Google Big Query 中的数据透视表

Question

我需要计算所有对 A、B、C、D 唯一的 ID。所以 "A"/"A" , "B"/"B","C"/"C", "D"/"D" - 应该给我 A,B,C,D 独有的 ID 计数。 而 "A"/"B" 和 "B"/"A"- 将是 ID，其位置为 A 和 B。类似地，"A"/"C" 和 "C"/"A" = 将是 ID，其具有放置为A和C。ID在两个地方的重叠。计数需要随着每次重叠而不断增加请有人建议。我有一张如下表

ID     Place
1       A
2       B
1       C
6       B
4       D
5       A
6       C
7       A
8       A
8       C

你能指导我想出以下输出吗

   A B C D
A  2 0 2 0
B  0 1 1 0
C  2 1 0 0
D  0 0 0 1

Answer 1

以下是 BigQuery 标准 SQL

#standardSQL
WITH self AS (
  SELECT arr[OFFSET(0)] place, COUNT(1) cnt
  FROM (
    SELECT ARRAY_AGG(place) arr, id
    FROM `project.dataset.table`
    GROUP BY id
    HAVING ARRAY_LENGTH(arr) = 1
  )
  GROUP BY place
), pairs AS (
  SELECT id, ARRAY_AGG(place) arr
  FROM `project.dataset.table` 
  GROUP BY id
), flat_matrix AS (
  SELECT place1, place2, COUNT(DISTINCT id) cnt
  FROM pairs, UNNEST(arr) place1, UNNEST(arr) place2
  WHERE place1 <> place2
  GROUP BY 1, 2
  UNION ALL
  SELECT place, place, cnt
  FROM self
)
SELECT place1 place,
  MAX(IF(place2 = 'A', cnt, 0)) AS A,
  MAX(IF(place2 = 'B', cnt, 0)) AS B,
  MAX(IF(place2 = 'C', cnt, 0)) AS C,
  MAX(IF(place2 = 'D', cnt, 0)) AS D 
FROM flat_matrix    

您可以使用您问题中的虚拟数据进行测试，如下所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT 1 id, 'A' place UNION ALL
  SELECT 2, 'B' UNION ALL
  SELECT 1, 'C' UNION ALL
  SELECT 6, 'B' UNION ALL
  SELECT 4, 'D' UNION ALL
  SELECT 5, 'A' UNION ALL
  SELECT 6, 'C' UNION ALL
  SELECT 7, 'A' UNION ALL
  SELECT 8, 'A' UNION ALL
  SELECT 8, 'C'   
), self AS (
  SELECT arr[OFFSET(0)] place, COUNT(1) cnt
  FROM (
    SELECT ARRAY_AGG(place) arr, id
    FROM `project.dataset.table`
    GROUP BY id
    HAVING ARRAY_LENGTH(arr) = 1
  )
  GROUP BY place
), pairs AS (
  SELECT id, ARRAY_AGG(place) arr
  FROM `project.dataset.table` 
  GROUP BY id
), flat_matrix AS (
  SELECT place1, place2, COUNT(DISTINCT id) cnt
  FROM pairs, UNNEST(arr) place1, UNNEST(arr) place2
  WHERE place1 <> place2
  GROUP BY 1, 2
  UNION ALL
  SELECT place, place, cnt
  FROM self
)
SELECT place1 place,
  MAX(IF(place2 = 'A', cnt, 0)) AS A,
  MAX(IF(place2 = 'B', cnt, 0)) AS B,
  MAX(IF(place2 = 'C', cnt, 0)) AS C,
  MAX(IF(place2 = 'D', cnt, 0)) AS D 
FROM flat_matrix
GROUP BY place1
-- ORDER BY place

结果为

Row place   A   B   C   D    
1   A       2   0   2   0    
2   B       0   1   1   0    
3   C       2   1   0   0    
4   D       0   0   0   1    

Answer 2

我想你基本上想要：

with t as (
      select t.*, row_number() over (order by id) as seqnum
      from t
     )
select t.place,
       max(case when t2.place = 'A' then 1 else 0 end) as A,
       max(case when t2.place = 'B' then 1 else 0 end) as B,
       max(case when t2.place = 'C' then 1 else 0 end) as C,
       max(case when t2.place = 'D' then 1 else 0 end) as D
from t join
     t t2
     on t.id = t2.id and t.seqnum <> t2.seqnum
group by t.place
order by t.place;

这不完全是您在问题中的输出，但它似乎在逻辑上处理了重叠。我看不出你是如何将“A”/“A”与 1 联系起来，而将“C”/“C”与 0 联系起来的。