这可以在纯sql中完成。在oracle中，可以使用regexp函数和正则表达式拆分分隔字符串，然后使用字符串聚合生成每个字符串的名称列表：

with cte (name, temperament, temp, cnt, lvl) as (
    select 
        name, 
        temperament, 
        regexp_substr (temperament, '[^, ]+', 1, 1) temp, 
        regexp_count(temperament, ',') cnt,
        1 lvl 
    from mytable
    union all
    select 
        name, 
        temperament, 
        regexp_substr (temperament, '[^, ]+', 1, lvl  + 1), 
        cnt,
        lvl + 1
    from cte
    where lvl <= cnt
)
select 
    temp temperament, 
    listagg(name, ', ') within group(order by name) name, 
    count(*) cnt
from cte
group by temp
order by 1

db小提琴演示：

TEMPERAMENT | NAME                  | CNT
:---------- | :-------------------- | --:
Caring      | Golden                |   1
Cute        | Golden, Husky, Poodle |   3
Loving      | Golden                |   1
Loyal       | Golden, Husky         |   2
Smart       | Husky, Poodle         |   2

如果有人需要答案：
var result=（from t in（（from t1 in db.mytables select new{tmp=t1.temp1}）.concat（from t2 in db.mytables select new{tmp=t2.temp2}）.concat（from t3 in db.mytables select new{tmp=t3.temp3}））将t.tmp按t.tmp分组到g select new{tmp=g.key，cnx=g.count（）}）.tolist（）；
希望它能帮助别人！