beg*_*ner 5 sql missing-data google-bigquery
我正在尝试创建一个 SQL,以便我可以在 Google Data Studio 中通过 BigQuery 的连接制作一个时间序列图表。你可以在下面看到我的 SQL。
WITH
CTE_1 AS
(SELECT ID, Date, Min_Predict, Max_Predict, Interval
,ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Date) AS row_num
FROM
table),
CTE_2 AS
(SELECT Date, Min_Predict, Max_Predict,
SUM(Min_Predict) OVER (ORDER BY Date) AS Min,
SUM(Max_Predict) OVER (ORDER BY Date) AS Max
FROM CTE_1
WHERE
row_num = 1 AND Interval = 'A')
SELECT Date, Min, Max
From CTE_2
GROUP BY Date, Min, Max
ORDER BY Date
Run Code Online (Sandbox Code Playgroud)
结果我得到了这张表。
Row ProgressDate EstMin EstMax
1 2017-07-21T00:00:00Z 0.125 0.25
2 2017-07-24T00:00:00Z 5.125 5.375
3 2017-07-25T00:00:00Z 8.75 10.25
4 2017-07-26T00:00:00Z 10.0 12.0
5 2017-07-27T00:00:00Z 10.5 12.75
6 2017-08-01T00:00:00Z 15.25 19.125
7 2017-08-02T00:00:00Z 15.5 19.375
8 2017-08-05T00:00:00Z 16.25 20.625
Run Code Online (Sandbox Code Playgroud)
如您所见,我缺少日期,例如在 21.07 和 24.07 之间。如何用前一天的数据填充那些缺失的日期?因为在数据工作室中,我在那些日子里丢失了数据,我可以将它们等于 0 但我不想要这个。
Mik*_*ant 13
以下是 BigQuery 标准 SQL 并基于您当前的结果
#standardSQL
WITH your_current_result AS (
......
), days AS (
SELECT day
FROM (
SELECT
MIN(DATE(TIMESTAMP(ProgressDate))) min_dt,
MAX(DATE(TIMESTAMP(ProgressDate))) max_dt
FROM your_current_result
), UNNEST(GENERATE_DATE_ARRAY(min_dt, max_dt)) day
)
SELECT day,
LAST_VALUE(EstMin IGNORE NULLS) OVER(ORDER BY day) EstMin,
LAST_VALUE(EstMax IGNORE NULLS) OVER(ORDER BY day) EstMax
FROM days
LEFT JOIN your_current_result
ON day = DATE(TIMESTAMP(ProgressDate))
-- ORDER BY day
Run Code Online (Sandbox Code Playgroud)
您可以使用问题中的输出示例来测试,玩上面
#standardSQL
WITH your_current_result AS (
SELECT '2017-07-21T00:00:00Z' ProgressDate, 0.125 EstMin, 0.25 EstMax UNION ALL
SELECT '2017-07-24T00:00:00Z', 5.125, 5.375 UNION ALL
SELECT '2017-07-25T00:00:00Z', 8.75, 10.25 UNION ALL
SELECT '2017-07-26T00:00:00Z', 10.0, 12.0 UNION ALL
SELECT '2017-07-27T00:00:00Z', 10.5, 12.75 UNION ALL
SELECT '2017-08-01T00:00:00Z', 15.25, 19.125 UNION ALL
SELECT '2017-08-02T00:00:00Z', 15.5, 19.375 UNION ALL
SELECT '2017-08-05T00:00:00Z', 16.25, 20.625
), days AS (
SELECT day
FROM (
SELECT
MIN(DATE(TIMESTAMP(ProgressDate))) min_dt,
MAX(DATE(TIMESTAMP(ProgressDate))) max_dt
FROM your_current_result
), UNNEST(GENERATE_DATE_ARRAY(min_dt, max_dt)) day
)
SELECT day,
LAST_VALUE(EstMin IGNORE NULLS) OVER(ORDER BY day) EstMin,
LAST_VALUE(EstMax IGNORE NULLS) OVER(ORDER BY day) EstMax
FROM days
LEFT JOIN your_current_result
ON day = DATE(TIMESTAMP(ProgressDate))
ORDER BY day
Run Code Online (Sandbox Code Playgroud)
结果
Row day EstMin EstMax
1 2017-07-21 0.125 0.25
2 2017-07-22 0.125 0.25
3 2017-07-23 0.125 0.25
4 2017-07-24 5.125 5.375
5 2017-07-25 8.75 10.25
6 2017-07-26 10.0 12.0
7 2017-07-27 10.5 12.75
8 2017-07-28 10.5 12.75
9 2017-07-29 10.5 12.75
10 2017-07-30 10.5 12.75
11 2017-07-31 10.5 12.75
12 2017-08-01 15.25 19.125
13 2017-08-02 15.5 19.375
14 2017-08-03 15.5 19.375
15 2017-08-04 15.5 19.375
16 2017-08-05 16.25 20.625
Run Code Online (Sandbox Code Playgroud)