How to find the nearest neighbor in Hive? Any windowing function? - hadoop

Question

Given a table
$cat data.csv
ID,State,City,Price,Flag
1,CA,A,95,0
2,CA,A,96,1
3,CA,A,195,1
4,NY,B,124,0
5,NY,B,128,1
6,NY,C,24,0
7,NY,C,27,1
8,NY,C,29,0
9,NY,C,39,1
Expected Result:
ID0, ID1
1,2
4,5
6,7
8,7
for each ID with Flag=0 above, we want to find another ID from Flag=1, with the same "State" and "City", and the nearest Price.
I have two rough stupid ideas:
Method 1.
Use a left outer join with the table itself on
(a.State=b.State and a.City=b.city and a.Flag=0 and b.Flag=1),
where a.Flag=0 and b.Flag=1,
and then use RANK() over (partitioned by a.State,a.City order by a.Price - b.Price) as rank
where rank=1
Method 2.
Use a left outer join with the table itself,
on
(a.State=b.State and a.City=b.city and a.Flag=0 and b.Flag=1),
where a.Flag=0 and b.Flag=1,
and then Use Distribute by a.State,a.City Sort by Price_Diff ASC limit 1
What's the best way to find the nearest neighbor in Hive?
Any valuable tips will be greatly appreciated!

Answer 1

select a.id, b.id , min(abs(b.price-a.price)) as delta
from data as a
inner join data as b
on a.country=b.country and
a.flag=0 and b.flag=1 and
a.city=b.city
group by a.id, b.id
order by delta asc;
This returns
1 2 1 <---
8 7 2 <---
6 7 3 <---
4 5 4 <---
8 9 10
6 9 15
1 3 100
The problem is that the last 3 rows have the same id used into the first 4.
select a.id as id0, b.id as id1, abs(b.price-a.price) as delta,
rank() over ( partition by a.country, a.city order by abs(b.price-a.price) )
from data as a
inner join data as b
on a.country=b.country and
a.flag=0 and b.flag=1 and
a.city=b.city;
This will return
id0 id1 prc rank
1 2 1 1 <---
1 3 100 2
4 5 4 1 <---
8 7 2 1 <---
6 7 3 2
8 9 10 3
6 9 15 4
We are missing 6,7 and this is somehow correct.
6,NY,C,24,0
7,NY,C,27,1
8,NY,C,29,0
9,NY,C,39,1
The lowest price difference for (6,7),(6,9),(8,7),(8,9) is in (8,7). (ambiguous join)
I think you will love this video about this topic : Big Data Analytics Using Window Functions