使用差距和羣島查找小時連續/ Dates- SQL/BigQuery的

我有一個表中的BigQuery，看起來像這樣：使用差距和羣島查找小時連續/ Dates- SQL/BigQuery的

Caller_Number | month | day| call_time 
--------------|--------|-----|---------- 
1    | 5  | 15 | 12:56:17

我想寫的BigQuery的SQL查詢，讓我來算至少進行一次呼叫的連續小時數（按呼叫者號碼排序）以及發生呼叫的至少連續10個小時的連續時間（按呼叫者號碼分類）。我一直在尋找差距和島嶼上的現有資源，但似乎無法弄清楚如何將它應用到連續的日期和時間。

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2016-03-21 argunaw

「按caller_number排序」使上述問題非常模糊。你應該提供更多的細節。你可能想分享預期結果的例子。沒有「通過caller_number排序」或具有「通過caller_number分區」將使其完全清楚目前存在的故事 –

好吧，抽樣結果是這樣的： Caller_Number |月| Day | Num_Consec_Hours 第二個樣本結果是這樣的： Caller_Number |月| Num_Consec_Days – argunaw

下面是工作示例連續小時
步驟是
1.「提取」從call_time

HOUR(TIMESTAMP(CURRENT_DATE() + ' ' + call_time))

2.Find前一小時組的

LAG([hour]) OVER(PARTITION BY Caller_Number, [month], [day] ORDER BY [hour])

3.Calculate啓動小時連續小時 - 1 - 啓動，0 - 組延續

IFNULL(INTEGER([hour] - prev_hour > 1), 1)

4.Assign組號碼各組

SUM(seq) OVER(PARTITION BY Caller_Number, [month], [day] ORDER BY [hour])

最後，即指否極泰來 - 逐組號和計數通話和小時

希望這給你良好的開端，爲實現連續幾天上類似的邏輯聖地亞哥小時結果的頂部

SELECT Caller_Number, [month], [day], seq_group, 
    EXACT_COUNT_DISTINCT([hour]) AS hours_count, COUNT(1) AS calls_count 
FROM (
    SELECT Caller_Number, [month], [day], [hour], 
    SUM(seq) OVER(PARTITION BY Caller_Number, [month], [day] 
        ORDER BY [hour]) AS seq_group 
    FROM (
    SELECT Caller_Number, [month], [day], [hour], 
     IFNULL(INTEGER([hour] - prev_hour > 1), 1) AS seq 
    FROM (
     SELECT Caller_Number, [month], [day], [hour], 
     LAG([hour]) OVER(PARTITION BY Caller_Number, [month], [day] 
         ORDER BY [hour]) AS prev_hour 
     FROM (
     SELECT Caller_Number, [month], [day], 
      HOUR(TIMESTAMP(CURRENT_DATE() + ' ' + call_time)) AS [hour] 
     FROM YourTable 
    ) 
    ) 
) 
) 
GROUP BY Caller_Number, [month], [day], seq_group

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2016-03-22 17:51:17

使用差距和羣島查找小時連續/ Dates- SQL/BigQuery的

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