如何根據計算對行進行分組？

Question

我在嘗試對時間戳之間的差異小於60秒的連續行（按時間戳排序）進行分組時遇到問題。

下面是一個示例表：

user video  timestamp     time_diff 
---- -------- ----------------------- --------- 
Bob  balldrop 2017-06-01 06:00:00 UTC  null 
Bob  balldrop 2017-06-01 06:00:10 UTC  -10 
Bob  balldrop 2017-06-01 06:00:30 UTC  -20 
Bob  balldrop 2017-06-01 06:00:45 UTC  -15 
Bob  balldrop 2017-06-01 06:04:00 UTC  -195 
Bob  balldrop 2017-06-01 06:04:30 UTC  -30 
Bob  bounce  2017-06-01 06:05:00 UTC  null 
Bob  bounce  2017-06-01 06:05:20 UTC  -20 

期望的結果：

user video  timestamp     group 
---- -------- ----------------------- --------- 
Bob  balldrop 2017-06-01 06:00:00 UTC  1 
Bob  balldrop 2017-06-01 06:00:10 UTC  1 
Bob  balldrop 2017-06-01 06:00:30 UTC  1 
Bob  balldrop 2017-06-01 06:00:45 UTC  1 
Bob  balldrop 2017-06-01 06:04:00 UTC  2 
Bob  balldrop 2017-06-01 06:04:30 UTC  2 
Bob  bounce  2017-06-01 06:05:00 UTC  3 
Bob  bounce  2017-06-01 06:05:20 UTC  3 

Answer 1

爲BigQuery的標準SQL - 下面玩：

#standardSQL 
WITH data AS (
    SELECT 'Bob' AS user, 'balldrop' AS video, TIMESTAMP '2017-06-01 06:00:00 UTC' AS ts UNION ALL 
    SELECT 'Bob', 'balldrop', TIMESTAMP '2017-06-01 06:00:10 UTC' UNION ALL 
    SELECT 'Bob', 'balldrop', TIMESTAMP '2017-06-01 06:00:30 UTC' UNION ALL 
    SELECT 'Bob', 'balldrop', TIMESTAMP '2017-06-01 06:00:45 UTC' UNION ALL 
    SELECT 'Bob', 'balldrop', TIMESTAMP '2017-06-01 06:04:00 UTC' UNION ALL 
    SELECT 'Bob', 'balldrop', TIMESTAMP '2017-06-01 06:04:30 UTC' UNION ALL 
    SELECT 'Bob', 'bounce', TIMESTAMP '2017-06-01 06:05:00 UTC' UNION ALL 
    SELECT 'Bob', 'bounce', TIMESTAMP '2017-06-01 06:05:20 UTC' 
) 
SELECT 
    user, video, ts, 
    SUM(diff) OVER(PARTITION BY user ORDER BY ts) AS group_number 
FROM (
    SELECT 
    user, video, ts, 
    IF(TIMESTAMP_DIFF(ts, LAG(ts) OVER(PARTITION BY user, video ORDER BY ts), SECOND) < 60, 0, 1) AS diff 
    FROM data 
) 
-- ORDER BY ts 

目前尚不清楚如何想爲不同的用戶編號組，以便PARTITION BY可以是廣告justed取決於它