Используйте UNION ALL с агрегацией.
Демо:
with your_table as (
select stack(4,
'tom' ,'A','A','B',
'tom' ,'B','A','C',
'peter','B','C','C',
'peter','A','B','C'
) as (name,chinese,math,english)
)
select name, 'chinese' as object,
count(case when chinese='A' then 1 end) as A,
count(case when chinese='B' then 1 end) as B,
count(case when chinese='C' then 1 end) as C
from your_table
group by name
UNION ALL
select name, 'math' as object,
count(case when math='A' then 1 end) as A,
count(case when math='B' then 1 end) as B,
count(case when math='C' then 1 end) as C
from your_table
group by name
UNION ALL
select name, 'english' as object,
count(case when english='A' then 1 end) as A,
count(case when english='B' then 1 end) as B,
count(case when english='C' then 1 end) as C
from your_table
group by name;
Результат:
name object a b c
peter chinese 1 1 0
tom chinese 1 1 0
peter math 0 1 1
tom math 2 0 0
peter english 0 0 2
tom english 0 1 1