我有一个BigQuery表,其地址包括纬度/经度和其他BQ表,这些表具有从普查shapefile导入的有效geom定义。对于地址表中的每一行,我正在尝试查找包含它的geom行。
以下查询是我查找的lat / lng个人作品正常吗?
SELECT SLDLST FROM `geographies.tl_2018_sldl_*` sldl WHERE ST_CONTAINS(sldl.geom, ST_GEOGPOINT(-95.221080, 38.974500));
但是当我尝试将抽象抽象为连接时
SELECT
address_id,
SLDLST
FROM `launchpad-239920.address_standardization.temp_delete_geo_match_sample` ssgolden
LEFT JOIN `geographies.tl_2018_sldl_*` sldl ON ST_CONTAINS(sldl.geom, ST_GEOGPOINT(ssgolden.longitude, ssgolden.latitude));
我得到一个错误: “如果没有连接两端的字段相等的条件,则不能使用LEFT OUTER JOIN。”
如何重组我的联接查询,以便能够提取每个地址的匹配地理位置?
答案 0 :(得分:2)
以下是用于BigQuery标准SQL
如果要在输出中保留不匹配的地址-可以在下面使用
#standardSQL
WITH matched_addresses AS (
SELECT
address_id,
SLDLST
FROM `launchpad-239920.address_standardization.temp_delete_geo_match_sample` ssgolden
JOIN `geographies.tl_2018_sldl_X` sldl
ON ST_CONTAINS(sldl.geom, ST_GEOGPOINT(ssgolden.longitude, ssgolden.latitude))
)
SELECT * FROM matched_addresses UNION ALL
SELECT address_id, NULL
FROM `launchpad-239920.address_standardization.temp_delete_geo_match_sample`
WHERE NOT address_id IN (SELECT address_id FROM matched_addresses)
但如果您只对匹配感兴趣,请在一个以下使用
#standardSQL
WITH matched_addresses AS (
SELECT
address_id,
SLDLST
FROM `launchpad-239920.address_standardization.temp_delete_geo_match_sample` ssgolden
JOIN `geographies.tl_2018_sldl_X` sldl
ON ST_CONTAINS(sldl.geom, ST_GEOGPOINT(ssgolden.longitude, ssgolden.latitude))
)
SELECT * FROM matched_addresses
答案 1 :(得分:0)
一种自动处理不匹配地址的解决方案,而无需Mikhail建议的UNION_ALL
(这样可以提高性能):
#standardSQL
WITH addresses AS (
SELECT *, GENERATE_UUID() uuid
FROM `bigquery-public-data.new_york_taxi_trips.tlc_yellow_trips_2015` ssgolden
WHERE DATE(ssgolden.pickup_datetime) = '2015-10-07'
), matched_addresses AS (
SELECT ARRAY_AGG(
IF(
ST_CONTAINS(sldl.zone_geom, SAFE.ST_GEOGPOINT(ssgolden.pickup_longitude, ssgolden.pickup_latitude))
, sldl.zone_name, null)
IGNORE NULLs LIMIT 1)[OFFSET(0)] zone_name
FROM addresses ssgolden
CROSS JOIN `bigquery-public-data.new_york_taxi_trips.taxi_zone_geom` sldl
GROUP BY uuid
)
SELECT zone_name, COUNT(*) c
FROM matched_addresses
GROUP BY 1
ORDER BY c DESC
现在,让我们针对一大组几何图形(74,133个-整个美国以及更多-回应Michael的评论)测试性能:
#standardSQL
WITH addresses AS (
SELECT *, GENERATE_UUID() uuid
FROM `bigquery-public-data.new_york_taxi_trips.tlc_yellow_trips_2015` ssgolden
WHERE DATE(ssgolden.pickup_datetime) = '2015-10-07'
), matched_addresses AS (
SELECT ARRAY_AGG(
IF(
ST_CONTAINS(sldl.tract_geom, SAFE.ST_GEOGPOINT(ssgolden.pickup_longitude, ssgolden.pickup_latitude))
, FORMAT('%s %s', sldl._table_suffix,sldl.lsad_name), null)
IGNORE NULLs LIMIT 1)[OFFSET(0)] zone_name
FROM addresses ssgolden
CROSS JOIN `bigquery-public-data.geo_census_tracts.census_tracts_*` sldl
GROUP BY uuid
)
SELECT zone_name, COUNT(*) c
FROM matched_addresses
GROUP BY 1
ORDER BY c DESC