SQL Query For Total Points Within Radius of a Location - mysql

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!

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))

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.

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

Related

How to get more precision for a computed distance in MySQL, without using geometric types?

I must compute the distance between an object (a city) and each of the several entries from a MySQL table I have (some restaurants). This city and the restaurants are located in a same country.
The computed distance is used in order to show all the restaurants which are close to this city ; the threshold distance is arbitrary. Moreover, this is a ranked list: the closest restaurants are shown first, and the farest are shown at end-of-list. My problem is about this ranking.
What I've done for now
So I made some researches and I succeeded in computing this distance.
$special_select_distance = "DEGREES(ACOS(COS(RADIANS(" . $oneVilles->__get('latitude')[app::getLang()] . ")) * COS(RADIANS(lat)) * COS(RADIANS(lon) - RADIANS(" . $oneVilles->__get('longitude')[app::getLang()] . ")) + SIN(RADIANS(" . $oneVilles->__get('latitude')[app::getLang()] . ")) * SIN(RADIANS(lat))))";
$restaurants = $restaurantsDAO->getAll(null, ['distance DESC'] , null, 'HAVING distance < 1.9' , null , '*, ' . $special_select_distance . " AS distance");
... where:
['distance DESC'] stands for the ranking by distance
'HAVING distance < 1.9' stands for the arbitrary threshold
'*, ' . $special_select_distance . " AS distance" is the selector
$oneVilles->__get('latitude')[app::getLang()] and $oneVilles->__get('longitude')[app::getLang()] are the city's coordinates lat and lon
lat and lon are the restaurant's coordinates (automatically taken into the table we are iterating on, i.e.: restaurants table, since we use the restaurants DAO)
Question
Actual and unexpected result
For each of the restaurants that are quite close between themselves, the computed distance with the city remains the same.
Example: assume that restaurants A and B are quite close. Then, the distance between A and the city is the same than B and the city, it's my actual and unexpected result.
This is not what I want. Indeed, in reality one of these restaurants is closest to the city than the other. I think there isn't enough precision in MySQL.
Expected result
Expected result: to make the restaurants ranking according to the distance to the city working. In other words, to get a more precise computed distance.
Example: assume that restaurants A and B are quite close. Then, the distance between A and the city is shorter than B and the city, it's my expected result.
Examples of computed distances
Between a restaurant and the city (the restaurant being far from the city): 1.933156948976873
Between a restaurant A and the city (A being close to the city): 1.6054631070094885
Between a restaurant B and the city (B being close to A): 1.6054631070094885
Distances in points 2. and 3. are the same and it's not normal. I would want to have more digits, in order to be able to rank my restaurants more efficiently.
Constraints
I wouldn't want to change the configuration of the MySQL Server.
In particular: I absolutely can't use MySQL geometric types (it's a firm's constraint)
The expected solution should simply change the SQL query I wrote and provided to you, in order to be more precise, if it's possible.
Other methods of calculating the distance are allowed, if necessary.
For long distances, use the Haversine formula for accuracy. For short distances, Pythagoras is twice as fast.
16 significant digits (data type DOUBLE) is ludicrous. You don't need to distinguish two different fleas on your dog.
With Pythagoras, be sure to divide the longitude by the cosine of the latitude -- One degree of longitude near Helsinki is half as far as one degree at the equator.
Some more details here: http://mysql.rjweb.org/doc.php/latlng
If 1.6054631070094885 is a latitude diff, then think about it this way: If you and I are at the same longitude, but our latitudes are 1.605463 and 1.605464, then, well, I don't know you well enough to be that close.
It is not practical to compare two floating point values without having a fudge factor:
If abs(a-b) < 0.00001, then treat them as equal.
More
I recommend FLOAT for lat, lng, and distance since you are talking about restaurants. If you are not talking about more than, say, 100 miles or km, then this expression is sufficiently precise:
SQRT( ($lat - lat) *
($lat - lat) +
(($lng - lng) * COS(RADIANS(lat))) *
(($lng - lng) * COS(RADIANS(lat))) ) * $factor
Where...
lat and lng are names of FLOAT columns in the table, in units of degrees.
$lat and $lng are values of the location you are starting from, also in degrees. (PHP uses $; other languages use other conventions.)
$factor is 69.172 for miles or 111.325 for kilometers.
I would not display the result with more than perhaps 1 decimal place. (Don't display "12.345678 miles"; "12.3 miles" is good enough.)
A comparison of Pythagoras and GCD:
Pyt GCD
To Rennes: 93.9407 93.6542
To Vannes: 95.6244 95.6241

SQL - Agg Func Manhattan Distance

SO Link doesn't answer the question. I can't figure out how to solve this query on Hackerspace. None of the solutions online seem to be working. Is this a bug or am I doing something wrong?
Consider P1(a,b) and P2(c,d) to be two points on a 2D plane.
a happens to equal the minimum value in Northern Latitude (LAT_N in STATION).
b happens to equal the minimum value in Western Longitude (LONG_W in STATION).
c happens to equal the maximum value in Northern Latitude (LAT_N in STATION).
d happens to equal the maximum value in Western Longitude (LONG_W in STATION).
Query the Manhattan Distance between points and and round it to a scale of decimal places.
Input Format
The STATION table is described as follows:
STATION Table
ID | Number
City | VarChar2(21)
State | VarChar2(2)
LAT_N | Number
LONG_W | Number
Database: MySQL
Source: https://www.hackerrank.com/challenges/weather-observation-station-18/problem
Link: distance between two longitude and latitude (Tried, but none of the answers provided work.)
SELECT ROUND(ABS(MIN(Station.LAT_N) - MIN(Station.LONG_W)) + ABS(MAX(Station.LAT_N) - MAX(Station.Long_W)), 4)
FROM Station;
The formula for manhattan distance is | a - c| + | b - d| where a and b are min lat and long and c and d are max lat and long respectively.
select
round(
abs(
min(lat_n)- max(lat_n)
) + abs(
min(long_w)- max(long_w)
), 4
)
from
station;
I got 25 hakker points! so can I get 25 points from you?
Without just writing the answer: you need to calculate the horizontal difference between the min and max longitude, and add the vertical difference between the min and max latitude.
Your code does something a bit different. If you update your code accordingly, then the rest is OK and will be marked as correct by hackerrank.
You are comparing latitude and longitude when instead you need to compare latitude with latitude and longitude with longitude. The Manhattan distance between (1,3) and (2,4) is |1-2|+|3-4|, not |1-4|+|2-3|.
It should also be pointed out that since you're taking the min and max of the same range, you don't actually need the absolute value function. round(max(x)-min(x)+max(y)-min(y), 4) works perfectly well - and is slightly faster.
My answer for MS SQL
SELECT CAST(
ABS(MAX(LAT_N) - MIN(LAT_N)) + ABS(MAX(LONG_W) - MIN(LONG_W))
AS DECIMAL(20, 4))
FROM STATION
select round((max(lat_n)-min(lat_n)),4)+round((max(long_w)-min(long_w)),4)
from station;
As we will get result from diff of max and min we don't need abs.
The above code works for Sql Problem
SELECT ROUND(ABS(MAX(Station.LAT_N) - MIN(Station.LONG_W)) + ABS(MIN(Station.LAT_N) - MAX(Station.Long_W)), 4)
FROM Station;enter image description here

distance calculation between two tables of lat/lon

I have the following two tables
cities
id,lat,lon
mountains
id,latitude,longitude
SELECT cities.id,
(SELECT id FROM mountains
WHERE SQRT(POW(69.1 * ( latitude - cities.lat ) , 2 ) +
POW( 69.1 * (cities.lon - longitude ) *
COS( latitude / 57.3 ) , 2 ) )<20 LIMIT 1) as mountain_id
FROM cities
(Query took 0.5060 seconds.)
I've removed some parts of the query (e.g. order by, where) for the complexity's sake. However it doesn't affect the execution time really.
The EXPLAIN below.
id select_type table type possible_keys key key_len ref rows Extra
1 PRIMARY cities ALL NULL NULL NULL NULL 478379
2 DEPENDENT SUBQUERY mountains ALL NULL NULL NULL NULL 15645 Using where
Using the SELECT itself is not my problem but when I try to use the given result... e.g.
id mountain_id
588437 NULL
588993 4269
589014 4201
589021 4213
589036 4952
589052 7625
589113 9235
589125 NULL
589176 1184
589210 4317
...to UPDATE a table everything gets awfully slow. I tried pretty much everything that I know of. I do know that a dependent sub-query isn't optimal but I don't know how to get rid of it.
Is there any way to improve my query. Maybe changing it into a JOIN?
The 2 tables itself have nothing really in common except latitude and longitude which are different and are only brought into relation when using calculations.
Spatial distance search (km,miles) in MariaDB seems not to be available yet.
The trick to making this sort of operation fast is to avoid doing all that computation on every possible pair of lat/lon points. To do that you should incorporate a bounding-box operation.
Let's start by using a JOIN. In pseudocode, you want something like this, but it doesn't matter if you catch a few extra pairs, as long as they are further apart than the others.
SELECT c.city_id, m.mountain_id
FROM cities c
JOIN mountains m ON distance_in_miles(c, m) < 20
So we need to figure out how to make that ON clause fast -- make it use indexes rather than rambling around all the cities and mountains (with apologies to Woody Guthrie).
Let's try this for the ON clause. It searches within square bounding boxes of +/- 20 miles for nearby pairs.
SELECT c.city_id, m.mountain_id
FROM cities c
JOIN mountains m
ON m.lat BETWEEN c.lat - (20.0 / 69.0)
AND c.lat + (20.0 / 69.0)
AND m.lon BETWEEN c.lon - (20.0 / (69.0 * COS(RADIANS(c.lat))))
AND c.lon + (20.0 / (69.0 * COS(RADIANS(c.lat))))
In this query, 20.0 is the comparison limit radius, and 69.0 is the constant defining statute miles per degree of latitude.
Then, put compound indexes on (lat, lon, id) on both tables, and your JOIN operation will be able to use index range scans to make the query more efficient.
Finally, you can augment that query with these sorts of clauses, in pseudocode
ORDER BY dist_in_miles (c,m) ASC
LIMIT 1
Here you actually need to use a distance formula. The cartesian-distance formula in your question is an approximation that works tolerably well unless you're near the pole. You may want to use a great circle formula instead. Those are called spherical cosine law, haversine, or Vincenty formulas.

MySQL Query To Select Closest City

I am trying to repeat the following query for all rows. Basically I am trying to map the closest city (based on the latitude and longitude) to the places latitude and longitude. I have a table places which contains the places that need to be mapped, and a table CityTable with the places to be matched to. I have the following query which works for a single row:
SELECT p.placeID, p.State, p.City, p.County, p.name,
SQRT(POW((69.1 * (p.lat - z.Latitude)), 2 )
+ POW((53 * (p.lng - z.Loungitude)), 2)) AS distance,
p.lat,p.lng,z.Latitude,z.Loungitude,z.City
FROM places p,CityTable z
WHERE p.placeID = 1
ORDER BY distance ASC
LIMIT 1;
This works for a single location. Obviously I would need to remove the WHERE constraints to apply it to the entire table.The problem that I am encountering is that it seems to want to make a copy to compare to every other element in the table. For example, if there are 100 rows in p and 100 rows in z, then the resulting table seems to be 10,000 rows. I need the table to be of size count(*) for p. Any ideas? Also, are there any more efficient ways to do this if my table p contains over a million rows? Thanks.
You can find the nearest city to a place using:
SELECT p.placeID, p.State, p.City, p.County, p.name,
(select z.City
from CityTable z
order by SQRT(POW((69.1 * (p.lat - z.Latitude)), 2 ) + POW((53 * (p.lng - z.Loungitude)), 2))
limit 1
) as City,
p.lat, p.lng
FROM places p
ORDER BY distance ASC;
(If you want additional city information, join the city table back in on City.)
This doesn't solve the problem of having to do the Cartesian product. It does, however, frame it in a different way. If you know that a city is within five degrees longitude/latitude of any place, then you can make the subquery more efficient:
(select z.City
from CityTable z
where z.lat >= p.lat + 5 and z.lat <= p.lat - 5 and
z.long <= p.long + 5 and z.long <= p.lat - 5
order by SQRT(POW((69.1 * (p.lat - z.Latitude)), 2 ) + POW((53 * (p.lng - z.Loungitude)), 2))
limit 1
) as City,
p.lat, p.lng;
This query will use an index on lat. It might even use an index on lat, long.
If this isn't sufficient, then you might consider another way of reducing the search space, by looking only at neighboring states (in the US) or countries.
Finally, you may want to consider the geospatial extensions to MySQL if you are often dealing with this type of data.

Slow SQL Query by Limit/Order dynamic field (coordinates from X point)

I'm trying to make a SQL query on a database of 7 million records, the database "geonames" have the "latitude" and "longitude" in decimal(10.7) indexed both, the problem is that the query is too slow:
SELECT SQL_NO_CACHE DISTINCT
geonameid,
name,
(6367.41 * SQRT(2 * (1-Cos(RADIANS(latitude)) * Cos(0.704231626533) * (Sin(RADIANS(longitude))*Sin(-0.0669560660943) + Cos(RADIANS(longitude)) * Cos(-0.0669560660943)) - Sin(RADIANS(latitude)) * Sin(0.704231626533)))) AS Distance
FROM geoNames
WHERE (6367.41 * SQRT(2 * (1 - Cos(RADIANS(latitude)) * Cos(0.704231626533) * (Sin(RADIANS(longitude)) * Sin(-0.0669560660943) + cos(RADIANS(longitude)) * Cos(-0.0669560660943)) - Sin(RADIANS(latitude)) * Sin(0.704231626533))) <= '10')
ORDER BY Distance
The problem is sort by the "Distance" field, which when created dynamically take long to seep into the condition "WHERE", if I remove the condition of the "WHERE ... <= 10" takes only 0.34 seconds, but the result is 7 million records and to transfer data from MySQL to PHP takes almost 120 seconds.
Can you think of any way to make the query to not lose performance by limiting the Distance field, given that the query will very often change the values?
This kind of query cannot use an index but must compute whether the lat/lon of each row falls within the specified distance. Therefore, it is typical that some form of preprocessing is used to limit the scan to a subset of rows. You could create tables corresponding to distance "bands" (2, 5, 8, 10, 20 miles/km -- whatever makes sense for your application requirements) and then populate these bands and keep them up to date. If you want only those medical providers, say, or hotels, or whatever, within 10 miles of a given location, there's no need to worry about the ones that are hundreds or thousands of miles away. With ad hoc queries you could inner join on the "within 10 miles" band, say, and thereby exclude from the comparison scan all rows where the computed distance > 10. When the location varies, the "elegant" way to handle this is to implement an RTREE, but you can define your encompassing region in any arbitrary way you like if you have access to additional data -- e.g. by using zipcodes or counties or states.
There are two things you can do:
Make sure the datatypes are the same on both sides of a comparison: ie compare with 10 (a number), not '10' (a char type) - it will make less work for the DB
In cases like this, I create a view, which means the calculation to be made just once, even if you refer to it more than once in the query
If these two points are incorporated into you code, you get:
CREATE VIEW geoNamesDistance AS
SELECT SQL_NO_CACHE DISTINCT
geonameid,
name,
(6367.41 * SQRT(2 * (1-Cos(RADIANS(latitude)) * Cos(0.704231626533) * (Sin(RADIANS(longitude))*Sin(-0.0669560660943) + Cos(RADIANS(longitude)) * Cos(-0.0669560660943)) - Sin(RADIANS(latitude)) * Sin(0.704231626533)))) AS Distance
FROM geoNames;
SELECT * FROM geoNamesDistance
WHERE Distance <= 10
ORDER BY Distance;
I came up with:
select * from retailer
where latitude is not null and longitude is not null
and pow(2*(latitude - ?), 2) + pow(longitude - ?, 2) < your_magic_distance_value
With this fast & easy flat-Earth code, Los Angeles is closer to Honolulu than San Fransisco, but i doubt customers will consider that when going that far to shop.