SQL：如何获取时间序列中两个相邻行之间的值变化？答案

【问题标题】：SQL: How to get change in value between two adjacent rows in a time series?SQL：如何获取时间序列中两个相邻行之间的值变化？
【发布时间】：2018-09-24 23:43:32
【问题描述】：

我有一个名为NEW_YORK_TEMPERATURES 的表，例如：

注意：添加间距以显示不同的位置和日期

datetime,           location, min_temp, max_temp

2018-01-01 12:00:00,  seneca,     76.1,     76.5
2018-01-01 12:10:00,  seneca,     76.1,     76.5
2018-01-01 12:20:00,  seneca,     76.2,     76.6
2018-01-01 12:30:00,  seneca,     76.1,     76.6
2018-01-01 12:40:00,  seneca,     76.1,     76.5

2018-01-02 12:00:00,  seneca,     76.6,     77.3
2018-01-02 12:10:00,  seneca,     76.6,     77.3
2018-01-02 12:20:00,  seneca,     76.6,     77.3
2018-01-02 12:30:00,  seneca,     76.6,     77.3
2018-01-02 12:40:00,  seneca,     76.6,     77.3

2018-01-01 12:00:00, conesus,     66.1,     66.5
2018-01-01 12:10:00, conesus,     66.1,     66.5
2018-01-01 12:20:00, conesus,     66.2,     66.6
2018-01-01 12:30:00, conesus,     66.1,     66.6
2018-01-01 12:40:00, conesus,     66.1,     66.5

2018-01-02 12:00:00, conesus,     66.4,     66.7
2018-01-02 12:10:00, conesus,     66.4,     66.8
2018-01-02 12:20:00, conesus,     66.4,     66.9
2018-01-02 12:30:00, conesus,     66.4,     66.9
2018-01-02 12:40:00, conesus,     66.4,     66.9   

2018-01-01 12:00:00, ontario,     63.1,     63.5
2018-01-01 12:10:00, ontario,     63.1,     63.5
2018-01-01 12:20:00, ontario,     63.2,     63.6
2018-01-01 12:30:00, ontario,     63.1,     63.6
2018-01-01 12:40:00, ontario,     63.1,     63.5

2018-01-02 12:00:00, ontario,     63.3,     63.8
2018-01-02 12:10:00, ontario,     63.3,     63.8    
2018-01-02 12:20:00, ontario,     63.3,     63.8
2018-01-02 12:30:00, ontario,     63.3,     63.7
2018-01-02 12:40:00, ontario,     63.3,     63.7

我需要一种方法来获取两个连续时间戳之间范围变化的差异。第一步是通过执行以下操作创建展开列：

select 
    datetime, 
    location, 
    min_temp, 
    max_temp, 
    max_temp - min_temp as range 
from NEW_YORK_TEMPERATURES 
order by datetime

获取如下表格：

datetime,           location, min_temp, max_temp, range

2018-01-01 12:00:00,  seneca,     76.1,     76.5,   0.4
2018-01-01 12:10:00,  seneca,     76.1,     76.5,   0.4
2018-01-01 12:20:00,  seneca,     76.2,     76.6,   0.4
2018-01-01 12:30:00,  seneca,     76.1,     76.6,   0.5
2018-01-01 12:40:00,  seneca,     76.1,     76.5,   0.4

2018-01-02 12:00:00,  seneca,     76.6,     77.3,   0.7
2018-01-02 12:10:00,  seneca,     76.6,     77.3,   0.7
2018-01-02 12:20:00,  seneca,     76.6,     77.3,   0.7
2018-01-02 12:30:00,  seneca,     76.6,     77.3,   0.7
2018-01-02 12:40:00,  seneca,     76.6,     77.3,   0.7

2018-01-01 12:00:00, conesus,     66.1,     66.5,   0.4
2018-01-01 12:10:00, conesus,     66.1,     66.5,   0.4
2018-01-01 12:20:00, conesus,     66.2,     66.6,   0.4
2018-01-01 12:30:00, conesus,     66.1,     66.6,   0.5
2018-01-01 12:40:00, conesus,     66.1,     66.5,   0.4

2018-01-02 12:00:00, conesus,     66.4,     66.7,   0.3
2018-01-02 12:10:00, conesus,     66.4,     66.8,   0.4
2018-01-02 12:20:00, conesus,     66.4,     66.9,   0.5
2018-01-02 12:30:00, conesus,     66.4,     66.9,   0.5
2018-01-02 12:40:00, conesus,     66.4,     66.9,   0.5

2018-01-01 12:00:00, ontario,     63.1,     63.5,   0.4
2018-01-01 12:10:00, ontario,     63.1,     63.5,   0.4
2018-01-01 12:20:00, ontario,     63.2,     63.6,   0.4
2018-01-01 12:30:00, ontario,     63.1,     63.6,   0.5
2018-01-01 12:40:00, ontario,     63.1,     63.5,   0.4

2018-01-02 12:00:00, ontario,     63.3,     63.8,   0.5
2018-01-02 12:10:00, ontario,     63.3,     63.8,   0.5   
2018-01-02 12:20:00, ontario,     63.3,     63.8,   0.5
2018-01-02 12:30:00, ontario,     63.3,     63.7,   0.4
2018-01-02 12:40:00, ontario,     63.3,     63.7,   0.4

但是我怎样才能在同一位置获得相邻条之间范围的变化，以便我的结果看起来像：

datetime,           location, min_temp, max_temp, range, change_in_range

2018-01-01 12:00:00,  seneca,     76.1,     76.5,   0.4              nan
2018-01-01 12:10:00,  seneca,     76.1,     76.5,   0.4              0.0
2018-01-01 12:20:00,  seneca,     76.2,     76.6,   0.4              0.0
2018-01-01 12:30:00,  seneca,     76.1,     76.6,   0.5              0.1
2018-01-01 12:40:00,  seneca,     76.1,     76.5,   0.4             -0.1

2018-01-02 12:00:00,  seneca,     76.6,     77.3,   0.7              0.0
2018-01-02 12:10:00,  seneca,     76.6,     77.3,   0.7              0.0
2018-01-02 12:20:00,  seneca,     76.6,     77.3,   0.7              0.0
2018-01-02 12:30:00,  seneca,     76.6,     77.3,   0.7              0.0
2018-01-02 12:40:00,  seneca,     76.6,     77.3,   0.7              0.0

2018-01-01 12:00:00, conesus,     66.1,     66.5,   0.4              nan
2018-01-01 12:10:00, conesus,     66.1,     66.5,   0.4              0.0
2018-01-01 12:20:00, conesus,     66.2,     66.6,   0.4              0.0
2018-01-01 12:30:00, conesus,     66.1,     66.6,   0.5              0.1
2018-01-01 12:40:00, conesus,     66.1,     66.5,   0.4              0.0

2018-01-02 12:00:00, conesus,     66.4,     66.7,   0.3             -0.1
2018-01-02 12:10:00, conesus,     66.4,     66.8,   0.4              0.1
2018-01-02 12:20:00, conesus,     66.4,     66.9,   0.5              0.1
2018-01-02 12:30:00, conesus,     66.4,     66.9,   0.5              0.0
2018-01-02 12:40:00, conesus,     66.4,     66.9,   0.5              0.0

2018-01-01 12:00:00, ontario,     63.1,     63.5,   0.4              nan
2018-01-01 12:10:00, ontario,     63.1,     63.5,   0.4              0.0
2018-01-01 12:20:00, ontario,     63.2,     63.6,   0.4              0.0
2018-01-01 12:30:00, ontario,     63.1,     63.6,   0.5              0.1
2018-01-01 12:40:00, ontario,     63.1,     63.5,   0.4             -0.1

2018-01-02 12:00:00, ontario,     63.3,     63.8,   0.5              0.1
2018-01-02 12:10:00, ontario,     63.3,     63.8,   0.5              0.0
2018-01-02 12:20:00, ontario,     63.3,     63.8,   0.5              0.0
2018-01-02 12:30:00, ontario,     63.3,     63.7,   0.4             -0.1
2018-01-02 12:40:00, ontario,     63.3,     63.7,   0.4              0.0

【问题讨论】：

“2018-01-02 12:00:00/seneca”的 0 没有意义。

标签： sql postgresql

【解决方案1】：

您可以将lag() 与横向连接一起使用：

select t.*, v.range,
       (range - lag(range) over (partition by location order by datetime)) as change_in_range
from NEW_YORK_TEMPERATURES t cross join lateral
     (values (max_temp - min_temp)) v(range)
order by location, datetime;

注意：这会将每个位置的第一个值表示为 NULL 而不是 nan。

另外，对于您想要的输出，您需要order by location, datetime。

【讨论】：

【解决方案2】：

一个选项使用LAG：

SELECT 
    datetime,
    location,
    min_temp,
    max_temp,
    max_temp - min_temp AS range,
    (max_temp - min_temp) - LAG(max_temp - min_temp) OVER
        (PARTITION BY location ORDER BY datetime) change_in_range
FROM NEW_YORK_TEMPERATURES 
ORDER BY
    location, datetime;

Demo

【讨论】：

【解决方案3】：

由于您似乎希望按日期（从 datetime 字段截断）和位置对“范围差异”进行分区，因此您需要将截断的日期添加到您的LAG 函数。使用 CTE，您会得到如下结果：

    WITH temps_with_ranges AS
    (
        SELECT *, 
            date_trunc('day', datetime) AS dt, 
            max_temp - min_temp AS "range"
        FROM NEW_YORK_TEMPERATURES
    )
SELECT * , "range" - LAG("range") OVER (PARTITION BY location, dt ORDER BY datetime)
FROM temps_with_ranges

【讨论】：

这个答案和我昨天给出的有什么不同？
@TimBiegeleisen：您的没有按位置和日期划分。另外，我认为 CTE 让事情变得更简洁。
仔细查看原始问题中的预期范围差异。没有按日期划分，只有按位置划分。
@TimBiegeleisen，我同意预期结果并未明确表明他想按日期分区，但 OP 在日期组和位置组之间也有视觉分离。因此，我认为至少将其作为一种选择是有价值的。此外，正如 Gordon Linoff 指出的那样，无论如何，预期的结果并不一致。