Question

我有下面的数据集，看起来像这样。

t               mean        max     min     std     data_id
4/14/2010 0:00  12.6941 12.6941 12.6941 12.6941          1
4/14/2010 0:00  12.3851 12.3851 12.3851 12.3851          2
4/14/2010 0:20  12.389  12.389  12.389  12.389           1
4/14/2010 0:20  12.1836 12.1836 12.1836 12.1836          2
4/14/2010 0:20  11.3887 11.3887 11.3887 11.3887          4

我想将数据转换为

t,str_agg
'2010-04-14 00:00:00','12.6941','12.6941','12.6941','12.6941','12.3851','12.3851','12.3851','12.3851',,,,
'2010-04-14 00:10:00','12.3890','12.3890','12.3890','12.3890','12.1836','12.1836','12.1836','12.1836','11.3887','11.3887','11.3887','11.3887

我尝试过以下查询： -

WITH dataset AS (
    SELECT *
    FROM
        (
            VALUES
            ('2010-04-14T00:00'::TIMESTAMP, 12.6941, 12.6941, 12.6941, 12.6941, 1),
            ('2010-04-14T00:00'::TIMESTAMP, 12.3851, 12.3851, 12.3851, 12.3851, 2),
            ('2010-04-14T00:20'::TIMESTAMP, 12.389, 12.389, 12.389, 12.389, 1),
            ('2010-04-14T00:20'::TIMESTAMP, 12.1836, 12.1836, 12.1836, 12.1836, 2),
            ('2010-04-14T00:20'::TIMESTAMP, 11.3887, 11.3887, 11.3887, 11.3887, 13)
        ) AS data(t, mean, max, min, std, data_id)
),
dataset_full AS (
    SELECT
        coalesce(t, time) AS t,
        mean,
        max,
        min,
        std,
        data_id
    FROM
        generate_series(
                (SELECT min(t) FROM dataset),
                (SELECT max(t) FROM dataset),
                '10 minutes')
            AS times(time)
        CROSS JOIN generate_series(
                       (SELECT min(data_id) FROM dataset),
                       (SELECT max(data_id) FROM dataset))
            AS data_id(id)
        LEFT JOIN dataset ON times.time = dataset.t AND data_id.id = dataset.data_id
)
SELECT
    t,
    string_agg(concat(mean, ',', max, ',', min, ',', std), ',')
FROM dataset_full
GROUP BY t
ORDER BY t;

我得到以下结果： -

'2010-04-14 00:00:00','12.6941,12.6941,12.6941,12.6941,12.3851,12.3851,12.3851,12.3851,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,'
'2010-04-14 00:10:00',',,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,'
'2010-04-14 00:20:00','12.389,12.389,12.389,12.389,12.1836,12.1836,12.1836,12.1836,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,11.3887,11.3887,11.3887,11.3887'

但我想要以下结果： -

'2010-04-14 00:00:00','12.6941,12.6941,12.6941,12.6941,12.3851,12.3851,12.3851,12.3851,,,,'
'2010-04-14 00:20:00','12.389,12.389,12.389,12.389,12.1836,12.1836,12.1836,12.1836,11.3887,11.3887,11.3887,11.3887'

任何人都可以帮我解决上述问题！

Answer 1

您的基本问题是您正在生成时间间隔和数据集ID，而不是从数据中读取它们。这会影响dataset_full CTE。您似乎只想要数据中的某些值。

因此：

with dataset as (
      select *
      from (values ('2010-04-14T00:00'::TIMESTAMP, 12.6941, 12.6941, 12.6941, 12.6941, 1),
                   ('2010-04-14T00:00'::TIMESTAMP, 12.3851, 12.3851, 12.3851, 12.3851, 2),
                   ('2010-04-14T00:20'::TIMESTAMP, 12.389, 12.389, 12.389, 12.389, 1),
                   ('2010-04-14T00:20'::TIMESTAMP, 12.1836, 12.1836, 12.1836, 12.1836, 2),
                   ('2010-04-14T00:20'::TIMESTAMP, 11.3887, 11.3887, 11.3887, 11.3887, 13)
           ) AS data(t, mean, max, min, std, data_id)
      ),
     dataset_full as (
       select t.t, d.data_id,
              ds.mean, ds.max, ds.min, ds.std
       from (select distinct t from dataset) t cross join
            (select distinct data_id from dataset) d left join
            dataset ds
            on ds.t = t.t and ds.data_id = d.data_id
     )
select t,string_agg(concat(mean, ',', max, ',', min, ',', std), ',' order by data_id)
from dataset_full
group by t
order by t;

Here是SQL小提琴。

另请注意order by中的string_agg()。据推测，您希望这些值的顺序为dataset_id。

Answer 2

您可以将LEFT JOIN替换为JOIN：

WITH dataset AS (
    SELECT *
    FROM(VALUES
     ('2010-04-14T00:00'::TIMESTAMP, 12.6941, 12.6941, 12.6941, 12.6941, 1),
     ('2010-04-14T00:00'::TIMESTAMP, 12.3851, 12.3851, 12.3851, 12.3851, 2),
     ('2010-04-14T00:20'::TIMESTAMP, 12.389, 12.389, 12.389, 12.389, 1),
     ('2010-04-14T00:20'::TIMESTAMP, 12.1836, 12.1836, 12.1836, 12.1836, 2),
     ('2010-04-14T00:20'::TIMESTAMP, 11.3887, 11.3887, 11.3887, 11.3887, 13)
     ) AS data(t, mean, max, min, std, data_id)
),
dataset_full AS (
    SELECT coalesce(t, time) AS t,
        mean,
        max,
        min,
        std,
        data_id
    FROM generate_series(
                (SELECT min(t) FROM dataset),
                (SELECT max(t) FROM dataset),
                '10 minutes')
            AS times(time)
        CROSS JOIN generate_series(
                       (SELECT min(data_id) FROM dataset),
                       (SELECT max(data_id) FROM dataset))
            AS data_id(id)
        JOIN dataset    -- here
          ON times.time = dataset.t 
         AND data_id.id = dataset.data_id
)
SELECT t,string_agg(concat(mean, ',', max, ',', min, ',', std), ',')
FROM dataset_full
GROUP BY t
ORDER BY t;

<强> DBFiddle Demo

修改

第一行中的,,,,被删除，我不想

... cte2 AS ( SELECT t, string_agg(concat(mean, ',', max, ',', min, ',', std), ',') AS s , COUNT(*) AS c FROM dataset_full GROUP BY t ) SELECT t, s|| REPEAT(',,,,', (MAX(c) OVER() - c)::int) FROM cte2 ORDER BY t;

<强> DBFiddle Demo2

输出：

┌─────────────────────┬─────────────────────────────────────────────────────────────────────────────────────────────┐ │ t │ result │ ├─────────────────────┼─────────────────────────────────────────────────────────────────────────────────────────────┤ │ 2010-04-14 00:00:00 │ 12.6941,12.6941,12.6941,12.6941,12.3851,12.3851,12.3851,12.3851,,,, │ │ 2010-04-14 00:20:00 │ 12.389,12.389,12.389,12.389,12.1836,12.1836,12.1836,12.1836,11.3887,11.3887,11.3887,11.3887 │ └─────────────────────┴─────────────────────────────────────────────────────────────────────────────────────────────┘

在postgres中显示非现有行的逗号分隔值

2 个答案: