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Question

在我的BigQuery資料表的BigQuery資料表蔓延秒稱爲[遊戲]（約5萬行），在那裏我有以下結構行：

user_id game_id game_play_time 
1234567 3444432 2017-05-30 15:26:57 UTC 
1234567 3444432 2017-05-30 15:26:58 UTC 
1234567 3444432 2017-05-30 15:26:59 UTC 
9876544 8586588 2017-05-30 23:26:11 UTC 
4638889 8698798 2017-05-30 15:26:58 UTC 
4638889 8698798 2017-05-30 15:27:58 UTC 

我需要刪除其擁有的行相同的user_id和game_id但後續遊戲之間的時間差等於或小於1秒（保持第一次出現）。

結果應該如下：

user_id game_id game_play_time 
1234567 3444432 2017-05-30 15:26:57 UTC 
9876544 8586588 2017-05-30 23:26:11 UTC 
4638889 8698798 2017-05-30 15:26:58 UTC 
4638889 8698798 2017-05-30 15:27:58 UTC 

Answer 1

下面是BigQuery的標準SQL

#standardSQL 
SELECT 
    user_id, 
    game_id, 
    MIN(game_play_time) AS game_play_time 
FROM (
    SELECT 
    user_id, 
    game_id, 
    game_play_time, 
    SUM(step) OVER(PARTITION BY user_id, game_id ORDER BY game_play_time) AS grp 
    FROM (
    SELECT 
     user_id, 
     game_id, 
     game_play_time, 
     CASE WHEN IFNULL(TIMESTAMP_DIFF(game_play_time, LAG(game_play_time) OVER(PARTITION BY user_id, game_id ORDER BY game_play_time), SECOND), 0) > 1 THEN 1 ELSE 0 END AS step 
    FROM YourTable 
) 
) 
GROUP BY user_id, game_id, grp 
-- ORDER BY user_id, game_id, grp 

你可以測試它下面的虛擬數據（從例如在你的問題+幾行，以使其更通用）

#standardSQL 
WITH YourTable AS(
    SELECT '1234567' AS user_id, '3444432' AS game_id, TIMESTAMP('2017-05-30 12:26:57') game_play_time UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 12:26:57') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 13:26:57') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 13:26:57') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 15:26:57') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 15:26:57') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 15:26:58') UNION ALL 
    SELECT '1234567', '3444432', TIMESTAMP('2017-05-30 15:26:59') UNION ALL 
    SELECT '9876544', '8586588', TIMESTAMP('2017-05-30 23:26:11') UNION ALL 
    SELECT '4638889', '8698798', TIMESTAMP('2017-05-30 15:26:58') UNION ALL 
    SELECT '4638889', '8698798', TIMESTAMP('2017-05-30 15:27:58') 
) 
SELECT 
    user_id, 
    game_id, 
    MIN(game_play_time) AS game_play_time 
FROM (
    SELECT 
    user_id, 
    game_id, 
    game_play_time, 
    SUM(step) OVER(PARTITION BY user_id, game_id ORDER BY game_play_time) AS grp 
    FROM (
    SELECT 
     user_id, 
     game_id, 
     game_play_time, 
     CASE WHEN IFNULL(TIMESTAMP_DIFF(game_play_time, LAG(game_play_time) OVER(PARTITION BY user_id, game_id ORDER BY game_play_time), SECOND), 0) > 1 THEN 1 ELSE 0 END AS step 
    FROM YourTable 
) 
) 
GROUP BY user_id, game_id, grp 
-- ORDER BY user_id, game_id, grp 

Answer 2

它是否適合你？

SELECT 
    user_id, 
    game_id, 
    MIN(game_play_time) game_play_time 
FROM(
    SELECT 
    user_id, 
    game_id, 
    game_play_time, 
    lead_time, 
    (UNIX_SECONDS(lead_time) - UNIX_SECONDS(game_play_time) <= 1) diff 
FROM(
    SELECT 
    user_id, 
    game_id, 
    game_play_time game_play_time, 
    LEAD(game_play_time,1) OVER(PARTITION BY user_id, game_id order by game_play_time) lead_time 
FROM data 
) 
) 
GROUP BY user_id,game_id, diff 
ORDER BY user_id, game_id, game_play_time 

，其中數據是輸入數據，我定義它像這樣：

WITH data AS(
select '1234567' as user_id, '3444432' as game_id, timestamp('2017-05-30 15:26:57') game_play_time union all 
select '1234567' as user_id, '3444432' as game_id, timestamp('2017-05-30 15:26:58') game_play_time union all 
select '1234567' as user_id, '3444432' as game_id, timestamp('2017-05-30 15:26:59') game_play_time union all 
select '9876544' as user_id, '8586588' as game_id, timestamp('2017-05-30 23:26:11') game_play_time union all 
select '4638889' as user_id, '8698798' as game_id, timestamp('2017-05-30 15:26:58') game_play_time union all 
select '4638889' as user_id, '8698798' as game_id, timestamp('2017-05-30 15:27:58') game_play_time 
) 

即使它似乎是在這裏工作，我不知道，如果仍然有一些角落情況下它贏得了」工作。也許數據中的結果可能會顯示一切正常。