我有一个很大的.csv数据集,包含10e7点,坐标(纬度、经度)代表游客的位置。我有另一个数据集,包含10e3点,坐标代表商店的位置。
我想用某种测地线公式,把最近的商店和每一位游客联系起来。
我想要一些非常快速和高效的东西,我可以在python (例如熊猫)或Google BigQuery上运行。
谁能给我个线索吗?
发布于 2016-11-16 20:18:40
为了补充费利佩的答案:
您可以使用SQL UDF和。
JS UDF有一些SQL UDF不具备的限制。
因此,与Felipe的其余代码一起使用的等效SQL UDF是
CREATE TEMPORARY FUNCTION distance(lat1 FLOAT64, lon1 FLOAT64, lat2 FLOAT64, lon2 FLOAT64)
RETURNS FLOAT64 AS ((
WITH constants AS (
SELECT 0.017453292519943295 AS p
)
SELECT 12742 * ASIN(SQRT(
0.5 - COS((lat2 - lat1) * p)/2 +
COS(lat1 * p) * COS(lat2 * p) *
(1 - COS((lon2 - lon1) * p))/2))
FROM constants
));我尽量保留各自JS的布局,这样您就可以看到它是如何创建的。
发布于 2016-11-16 13:19:06
这是一个快速解决方案,找到最近的NOAA气象站在21,221个城市在DBpedia (v2014)。
#standardSQL
CREATE TEMPORARY FUNCTION distance(lat1 FLOAT64, lon1 FLOAT64, lat2 FLOAT64, lon2 FLOAT64)
RETURNS FLOAT64
LANGUAGE js AS """
var p = 0.017453292519943295; // Math.PI / 180
var c = Math.cos;
var a = 0.5 - c((lat2 - lat1) * p)/2 +
c(lat1 * p) * c(lat2 * p) *
(1 - c((lon2 - lon1) * p))/2;
return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km
""";
SELECT *
FROM (
SELECT city, country_label, distance, name weather_station, country,
RANK() OVER(PARTITION BY city ORDER BY distance DESC) rank
FROM (
SELECT city, a.country_label, distance(a.lat,a.lon,b.lat,b.lon) distance, b.name, b.country
FROM (
SELECT rdf_schema_label city, country_label, country,
CAST(REGEXP_EXTRACT(point, r'(-?\d*\.\d*)') as FLOAT64) lat,
CAST(REGEXP_EXTRACT(point, r' (-?\d*\.\d*)') as FLOAT64) lon
FROM `fh-bigquery.dbpedia2014temp.City`
WHERE point!='NULL'
) a
JOIN (
SELECT name, country, usaf, wban, lat, lon
FROM `bigquery-public-data.noaa_gsod.stations`
WHERE lat != 0.0 AND lon !=0.0
) b
ON CAST(a.lat as INT64)=CAST(b.lat as INT64)
AND CAST(a.lon as INT64)=CAST(b.lon as INT64)
)
)
WHERE rank=1注意事项:

https://stackoverflow.com/questions/40492187
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