如何在大查询中使用 ST_CONTAINS 加入地理列

Question

我有一个 BigQuery 表，其地址包括 Lat/Lng 和其他 BQ 表，其中包含从 Census shapefile 导入的工作几何定义。对于地址表中的每一行，我都在尝试查找包含它的 geom 行。

以下查询是我查找 INDIVIDUAL lat/lng 工作正常：

SELECT SLDLST FROM `geographies.tl_2018_sldl_*` sldl WHERE ST_CONTAINS(sldl.geom, ST_GEOGPOINT(-95.221080, 38.974500));

但是当我尝试抽象成一个像这样的连接时

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));

我收到一个错误： “如果没有连接两边字段相等的条件，就不能使用左外连接。”

如何重组我的联接查询以便能够为每个地址提取匹配的地理位置？

Answer 1

以下是 BigQuery 标准 SQL

如果您想在输出中保留不匹配的地址 - 您可以在下面使用

#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 - 请使用以下一个

#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  

Answer 2

一种自动处理不匹配地址的解决方案，不需要 Mikhail 建议的 UNION_ALL（因此性能可能会更好）：

#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，让我们针对一组更大的几何图形（74,133 个 - 整个美国及更多 - 响应 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