SQL Query For Total Points Within Radius of a Loca

2019-04-02 11:45发布

问题:

I have a database table of all zipcodes in the US that includes city,state,latitude & longitude for each zipcode. I also have a database table of points that each have a latitude & longitude associated with them. I'd like to be able to use 1 MySQL query to provide me with a list of all unique city/state combinations from the zipcodes table with the total number of points within a given radius of that city/state. I can get the unique city/state list using the following query:

select city,state,latitude,longitude
from zipcodes 
group by city,state order by state,city;

I can get the number of points within a 100 mile radius of a specific city with latitude '$lat' and longitude '$lon' using the following query:

select count(*) 
from points 
where (3959 * acos(cos(radians($lat)) * cos(radians(latitude)) * cos(radians(longitude) - radians($lon)) + sin(radians($lat)) * sin(radians(latitude)))) < 100;

What I haven't been able to do is figure out how to combine these queries in a way that doesn't kill my database. Here is one of my sad attempts:

select city,state,latitude,longitude,
    (select count(*) from points
     where status="A" AND 
          (3959 * acos(cos(radians(zipcodes.latitude)) * cos(radians(latitude)) * cos(radians(longitude) - radians(zipcodes.longitude)) + sin(radians(zipcodes.latitude)) * sin(radians(latitude)))) < 100) as 'points' 
from zipcodes 
group by city,state order by state,city;

The tables currently have the following indexes:

Zipcodes - `zip` (zip)
Zipcodes - `location` (state,city)
Points - `status_length_location` (status,length,longitude,latitude)

When I run explain before the previous MySQL query here is the output:

+----+--------------------+----------+------+------------------------+------------------------+---------+-------+-------+---------------------------------+
| id | select_type        | table    | type | possible_keys          | key                    | key_len | ref   | rows  | Extra                           |
+----+--------------------+----------+------+------------------------+------------------------+---------+-------+-------+---------------------------------+
|  1 | PRIMARY            | zipcodes | ALL  | NULL                   | NULL                   | NULL    | NULL  | 43187 | Using temporary; Using filesort | 
|  2 | DEPENDENT SUBQUERY | points   | ref  | status_length_location | status_length_location | 2       | const | 16473 | Using where; Using index        | 
+----+--------------------+----------+------+------------------------+------------------------+---------+-------+-------+---------------------------------+

I know I could loop through all the zipcodes and calculate the number of matching points within a given radius but the points table will be growing all the time and I'd rather not have stale point totals in the zipcodes database. I'm hoping a MySQL guru out there can show me the error of my ways. Thanks in advance for your help!

回答1:

MySQL Guru or not, the problem is that unless you find a way of filtering out various rows, the distance needs to be calculated between each point and each city...

There are two general approaches that may help the situation

  • make the distance formula simpler
  • filter out unlikely candidates to the 100k radius from a given city

Before going into these two avenue of improvement, you should decide on the level of precision desired with regard to this 100 miles distance, also you should indicate which geographic area is covered by the database (is this just continental USA etc.

The reason for this is that while more precise numerically, the Great Circle formula, is very computationally expensive. Another avenue of performance improvement would be to store "Grid coordinates" of sorts in addtion (or instead of) the Lat/Long coordinates.

Edit:
A few ideas about a simpler (but less precise) formula:
Since we're dealing with relatively small distances, (and I'm guessing between 30 and 48 deg Lat North), we can use the euclidean distance (or better yet the square of the euclidean distance) rather than the more complicated spherical trigonometry formulas.
depending on the level of precision expected, it may even be acceptable to have one single parameter for the linear distance for a full degree of longitude, taking something average over the area considered (say circa 46 statute miles). The formula would then become

  LatDegInMi = 69.0
  LongDegInMi = 46.0
  DistSquared = ((Lat1 - Lat2) * LatDegInMi) ^2 + ((Long1 - Long2) * LongDegInMi) ^2

On the idea of a columns with grid info to filter to limit the number of rows considered for distance calculation.
Each "point" in the system, be it a city, or another point (?delivery locations, store locations... whatever) is assigned two integer coordinate which define the square of say 25 miles * 25 miles where the point lies. The coordinates of any point within 100 miles from the reference point (a given city), will be at most +/- 4 in the x direction and +/- 4 in the y direction. We can then write a query similar to the following

SELECT city, state, latitude, longitude, COUNT(*)
FROM zipcodes Z
JOIN points P 
  ON P.GridX IN (
    SELECT GridX - 4, GridX - 3, GridX - 2, GridX - 1, GridX, GridX +1, GridX + 2 GridX + 3, GridX +4
   FROM zipcode ZX WHERE Z.id = ZX.id)
  AND
   P.GridY IN (
    SELECT GridY - 4, GridY - 3, GridY - 2, GridY - 1, GridY, GridY +1, GridY + 2 GridY + 3, GridY +4
   FROM zipcode ZY WHERE Z.id = ZY.id)
WHERE P.Status = A
   AND ((Z.latitude - P.latitude) * LatDegInMi) ^2 
      + ((Z.longitude - P.longitude) * LongDegInMi) ^2 < (100^2)
GROUP BY city,state,latitude,longitude;

Note that the LongDegInMi could either be hardcoded (same for all locations within continental USA), or come from corresponding record in the zipcodes table. Similarly, LatDegInMi could be hardcoded (little need to make it vary, as unlike the other it is relatively constant).

The reason why this is faster is that for most records in the cartesian product between the zipcodes table and the points table, we do not calculate the distance at all. We eliminate them on the basis of a index value (the GridX and GridY).

This brings us to the question of which SQL indexes to produce. For sure, we may want: - GridX + GridY + Status (on the points table) - GridY + GridX + status (possibly) - City + State + latitude + longitude + GridX + GridY on the zipcodes table

An alternative to the grids is to "bound" the limits of latitude and longitude which we'll consider, based on the the latitude and longitude of the a given city. i.e. the JOIN condition becomes a range rather than an IN :

JOIN points P 
  ON    P.latitude > (Z.Latitude - (100 / LatDegInMi)) 
    AND P.latitude < (Z.Latitude + (100 / LatDegInMi)) 
    AND P.longitude > (Z.longitude - (100 / LongDegInMi)) 
    AND P.longitude < (Z.longitude + (100 / LongDegInMi)) 


回答2:

When I do these type of searches, my needs allow some approximation. So I use the formula you have in your second query to first calculate the "bounds" -- the four lat/long values at the extremes of the allowed radius, then take those bounds and do a simple query to find the matches within them (less than the max lat, long, more than the minimum lat, long). So what I end up with is everything within a square sitting inside the circle defined by the radius.



回答3:

SELECT * FROM tblLocation 
    WHERE 2 > POWER(POWER(Latitude - 40, 2) + POWER(Longitude - -90, 2), .5)

where the 2 > part would be the number of parallels away and 40 and -90 are lat/lon of the test point

Sorry I didn't use your tablenames or structures, I just copied this out of one of my stored procedures I have in one of my databases.

If I wanted to see the number of points in a zip code I suppose I would do something like this:

SELECT 
    ParcelZip, COUNT(LocationID) AS LocCount 
FROM 
    tblLocation 
WHERE 
    2 > POWER(POWER(Latitude - 40, 2) + POWER(Longitude - -90, 2), .5)
GROUP BY 
    ParcelZip

Getting the total count of all locations in the range would look like this:

SELECT 
    COUNT(LocationID) AS LocCount 
FROM 
    tblLocation 
WHERE 
    2 > POWER(POWER(Latitude - 40, 2) + POWER(Longitude - -90, 2), .5)

A cross join may be inefficient here since we are talking about a large quantity of records but this should do the job in a single query:

SELECT 
    ZipCodes.ZipCode, COUNT(PointID) AS LocCount 
FROM
    Points
CROSS JOIN 
    ZipCodes
WHERE 
    2 > POWER(POWER(Points.Latitude - ZipCodes.Latitude, 2) + POWER(Points.Longitude - ZipCodes.Longitude, 2), .5)
GROUP BY 
    ZipCodeTable.ZipCode