如何在 SQL 雪花中找到数组中以下行之间的差异？

Question

我正在尝试比较同一列中以下行（分组依据）的两个 arrays 并返回不同列中的数组和总数。 我想在此链接上提到类似的结果比较两个 arrays 并计算相同字符串的数量。 但是在这里 arrays 在列之间进行比较，但我想在行之间进行比较。

所以，我的初始数据集可能看起来像这样

|Id.       |      column 1       | 
|----------|---------------------|
|1         |   [cat, dog, bird]  |
|1         |.  [cat, bird]       |
|1         |   [cat, bear, tiger]| 
|1         |.  [cat, tiger]      |
|2         |   [cat, tiger]      |
|2         |   [cat, bear, tiger]|  
|2         |   [cat, bird]       |
|3         |.  [tiger]           |
|3         |.  [cat, bird]       |

所以，我的最终数据集可能看起来像这样。

|Id.       |      column 1       |     column 2     |     column 3     |
|----------|---------------------|------------------|------------------|
|1         |  [cat, dog, bird]   |  [dog]           |          1       |
|1         |. [cat, bird]        |  [bird]          |          1       |
|1         |  [cat, bear, tiger] |  [bear]          |          1       |
|1         |  [cat, tiger]       |. [cat, tiger ]   |                  |
|2         |  [cat, tiger]       |                  |          0       |
|2         |  [cat, bear, tiger] |  [bear, tiger].  |          2       |    
|2         |  [cat, bird]        |  [cat, bird]     |                  |
|3         |  [tiger]            |  [tiger]         |          1       |
|3         |  [cat, bird]        |  [cat, bird      |                  |

第 2 列包含有关第一个数组但不属于第二个数组的信息，第 3 列包含有关第 2 列中有多少元素的信息。

谢谢你

Answer 1

所以 LEAD 示例：

with data(id, _order, col1) as (
    select column1, column2, parse_json(column3)
    from values
        (1, 1, '["cat", "dog", "bird"]'),
        (1, 2, '["cat", "bird"]'),
        (1, 3, '["cat", "bear", "tiger"]'),
        (1, 4, '["cat", "tiger"]'),
        (2, 5, '["cat", "tiger"]'),
        (2, 6, '["cat", "bear", "tiger"]'),
        (2, 7, '["cat", "bird"]'),
        (3, 8, '["tiger"]'),
        (3, 9, '["cat", "bird"]')
)
select *
    ,lead(col1) over (partition by id order by _order ) as lead_col1
from data
order by _order;

ID	_命令	COL1	LEAD_COL1
1	1	[“猫”、“狗”、“鸟”]	[“猫”，“鸟”]
1	2	[“猫”，“鸟”]	[“猫”、“熊”、“老虎”]
1	3	[“猫”、“熊”、“老虎”]	[“猫”，“老虎”]
1	4	[“猫”，“老虎”]	null
2	5	[“猫”，“老虎”]	[“猫”、“熊”、“老虎”]
2	6	[“猫”、“熊”、“老虎”]	[“猫”，“鸟”]
2	7	[“猫”，“鸟”]
3	8	[ “老虎” ]	[“猫”，“鸟”]
3	9	[“猫”，“鸟”]	null

无论如何：这是我将如何做的代码：

with data(id, col1) as (
    select column1, parse_json(column2)
    from values
        (1, '["cat", "dog", "bird"]'),
        (1, '["cat", "bird"]'),
        (1, '["cat", "bear", "tiger"]'),
        (1, '["cat", "tiger"]'),
        (2, '["cat", "tiger"]'),
        (2, '["cat", "bear", "tiger"]'),
        (2, '["cat", "bird"]'),
        (3, '["tiger"]'),
        (3, '["cat", "bird"]')
), flatten as (
    select
        d.id, 
        dense_rank() over(order by f.seq) as seq, 
        f.index, 
        f.value as val
    from data as d,
    table(flatten(input=>d.col1)) f
)
select 
    a.id
    ,a.seq
    ,array_agg(a.val) within group (order by a.index) as col1
    ,array_agg(nvl2(a.val, nvl2(b.val, null, a.val), b.val)) within group (order by a.index) as col2
    ,array_size(col2 )as col3
from flatten as a
full outer join flatten as b 
    on a.id = b.id and a.seq + 1 = b.seq and a.val = b.val
where a.id is not null
group by a.id, a.seq
order by a.seq;
    

ID	序列	COL1	COL2	COL3
1	1	[“猫”、“狗”、“鸟”]	[ “狗” ]	1
1	2	[“猫”，“鸟”]	[ “鸟” ]	1
1	3	[“猫”、“熊”、“老虎”]	[ “熊” ]	1
1	4	[“猫”，“老虎”]	[“猫”，“老虎”]	2
2	5	[“猫”，“老虎”]	[]	0
2	6	[“猫”、“熊”、“老虎”]	[“熊”，“老虎”]	2
2	7	[“猫”，“鸟”]	[“猫”，“鸟”]	2
3	8	[ “老虎” ]	[ “老虎” ]	1
3	9	[“猫”，“鸟”]	[“猫”，“鸟”]	2