按一年中的 52 周透视数据，并使用之前的值填充缺失值
问题描述 投票：0回答：1
我试图将一年中的数据值按 52 周进行透视（在示例中我只使用了 13 周），只要有空白值，我就想用以前的值填充它。
测试数据：
CREATE TABLE testdata (
[year] INT,
id1 VARCHAR(10),
id2 VARCHAR(10),
[week] INT,
date1 DATE,
date2 DATE
);
INSERT INTO testdata ([year], id1, id2, [week], date1, date2) VALUES
(2023, 'p1', 'p001', 1, '2023-01-01', '2023-01-01'),
(2023, 'p2', 'p002', 4, '2023-01-02', '2023-01-02'),
(2023, 'p3', 'p003', 7, '2023-01-03', '2023-01-03'),
(2023, 'p4', 'p004', 10, '2023-01-04', '2023-01-04'),
(2023, 'p5', 'p005', 13, '2023-01-05', '2023-01-05'),
(2024, 'p1', 'p001', 1, '2024-02-01', '2024-02-01'),
(2024, 'p2', 'p002', 2, '2024-02-02', '2024-02-02'),
(2024, 'p3', 'p003', 7, '2024-02-03', '2024-02-03'),
(2024, 'p4', 'p004', 10, '2024-02-04', '2024-02-04'),
(2024, 'p5', 'p005', 13, '2024-02-05', '2024-02-05');
我已经进行了多次尝试，但无法用以前的值填充空白。
WITH WeeklyData AS (
SELECT
[year],
[week],
date1
FROM testdata
)
, PivotedData AS (
SELECT
[year],
MAX(CASE WHEN [week] = 1 THEN date1 END) AS week_1_date1,
MAX(CASE WHEN [week] = 2 THEN date1 END) AS week_2_date1,
MAX(CASE WHEN [week] = 3 THEN date1 END) AS week_3_date1,
MAX(CASE WHEN [week] = 4 THEN date1 END) AS week_4_date1,
MAX(CASE WHEN [week] = 5 THEN date1 END) AS week_5_date1,
MAX(CASE WHEN [week] = 6 THEN date1 END) AS week_6_date1,
MAX(CASE WHEN [week] = 7 THEN date1 END) AS week_7_date1,
MAX(CASE WHEN [week] = 8 THEN date1 END) AS week_8_date1,
MAX(CASE WHEN [week] = 9 THEN date1 END) AS week_9_date1,
MAX(CASE WHEN [week] = 10 THEN date1 END) AS week_10_date1,
MAX(CASE WHEN [week] = 11 THEN date1 END) AS week_11_date1,
MAX(CASE WHEN [week] = 12 THEN date1 END) AS week_12_date1,
MAX(CASE WHEN [week] = 13 THEN date1 END) AS week_13_date1
FROM WeeklyData
GROUP BY [year]
)
SELECT
[year],
COALESCE(week_1_date1, LAG(week_1_date1) OVER (ORDER BY [year])) AS week_1_date1,
COALESCE(week_2_date1, LAG(week_2_date1) OVER (ORDER BY [year])) AS week_2_date1,
COALESCE(week_3_date1, LAG(week_3_date1) OVER (ORDER BY [year])) AS week_3_date1,
COALESCE(week_4_date1, LAG(week_4_date1) OVER (ORDER BY [year])) AS week_4_date1,
COALESCE(week_5_date1, LAG(week_5_date1) OVER (ORDER BY [year])) AS week_5_date1,
COALESCE(week_6_date1, LAG(week_6_date1) OVER (ORDER BY [year])) AS week_6_date1,
COALESCE(week_7_date1, LAG(week_7_date1) OVER (ORDER BY [year])) AS week_7_date1,
COALESCE(week_8_date1, LAG(week_8_date1) OVER (ORDER BY [year])) AS week_8_date1,
COALESCE(week_9_date1, LAG(week_9_date1) OVER (ORDER BY [year])) AS week_9_date1,
COALESCE(week_10_date1, LAG(week_10_date1) OVER (ORDER BY [year])) AS week_10_date1,
COALESCE(week_11_date1, LAG(week_11_date1) OVER (ORDER BY [year])) AS week_11_date1,
COALESCE(week_12_date1, LAG(week_12_date1) OVER (ORDER BY [year])) AS week_12_date1,
COALESCE(week_13_date1, LAG(week_13_date1) OVER (ORDER BY [year])) AS week_13_date1
FROM PivotedData
ORDER BY [year];
预期结果：红色的值是必须插入空白字段的值。
如果您有任何建议，我们将不胜感激。
1个回答
0
投票
投票
您可以通过以下方式实现您的目标：
- 确定所需的年份和周范围。
- 生成涵盖这些年份和周范围的日历表。新的
功能对此有很大帮助。GENERATE_SERIES() - 将生成的日历与源数据左连接，并计算每年和每周组合的
值。这可能为空。MAX(date1) - 使用
IGNORE NULLS` 选项用最新的先前可用值填充任何空值。LAST_VALUE() window function with the - 使用 条件聚合 或
将数据转换为最终的列形式。PIVOT
请注意，`LAST_VALUE() 窗口函数的
GENERATE_SERIES()IGNORE NULLS结果将类似于以下内容（使用条件聚合）：
SELECT
S.year,
MAX(CASE WHEN S.[week] = 1 THEN S.filled_date1 END) AS week_1_date1,
MAX(CASE WHEN S.[week] = 2 THEN S.filled_date1 END) AS week_2_date1,
MAX(CASE WHEN S.[week] = 3 THEN S.filled_date1 END) AS week_3_date1,
MAX(CASE WHEN S.[week] = 4 THEN S.filled_date1 END) AS week_4_date1,
MAX(CASE WHEN S.[week] = 5 THEN S.filled_date1 END) AS week_5_date1,
MAX(CASE WHEN S.[week] = 6 THEN S.filled_date1 END) AS week_6_date1,
MAX(CASE WHEN S.[week] = 7 THEN S.filled_date1 END) AS week_7_date1,
MAX(CASE WHEN S.[week] = 8 THEN S.filled_date1 END) AS week_8_date1,
MAX(CASE WHEN S.[week] = 9 THEN S.filled_date1 END) AS week_9_date1,
MAX(CASE WHEN S.[week] = 10 THEN S.filled_date1 END) AS week_10_date1,
MAX(CASE WHEN S.[week] = 11 THEN S.filled_date1 END) AS week_11_date1,
MAX(CASE WHEN S.[week] = 12 THEN S.filled_date1 END) AS week_12_date1,
MAX(CASE WHEN S.[week] = 13 THEN S.filled_date1 END) AS week_13_date1
FROM (
SELECT
S.year, S.week,
LAST_VALUE(S.MaxDate1) IGNORE NULLS OVER(ORDER BY S.year, S.week) AS filled_date1
FROM (
SELECT Y.value AS year, W.value as week, MAX(T.date1) AS MaxDate1
FROM (
SELECT
MIN(T.year) AS MinYear,
MAX(T.year) AS MaxYear,
GREATEST(MAX(T.week), 13) AS MaxWeek -- 13 weeks for demo purposes
FROM testdata T
) RNG
CROSS APPLY GENERATE_SERIES(RNG.MinYear, RNG.MaxYear) Y
CROSS APPLY GENERATE_SERIES(1, RNG.MaxWeek) W
LEFT JOIN testdata T
ON T.year = Y.value
AND T.week = W.value
GROUP BY Y.value, W.value
) S
) S
GROUP BY S.year
ORDER BY S.year
或以下内容（使用
PIVOTSELECT
PVT.year,
PVT.[1] AS week_1_date1,
PVT.[2] AS week_2_date1,
PVT.[3] AS week_3_date1,
PVT.[4] AS week_4_date1,
PVT.[5] AS week_5_date1,
PVT.[6] AS week_6_date1,
PVT.[7] AS week_7_date1,
PVT.[8] AS week_8_date1,
PVT.[9] AS week_9_date1,
PVT.[10] AS week_10_date1,
PVT.[11] AS week_11_date1,
PVT.[12] AS week_12_date1,
PVT.[13] AS week_13_date1
FROM (
SELECT
S.year, S.week,
LAST_VALUE(S.MaxDate1) IGNORE NULLS OVER(ORDER BY S.year, S.week) AS FilledDate1
FROM (
SELECT Y.value AS year, W.value as week, MAX(T.date1) AS MaxDate1
FROM (
SELECT
MIN(T.year) AS MinYear,
MAX(T.year) AS MaxYear,
GREATEST(MAX(T.week), 13) AS MaxWeek -- 13 weeks for demo purposes
FROM testdata T
) RNG
CROSS APPLY GENERATE_SERIES(RNG.MinYear, RNG.MaxYear) Y
CROSS APPLY GENERATE_SERIES(1, RNG.MaxWeek) W
LEFT JOIN testdata T
ON T.year = Y.value
AND T.week = W.value
GROUP BY Y.value, W.value
) S
) S
PIVOT (
MAX(S.FilledDate1)
FOR week IN ( -- 13 weeks for demo purposes
[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13]
)
) PVT
ORDER BY PVT.year
结果（额外一行显示结转到新的一年）：
| 年 | week_1_date1 | week_2_date1 | week_3_date1 | week_4_date1 | week_5_date1 | week_6_date1 | week_7_date1 | week_8_date1 | week_9_date1 | week_10_date1 | week_11_date1 | week_12_date1 | week_13_date1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2023 | 2023-01-01 | 2023-01-01 | 2023-01-01 | 2023-01-02 | 2023-01-02 | 2023-01-02 | 2023-01-03 | 2023-01-03 | 2023-01-03 | 2023-01-04 | 2023-01-04 | 2023-01-04 | 2023-01-05 |
| 2024 | 2024-02-01 | 2024-02-02 | 2024-02-02 | 2024-02-02 | 2024-02-02 | 2024-02-02 | 2024-02-03 | 2024-02-03 | 2024-02-03 | 2024-02-04 | 2024-02-04 | 2024-02-04 | 2024-02-05 |
| 2025 | 2024-02-05 | 2024-02-05 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 | 2025-03-01 |
请参阅 此数据库<>fiddle 进行演示
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