问题
Following the my question I have the following tables the first (Range) includes range of values and additional columns:
row | From | To | Country ....
-----|--------|---------|---------
1 | 1200 | 1500 |
2 | 2200 | 2700 |
3 | 1700 | 1900 |
4 | 2100 | 2150 |
...
The From and To are bigint and are exclusive. The Range table includes 1.8M records. Additional table (Values) contains 2.7M records and looks like:
row | Value | More columns....
--------|--------|----------------
1 | 1777 |
2 | 2122 |
3 | 1832 |
4 | 1340 |
...
I would like to create one table as followed:
row | Value | From | To | More columns....
--------|--------|--------|-------|---
1 | 1777 | 1700 | 1900 |
2 | 2122 | 2100 | 2150 |
3 | 1832 | 1700 | 1900 |
4 | 1340 | 1200 | 1500 |
...
I used the left outer join in the following code:
set n=1000;
select v.id
,v.val
,r.from_val
,r.to_val
from val v
left outer join
(select r.*
,floor(from_val/${hiveconf:n}) + pe.i as match_val
from val_range r
lateral view posexplode
(
split
(
space
(
cast
(
floor(to_val/${hiveconf:n})
- floor(from_val/${hiveconf:n})
as int
)
)
,' '
)
) pe as i,x
) r
on floor(v.val/${hiveconf:n}) =
r.match_val
where v.val between r.from_val and r.to_val
order by v.id
;
However there is a vast reduction in the number of records of the new table ~31k records out of 2.7M. How can it be if I use the left outer join? How can I fix it?
回答1:
Assuming we have a v.id
set n=1000;
select v.id
,r.from_val
,r.to_val
from val v
left join (select v.id
,r.from_val
,r.to_val
from val v
join (...) r
on floor(v.val/${hiveconf:n}) =
r.match_val
where v.val between r.from_val and r.to_val
) r
on r.id =
v.id
order by v.id
As for the OP request, here is the full query:
set n=1000;
select v.id
,r.from_val
,r.to_val
from val v
left join (select v.id
,r.from_val
,r.to_val
from val v
join (select r.*
,floor(from_val/${hiveconf:n}) + pe.i as match_val
from val_range r
lateral view posexplode
(
split
(
space
(
cast
(
floor(to_val/${hiveconf:n})
- floor(from_val/${hiveconf:n})
as int
)
)
,' '
)
) pe as i,x
) r
on floor(v.val/${hiveconf:n}) =
r.match_val
where v.val between r.from_val and r.to_val
) r
on r.id =
v.id
order by v.id
来源：https://stackoverflow.com/questions/44403489/reduction-in-the-number-of-records-using-range-join