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如何用什么方法在一栏左右 p积雪花数据?
原标题:What would be a good way to pivot a Snowpark DataFrame around a column?

我先读到Snow Park(Python) docs,有pivot用于数据Frames的功能,但这不是我所期望的。 我想能够用一栏 around一个数据框架,每栏有不同价值,一栏是具有这一价值的各行。 我要说的是,我有这一数据框架:

| a    | b              | c                      |
| ---- | -------------- | ---------------------- |
| 1    | X              | { "prop": 9 }          |
| 2    | Y              | { "prop": 10 }         |
| 3    | Z              | { "prop": 10 }         |
| 4    | Z              | { "prop": 11 }         |
| 5    | X              | { "prop": 9 }          |

我要说的是,我想用围绕“c.prop”的“数据框架”,来研究我需要做的变革,以获得这一数据框架:

| c.prop    | rows                       
| --------- | ----------------------------------------------------------------------------------- |
| 9         | [{ "a": 1, "b": "X" c: { "prop": 9 }}, { "a": 5, "b": "X" c: { "prop": 9 }}]         |
| 10        | [{ "a": 2, "b": "Y" c: { "prop": 10 }}, { "a": 3, "b": "Z" c: { "prop": 10 }}]       |
| 11        | [{ "a": 4, "b": "Z" c: { "prop": 11 }}]                                            |

现在,最接近的我必须找到一种解决办法,即“重新确认”数据范围,然后,就我所希望排出的每个不同价值而言,过滤原始数据框架,只保留该栏具有这一特定价值的各行。

问题回答

我请各位回答,因为我很容易照搬样本情景,只用<条码>、和<条码>标的>解决。 请注意,这在技术上不是“一纸空文”,而是你要求的结果:

with data_raw(x, a, b, c) as (
    select split(value,  | ) x, x[1]::int, trim(x[2]::string), parse_json(x[3])
    from table(split_to_table(
    $$| 1    | X              | { "prop": 9 }          |
    | 2    | Y              | { "prop": 10 }         |
    | 3    | Z              | { "prop": 10 }         |
    | 4    | Z              | { "prop": 11 }         |
    | 5    | X              | { "prop": 9 }          |$$,  
 ))
), data as (
    select * exclude(x)
    from data_raw
)

select "c.prop", array_agg(o) "rows"
from (
    select c:prop "c.prop", object_construct(*) o
    from data
)
group by 1
order by 1

“entergraph





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