Emulating an interval join in hive - join

Question

I am using hive 0.13.
I have two tables:
data table. columns: id, time. 1E10 rows.
mymap table. columns: id, name, start_time, end_time. 1E6 rows.
For each row in the data table I want to get the name from the mymap table matching the id and the time interval. So I want to do a join like:
select data.id, time, name from data left outer join mymap on data.id = mymap.id and time>=start_time and time<end_time
It is known that for every row in data there are 0 or 1 matches in mymap.
The above query is not supported in hive as it is a non-equi-join. Moving the inequality conditions into a where filter does not work cause the join explodes before the filter is applied:
select data.id, time, name from data left outer join mymap on data.id = mymap.id where mymap.id is null or (time>=start_time and time<end_time)
(I am aware that the queries are not exactly equivalent due to cases where there is a match for id but no matching interval. This can be solved as I describe here: Hive: work around for non equi left join)
How can I go about this?

Answer 1

You could perform your join and then query from that table. I didn't test this code, but it would read something like
select id
,time
,name
from (
select d.id
,d.time
,m.name
,m.start_time
,m.end_time
from data as d LEFT OUTER JOIN mymap as m
ON d.id = m.id
) x
where time>=start_time
AND time<end_time

Answer 2

You could potentially get around this issue by flattening out the data structure in table2 and using a UDF to process the joined records.
select
id,
time,
nameFinderUDF(b.name_list, time) as name
from
data a
LEFT OUTER JOIN
(
select
id,
collect_set(array(name,cast(start_time as string),cast(end_time as string))) as name_list
from
mymap
group by
id
) b
ON (a.id=b.id)
With a UDF that does something like:
public String evaluate(ArrayList<ArrayList<String>> name_list,Long time) {
for (int i;i<name_list.length;i++) {
if (time >= Long.parseLong(name_list[i][1]) && time <= Long.parseLong(name_list[i][2])) {
return name_list[i][0]
return null;
}
This approach should make the merge 1 to 1, but it could create a fairly large data structure repeated many times. It is still quite a bit more efficient than a straight join.