Sql 按ID BigQuery计算重叠间隔

我想根据ID计算我有多少重叠间隔

WITH table AS (
  SELECT 1001 as id, 1 AS start_time, 10 AS end_time UNION ALL
  SELECT 1001, 2, 5 UNION ALL
  SELECT 1002, 3, 4 UNION ALL
  SELECT 1003, 5, 8 UNION ALL
  SELECT 1003, 6, 8 UNION ALL
  SELECT 1001, 6, 20 
)

在这种情况下，预期结果应为：

2 overlapping for ID=1001
1 overlapping for ID=1003
0 overlapping for ID=1002
TOT OVERLAPPING = 3

每当有重叠（甚至部分重叠）时，我都需要将其计算为重叠

---

如何在BigQuery中实现这一点？

下面是针对BigQuery标准SQL的，并且是简单而直接的自连接、检查和计算重叠

#standardSQL
SELECT a.id, 
  COUNTIF(
    a.start_time BETWEEN b.start_time AND b.end_time
    OR a.end_time BETWEEN b.start_time AND b.end_time
    OR b.start_time BETWEEN a.start_time AND a.end_time
    OR b.end_time BETWEEN a.start_time AND a.end_time
  ) overlaps
FROM `project.dataset.table` a
LEFT JOIN `project.dataset.table` b
ON a.id = b.id AND TO_JSON_STRING(a) < TO_JSON_STRING(b)
GROUP BY id

另一个选项（为了避免使用分析功能而进行自连接）

---

显然，与上一版本相同的结果/输出下面的是BigQuery标准SQL，并且是简单而直接的自连接、检查和计算重叠

#standardSQL
SELECT a.id, 
  COUNTIF(
    a.start_time BETWEEN b.start_time AND b.end_time
    OR a.end_time BETWEEN b.start_time AND b.end_time
    OR b.start_time BETWEEN a.start_time AND a.end_time
    OR b.end_time BETWEEN a.start_time AND a.end_time
  ) overlaps
FROM `project.dataset.table` a
LEFT JOIN `project.dataset.table` b
ON a.id = b.id AND TO_JSON_STRING(a) < TO_JSON_STRING(b)
GROUP BY id

另一个选项（为了避免使用分析功能而进行自连接）

显然，对于与前一版本相同的结果/输出，所有重叠的逻辑比较开始和结束时间：

SELECT t1.id, 
       COUNTIF(t1.end_time > t2.start_time AND t2.start_time < t1.end_time) as num_overlaps
FROM `project.dataset.table` t1 LEFT JOIN
     `project.dataset.table` t2
     ON t1.id = t2.id 
GROUP BY t1.id;

所有重叠的逻辑比较开始和结束时间：

SELECT t1.id, 
       COUNTIF(t1.end_time > t2.start_time AND t2.start_time < t1.end_time) as num_overlaps
FROM `project.dataset.table` t1 LEFT JOIN
     `project.dataset.table` t2
     ON t1.id = t2.id 
GROUP BY t1.id;

---

谢谢我现在正在我的真实数据（约20GB）上测试它，但这需要时间。谢谢米哈伊尔，它工作起来很有魅力！我对

TO_JSON_STRING（a）

有点困惑。这相当于

a.start\u time

，还是我不明白的更多？@David-如果有不同的开始时间，这类似于a.start\u time到\u JSON\u STRING（a）<到\u JSON\u STRING（b）保证不会将两行连接到两行，感谢您的澄清！当然，如果还没有投票，请考虑：O）谢谢！我现在正在我的真实数据（约20GB）上测试它，但这需要时间。谢谢米哈伊尔，它工作起来很有魅力！我对

TO_JSON_STRING（a）

有点困惑。这相当于

a.start\u time

，还是我不明白的更多？@David-如果有不同的开始时间，这类似于a.start\u time到\u JSON\u STRING（a）<到\u JSON\u STRING（b）保证不会将两行连接到两行，感谢您的澄清！当然，如果还没有答案，请考虑投票：O）

with t as (
      select t.*, row_number() over (partition by id order by start_time) as seqnum
      from `project.dataset.table` t
     )
SELECT t1.id, 
       COUNTIF(t1.end_time > t2.start_time AND t2.start_time < t1.end_time) as num_overlaps
FROM t t1 LEFT JOIN
     t t2
     ON t1.id = t2.id AND t1.seqnum < t2.seqnum
GROUP BY t1.id;