postgresql - 多列的前5個值，以及值的總和

Question

我正在尋找關於計算多列的幅度前5個值的最佳（最有效）方法的建議，我還需要計算另一列的總和。

假設我有標題的數據:(人，日，月，每日，每日最大1，每日最大2），每個人的每一天我有總數量，最大數量（措施1）和最大數量（措施2） 。

我想要做的是計算每個人， 每個月 （1）每天的總和，（2）dailymax1的前5個值（3）dailymax2的前5個值。 可能是一個月內甚至沒有5個每日值，在這種情況下我希望返回null。

沒有連接，我想不出怎么做，因為我是一個sql新手。 我知道，對於組中前5個值中的每個值，將重復dailyqty的總和 - 這沒關系。

一些虛擬數據：

CREATE TABLE test (
    person varchar(50),
    daydate date,
    month integer,
    dailyqty double precision,
    dailymax1 double precision,
    dailymax2 double precision
);

INSERT INTO test(person, daydate, month, dailyqty, dailymax1, dailymax2)
VALUES
    ('A', '2015-01-01', 1, 5, 0.5, 4),
    ('A', '2015-01-02', 1, 8, 3, 4),
    ('A', '2015-01-03', 1, 7, 1, 3),
    ('A', '2015-01-04', 1, 1, 2, 2),
    ('A', '2015-01-05', 1, 9, 6, 8),
    ('A', '2015-01-06', 1, 7, 2.5, 7),
    ('A', '2015-01-07', 1, 2, 4, 7),
    ('A', '2015-01-08', 1, 5, 1, 3),
    ('B', '2015-01-01', 1, 20, 8, 1),
    ('B', '2015-01-02', 1, 22, 9, 2)

期望的結果

謝謝！ 一種

Answer 1

此查詢復制問題中發布的所需結果：

SELECT xt1.person, xt1.month, xt1.monthlyqty, xt3.max1, xt4.max2
FROM (
  SELECT SUM(COALESCE(t.dailyqty, 0)) as monthlyqty, t.person, t.month
  FROM test t
  GROUP by t.person, t.month
) xt1
CROSS JOIN (
  VALUES (1), (2), (3), (4), (5)
) xt2
LEFT OUTER JOIN (
  SELECT t.person, t.month, t.dailymax1 as max1
  , ROW_NUMBER() OVER (PARTITION BY t.person, t.month ORDER BY t.dailymax1 DESC NULLS LAST) as colnumber
  FROM test t
) xt3 ON xt2.column1 = xt3.colnumber AND xt1.person = xt3.person AND xt1.month = xt3.month
LEFT OUTER JOIN (
  SELECT t.person, t.month, t.dailymax2 as max2
  , ROW_NUMBER() OVER (PARTITION BY t.person, t.month ORDER BY t.dailymax2 DESC NULLS LAST) as colnumber
  FROM test t
) xt4 ON xt2.column1 = xt4.colnumber AND xt1.person = xt4.person AND xt1.month = xt4.month;

需要考慮的一些事情可能會改變查詢......首先，您可以考慮一下dailyqty，dailymax1和dailymax2這些列是否真的可以為空（就像在您的表定義中一樣）。 如果不是，您可以將COALESCE(t.dailyqty, 0)簡化為t.dailyqty ，將兩個DESC NULLS LAST簡化為DESC 。

其次，你可以考慮更換CROSS JOIN到xt2與JOIN來調用generate_series ，如： CROSS JOIN generate_series (1, 5) xt2然后更換xt2.column1出場只有xt2 。 我不確定哪種方法會更有效，也許兩者都做類似的事情，但如果存在顯着差異，則值得檢查一下您的實際數據。

最后，您說您想為每個人和每月計算，但“月”可以指“月”列或“白天”列中的月份。 我選擇了第一個選項，因為它更容易編寫:)，但修改了一些事情，查詢可以適用於其他列。

Answer 2

此查詢提供您需要的輸出：

WITH FilledData AS
(
    WITH Filler AS
    (
        WITH t1 AS
        (
            SELECT DISTINCT person, month FROM test
        ),
        t2 AS
        (
            SELECT generate_series as order FROM generate_series(1, 4)
        )
        SELECT t1.person, t1.month, CAST(NULL AS double precision) AS dailyqty, CAST(NULL AS double precision) AS dailymax1, CAST(NULL AS double precision) AS dailymax2 FROM t1 CROSS JOIN t2
    )
    SELECT person, month, dailyqty, dailymax1, dailymax2 FROM test UNION ALL
    SELECT person, month, dailyqty, dailymax1, dailymax2 FROM Filler ORDER BY person, month
),
monthlyqty AS
(
    SELECT person, month, SUM(dailyqty) AS monthlyqty FROM test GROUP BY person, month
),
dailymax1_table AS
(
    SELECT person, month, dailymax1, dailymax1_order
    FROM (
       SELECT *, row_number() over (partition by person, month order by dailymax1 desc NULLS LAST) as dailymax1_order
       FROM FilledData
    ) t1 WHERE dailymax1_order <= 5
),
dailymax2_table AS
(
    SELECT person, month, dailymax2, dailymax2_order
    FROM (
       SELECT *, row_number() over (partition by person, month order by dailymax2 desc NULLS LAST) as dailymax2_order
       FROM FilledData
    ) t2 WHERE dailymax2_order <= 5
)

SELECT dailymax1_table.person, dailymax1_table.month, monthlyqty.monthlyqty, dailymax1_table.dailymax1 as max1, dailymax2_table.dailymax2 as max2
FROM dailymax1_table JOIN monthlyqty
ON monthlyqty.person = dailymax1_table.person AND
   monthlyqty.month = dailymax1_table.month
JOIN dailymax2_table ON
   dailymax1_table.person = dailymax2_table.person AND
   dailymax1_table.month = dailymax2_table.month AND
   dailymax1_table.dailymax1_order = dailymax2_table.dailymax2_order;

Answer 3

您可以使用窗口函數將它們組合在一起：

select . . .
from (select t.*,
             sum(dailyqty) over (partition by person, date_trunc('month', datecol)) as monthqty,
             row_number() over (partition by person, date_trunc('month', datecol) order by dailyqty desc) as seqnum
      from t
     ) t
where seqnum <= 5;

您可以從子查詢中提取所需的列。