您绝对可以在没有外部GROUP BY的情况下执行此操作：

CREATE OR REPLACE FUNCTION `dataset.sumdistinct` (values array<struct<id int64,val int64>>) as (
  (
    select 
      struct(
       count(distinct x.id) as col1,sum(distinct x.val) as col2
      ) from unnest(values) as x
  )
);

select sumdistinct(array_agg(struct(id as id,val as val))) from `dataset.table`

以下是用于BigQuery标准SQL

#standardSQL
CREATE TEMP FUNCTION SumDistinct(arr ANY TYPE) AS ((
  SELECT AS STRUCT 
    COUNT(DISTINCT id) unique_ids,SUM(val) total_value
  FROM (
    SELECT ANY_VALUE(t).*
    FROM UNNEST(arr) t
    GROUP BY FORMAT('%t',t)
  )
));
SELECT SumDistinct(ARRAY_AGG(STRUCT(id,val))).*
FROM `project.dataset.data`   

如果要应用于您的问题的样本数据-结果为

Row unique_ids  total_value  
1   3           600 

根据您要如何将不同的val解析为相同的id，可以在下面的sql中调整聚合函数（max(val)）：

with data as 
(select 1 as id,100 as val union all
select 1,100 union all
select 1,100 union all
select 2,200 union all
select 2,200 union all
select 3,300 union all
select 3,300
)
SELECT count(1) as unique_ids,sum(val) as total_value
FROM (
SELECT id,max(val) val
FROM data
GROUP BY id
)

基于注释，看来您希望UDF能够调用内部的聚合函数。也许您正在寻找用户定义的聚合函数，BigQuery不支持该函数，但可以采用以下形式：

输出不是您期望的，因为UDF无法输出2列（如您的示例中所示），希望您有一个想法，需要对array_agg（）字段进行操作，并且您的UDF在内部进行UNNEST（）并能够使用系统集合函数，例如SUM（）：

CREATE TEMP FUNCTION sumdistinct (unique_key INT64,val_array ARRAY<INT64>) AS (
 (SELECT COALESCE(ROUND(COALESCE(CAST((SUM(DISTINCT (CAST(ROUND(COALESCE(safe_cast(val_to_sum as float64),0)*(1/1000*1.0),9) AS NUMERIC) + (cast(cast(concat('0x',substr(to_hex(md5(CAST(unique_key  AS STRING))),1,15)) as int64) as numeric) * 4294967296 + cast(cast(concat('0x',16,8)) as int64) as numeric)) * 0.000000001 )) - SUM(DISTINCT (cast(cast(concat('0x',8)) as int64) as numeric)) * 0.000000001) )  / (1/1000*1.0) AS FLOAT64),0),6),0)
 FROM unnest(val_array) val_to_sum)
);
with data as 
(select 1 as id,300
)
SELECT sumdistinct(id,array_agg(val))
FROM data
GROUP BY id