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 0:10  12.389  12.389  12.389  12.389           1
4/14/2010 0:10  12.1836 12.1836 12.1836 12.1836          2
4/14/2010 0:10  11.3887 11.3887 11.3887 11.3887          3

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'

因此，如果您查看结果，则对于t = 4/14/2010 0:00，data_id 3没有数据，因此结果将没有以逗号分隔的值。

我希望这个结果在postgres中。我尝试过以下查询： -

select t,string_agg(mean||','||max||','||min||','||std,',') within group(order by t)  
from table_name 
group by t  
order by t;

但这给了我以下结果： -

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'

Answer 1

WITH dataset AS (
    SELECT *
    FROM
        (
            VALUES
            ('2010-04-14T00:00'::TIMESTAMP, 1, 1, 1, 1, 1),
            ('2010-04-14T00:00'::TIMESTAMP, 2, 2, 2, 2, 2),
            ('2010-04-14T00:20'::TIMESTAMP, 3, 3, 3, 3, 1),
            ('2010-04-14T00:20'::TIMESTAMP, 4, 4, 4, 4, 2),
            ('2010-04-14T00:20'::TIMESTAMP, 5, 5, 5, 5, 3)
        ) AS data(t, mean, max, min, std, data_id)
),
timestamps AS (
    SELECT t FROM dataset GROUP BY t
),
data_id AS (
    SELECT data_id AS id FROM dataset GROUP BY data_id
),
dataset_full AS (
    SELECT
        coalesce(dataset.t, ts.t) AS t,
        mean,
        max,
        min,
        std,
        data_id
    FROM
--         generate_series(
--                 (SELECT min(t) FROM dataset),
--                 (SELECT max(t) FROM dataset),
--                 '10 minutes')
--             AS ts(t)
        timestamps AS ts
--        CROSS JOIN generate_series(
--                       (SELECT min(data_id) FROM dataset),
--                       (SELECT max(data_id) FROM dataset))
--            AS data_id(id)
        CROSS JOIN data_id
        LEFT JOIN dataset ON ts.t = 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;

数据集 CTE（公用表表达式）就在那里代替您的表。
dataset_full 通过为10m间隔和data_id值的每个组合生成一行来添加所有缺失的行。 dataset 然后LEFT JOIN到它，这意味着现在有那些以前不存在的行的NULL值。然后将NULL值转换为string_agg中的空字符串，从而产生您想要的结果。

修改
我在评论中根据请求更改了它，以便它只返回原始数据集中存在时间戳的行。

编辑2
我根据OP的另一个请求更改了它，只使用数据集中的data_ids。

显示表中非现有行的逗号分隔值

1 个答案: