英文:
percentile_cont signature error when calculating median
问题
我有一个名为“masterdata”的表，其结构如下：
user_eng_time，day，user_pseudo_id
我正在运行以下代码：
select percentile_cont(x,0.5) over () as median_time_in_seconds
from(
select round(user_eng_time/1000) from masterdata
) as x
我得到了以下错误：
聚合函数PERCENTILE_CONT没有匹配的签名，参数类型为: STRUCT<FLOAT64>，FLOAT64。支持的签名有: PERCENTILE_CONT(FLOAT64, FLOAT64); PERCENTILE_CONT(NUMERIC, NUMERIC); PERCENTILE_CONT(BIGNUMERIC, BIGNUMERIC) 在 [55:8] 处
英文:
Hi I have a table "masterdata" with the following structure:
user_eng_time, day, user_pseudo_id
I am running the following code:
select percentile_cont(x,0.5) over () as median_time_in_seconds
from(
select round(user_eng_time/1000) from masterdata
) as x
I am getting the following error:
No matching signature for aggregate function PERCENTILE_CONT for argument types: STRUCT<FLOAT64>, FLOAT64. Supported signatures: PERCENTILE_CONT(FLOAT64, FLOAT64); PERCENTILE_CONT(NUMERIC, NUMERIC); PERCENTILE_CONT(BIGNUMERIC, BIGNUMERIC) at [55:8]
答案1
得分: 2
percentile_cont 期望的第一个参数是一个浮点数值，而不是一个表格。在您的情况下，您可以使用 x.col 来引用表格 x 和列 col。
With masterdata as (
Select 10000*rand() as user_eng_time
from unnest(generate_array(1,100))
)
select *,percentile_cont(x.col,0.5) over () as median_time_in_seconds
from(
select round(user_eng_time/1000) as col from masterdata
) as x
英文:
percentile_cont expects as first entry a float value and not a table. In your case, you can use x.col to reference on table x and the column col.
With masterdata as (
Select 10000*rand() as user_eng_time
from unnest(generate_array(1,100))
)
select *,percentile_cont(x.col,0.5) over () as median_time_in_seconds
from(
select round(user_eng_time/1000) as col from masterdata
) as x
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