +/- 10 km, @lon +/- 10km)框中选择所有点.

  This will select all points approximately within the box (@lat +/- 10 km, @lon +/- 10km).

  这实际上不是一个盒子，而是一个球形矩形:球体的经度和纬度绑定段.这可能与 Franz Joseph Land 上的普通矩形不同，但在大多数人居住的地方都非常接近.

  This actually is not a box, but a spherical rectangle: latitude and longitude bound segment of the sphere. This may differ from a plain rectangle on the Franz Joseph Land, but quite close to it on most inhabited places.

  • 应用其他过滤条件以选择圆内的所有内容(而不是正方形)

  • Apply additional filtering to select everything inside the circle (not the square)

  可能会应用其他精细过滤来考虑大圆距(对于大距离)

  Possibly apply additional fine filtering to account for the big circle distance (for large distances)

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