Thread: Latitude / Longitude
Hi Folks - I'm working on a global weather database for a client and have hit an issue which I'm sure has been solved before. Unfortunately, the site search doesn't pull up anything. Essentially, I've got two tables, one with city/county/lat/long and another with lat/long/weather data. None of the lat/longs in the two tables match up directly, so I can't do a simple join of the two tables. I need to join on closest proximity on the lat/long fields. Any suggestions? It seems to me this will be pretty expensive on CPU resources unless there's a really elegant trick uncovered. Tim -- Founder - PHPBuilder.com / Geocrawler.com / SourceForge GPG Public Key: http://www.perdue.net/personal/pgp.php Perdue, Inc. / Immortal LLC 515-554-9520
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Look at contrib/earthdistance, I **think** it does what you need. LER On Thu, 2002-09-12 at 12:44, Tim Perdue wrote: > Hi Folks - > > I'm working on a global weather database for a client and have hit an issue > which I'm sure has been solved before. Unfortunately, the site search doesn't > pull up anything. > > Essentially, I've got two tables, one with city/county/lat/long and another > with lat/long/weather data. > > None of the lat/longs in the two tables match up directly, so I can't do a > simple join of the two tables. I need to join on closest proximity on the > lat/long fields. > > Any suggestions? It seems to me this will be pretty expensive on CPU resources > unless there's a really elegant trick uncovered. > > Tim > > -- > Founder - PHPBuilder.com / Geocrawler.com / SourceForge > GPG Public Key: http://www.perdue.net/personal/pgp.php > Perdue, Inc. / Immortal LLC > 515-554-9520 -- Larry Rosenman http://www.lerctr.org/~ler Phone: +1 972-414-9812 E-Mail: ler@lerctr.org US Mail: 1905 Steamboat Springs Drive, Garland, TX 75044-6749
On Thu, Sep 12, 2002 at 12:49:21 -0500, Larry Rosenman <ler@lerctr.org> wrote: > Look at contrib/earthdistance, I **think** it does what you need. There isn't an index for the distance operator. You can use boxes to limit the candidates if there is a small bound on how far apart points can be before they are no longer candidates for a match even if they are a nearest neighbor pair.
On Thu, Sep 12, 2002 at 12:49:21PM -0500, Larry Rosenman wrote: > Look at contrib/earthdistance, I **think** it does what you need. Thanks, yeah this starts to get me into the realm of what I need. It still has to run every possible permutation to figure out which one is closest for each postalcode. What I'll do is run that once and build a 3rd table which can be used to join the other two together using a view. Thanks, Tim -- Founder - PHPBuilder.com / Geocrawler.com / SourceForge GPG Public Key: http://www.perdue.net/personal/pgp.php
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Tim Perdue wrote:
> Hi Folks -
>
> I'm working on a global weather database for a client and have hit an issue
> which I'm sure has been solved before. Unfortunately, the site search doesn't
> pull up anything.
>
> Essentially, I've got two tables, one with city/county/lat/long and another
> with lat/long/weather data.
>
> None of the lat/longs in the two tables match up directly, so I can't do a
> simple join of the two tables. I need to join on closest proximity on the
> lat/long fields.
>
> Any suggestions? It seems to me this will be pretty expensive on CPU resources
> unless there's a really elegant trick uncovered.
I see you've gotten some other help, but in case you're interested, I'll give
you another alternative. Here's a plpgsql function I wrote a while ago based
on the Haversine formula:
CREATE FUNCTION "zipdist" (float8,float8,float8,float8 ) RETURNS float8 AS '
DECLARE
lat1 ALIAS FOR $1;
lon1 ALIAS FOR $2;
lat2 ALIAS FOR $3;
lon2 ALIAS FOR $4;
dist float8;
BEGIN
dist := 0.621 * 6371.2 * 2 *
atan2( sqrt(abs(0 + pow(sin(radians(lat2)/2 -
radians(lat1)/2),2) + cos(radians(lat1)) * cos(radians(lat2)) *
pow(sin(radians(lon2)/2 - radians(lon1)/2),2))),sqrt(abs(1 -
pow(sin(radians(lat2)/2 - radians(lat1)/2),2) + cos(radians(lat1)) *
cos(radians(lat2)) * pow(sin(radians(lon2)/2 - radians(lon1)/2),2))));
return dist;
END;
' LANGUAGE 'plpgsql';
I used the following PHP code to start looking for a match in a small circle,
and then expand it if no matches were found:
$dist = INIT_DIST;
$cnt = 0;
$cntr = 0;
do {
if ((! $zip == "") && (! $dist <= 0)) {
$sql = get_zip_sql($lon1d,$lat1d,$dist,$numtoshow);
$rs = connexec($conn,$sql);
$rsf = rsfetchrs($rs);
$dist *= 2;
$cntr++;
} else {
$cntr = 10;
}
} while (count($rsf) < $numadvisorstoshow && $cntr < 10);
Hopefully you get the idea.
As was suggested, you can narrow the results using a box to make the query
perform better, and then sort by distance to get the closest alternative.
Here's the related part of get_zip_sql():
function get_zip_sql($lon1d,$lat1d,$dist,$numtoshow)
{
$sql = "
SELECT DISTINCT <fields>
FROM tbl_a AS a
,tbl_d AS d
,tbl_a_zipcodes AS az
,tbl_zipcodes as z
WHERE
abs(z.lat - $lat1d) * 60 * 1.15078 <= $dist
and abs(z.long - $lon1d) * 60 * 1.15078 <= $dist
and zipdist($lat1d,$lon1d,lat,long) <= $dist
and z.zip = az.zipcode
<other criteria>
ORDER BY
LIMIT $numtoshow;
";
return $sql;
}
The "X * 60 * 1.15078" converts differences in degrees lat/long into rough
distances in miles. The zipdist function returns a more-or-less exact distance
using the Haversine formula.
Hope this helps. Let me know if you want/need more explanation of any of this.
Joe
Tim Perdue wrote: > > On Thu, Sep 12, 2002 at 12:49:21PM -0500, Larry Rosenman wrote: > > Look at contrib/earthdistance, I **think** it does what you need. > > Thanks, yeah this starts to get me into the realm of what I need. It still > has to run every possible permutation to figure out which one is closest for > each postalcode. What I'll do is run that once and build a 3rd table which can > be used to join the other two together using a view. That third table should be maintained by triggers and constraints, no? Jan -- #======================================================================# # It's easier to get forgiveness for being wrong than for being right. # # Let's break this rule - forgive me. # #================================================== JanWieck@Yahoo.com #
Tim Perdue wrote: > Hi Folks - > > I'm working on a global weather database for a client and have hit an issue > which I'm sure has been solved before. Unfortunately, the site search doesn't > pull up anything. > > Essentially, I've got two tables, one with city/county/lat/long and another > with lat/long/weather data. > > None of the lat/longs in the two tables match up directly, so I can't do a > simple join of the two tables. I need to join on closest proximity on the > lat/long fields. > > Any suggestions? It seems to me this will be pretty expensive on CPU resources > unless there's a really elegant trick uncovered. I guess that all depends on how you define expensive on CPU resources. I think the way I would do it is to define an sql function. For example, assume the following table structure: create table a (city text, city_location point, city_weather text); create table b (weather text, weather_location point); Create a function something like: create function closest_weather (point) returns text as 'select b.weather from b order by $1 <-> b.weather_location limit 1;' language sql; Then you could do something like: update a set city_weather = closest_weather(city_location); And if you had the lat/long as seperate numbers, just cast them as a point (i.e., point(long,lat) will return a point type). It's going to do a table scan for each city it updates, but that may or may not be a problem for you. -- Jeff Hoffmann PropertyKey.com
On Thursday 12 September 2002 15:49, Tim Perdue wrote: > On Thu, Sep 12, 2002 at 12:49:21PM -0500, Larry Rosenman wrote: > > Look at contrib/earthdistance, I **think** it does what you need. > > Thanks, yeah this starts to get me into the realm of what I need. It still > has to run every possible permutation to figure out which one is closest > for each postalcode. What I'll do is run that once and build a 3rd table > which can be used to join the other two together using a view. If you are looking for everything within a certain radius of a zip code, what I have done in the past is calculate the highest/lowest longitude and latitude and then select anything that falls between them and only check those ones to see if they fall within the radius. That's saves a lot of time over checking the exact distance for every zip code. - James