hive rank over grouped value - hive

Question

gurus, i stumbled in hive rank process, i woluld like to rank transaction in each day (with no repeating rank value for the same trx value)
date hour trx rnk
18/03/2018 0 1 24
18/03/2018 1 2 23
18/03/2018 2 3 22
18/03/2018 3 4 21
18/03/2018 4 5 20
18/03/2018 5 6 19
18/03/2018 6 7 18
18/03/2018 7 8 17
18/03/2018 8 9 16
18/03/2018 9 10 15
18/03/2018 10 11 14
18/03/2018 11 12 13
18/03/2018 12 13 12
18/03/2018 13 14 11
18/03/2018 14 15 10
18/03/2018 15 16 9
18/03/2018 16 17 8
18/03/2018 17 18 7
18/03/2018 18 19 6
18/03/2018 19 20 5
18/03/2018 20 21 4
18/03/2018 21 22 3
18/03/2018 22 23 2
18/03/2018 23 24 1
17/03/2018 0 1 24
17/03/2018 1 2 23
17/03/2018 2 3 22
17/03/2018 3 4 21
17/03/2018 4 5 20
17/03/2018 5 6 19
17/03/2018 6 7 18
17/03/2018 7 8 17
17/03/2018 8 9 16
17/03/2018 9 10 15
17/03/2018 10 11 14
17/03/2018 11 12 13
17/03/2018 12 13 12
17/03/2018 13 14 11
17/03/2018 14 15 10
17/03/2018 15 16 9
17/03/2018 16 17 8
17/03/2018 17 18 7
17/03/2018 18 19 6
17/03/2018 19 20 5
17/03/2018 20 21 4
17/03/2018 21 22 3
17/03/2018 22 23 2
17/03/2018 23 24 1
here is my code
select a.date, a.hour, trx, rank() over (order by a.trx) as rnk from(
select date,hour, count(*) as trx from smy_tb
group by date, hour
)a
limit 100;
the problem is:
1. rank value repeated with the same trx value
2. rank value continued to next date (it should be grouped for date and hour, so each date will only return 24 rank value)
need advice,
thank you

Answer 1

You should partition by date column and use a specific ordering.
rank() over (partition by a.date order by a.hour desc)

Answer 2

as explained by #BKS
this is the resolved code
select a.date, a.hour, trx, row_number() over (partition by a.date order by a.trx desc) as rnk from(
select date,hour, count(*) as trx from smy_tb
group by date, hour
)a
limit 100;