'Inner join with special conditions

Given an hourly table A with full records, e.g.:

User    Hour  Purchase 
Joe      1       0
Joe      2       0
Joe      3       0
Joe      4       1
Joe      5       0
Joe      6       0
Joe      7       0
Joe      8       1
Joe      9       1
Joe     10       0

And a subset B of it, e.g.

User    Hour    Purchase 
Joe      3         0
Joe      9         1
Joe     10         0

I want to keep only those records from A that are in B or at most 2hr behind the B subset, without duplication, e.g.

User    Hour Purchase 
Joe      1    0
Joe      2    0
Joe      3    0
Joe      7    0
Joe      8    0
Joe      9    1 
Joe     10    0

How can the result be achieved with an inner join, without duplication (in this case the hours 8&9) and preserving the right purchase values for the hours in B? (This is an MWE, assume multiple users and timestamps instead of hours)

sqljoingoogle-bigquery


Solution 1:[1]

Consider below simple approach

select * from tableA a
where exists (
  select 1 from tableB b
  where a.hour between b.hour - 2 and b.hour
  and a.user = b.user
)

if applied to sample in your question - output is

enter image description here

I hope in your real case you have datetime or timestamp instead of hour column, so in this case you will need slightly modify where a.hour between b.hour - 2 and b.hour part of above. it will look something like

where a.datetime between datetime_sub(b.datetime, interval 2 hour) and b.datetime

Solution 2:[2]

Try this

with _data as 
(
select 'Joe' as _user, 1 as _hour,0 as purchase union all
select 'Joe' as _user, 2 as _hour,0 as purchase union all
select 'Joe' as _user, 3 as _hour,0 as purchase union all
select 'Joe' as _user, 4 as _hour,1 as purchase union all
select 'Joe' as _user, 5 as _hour,0 as purchase union all
select 'Joe' as _user, 6 as _hour,0 as purchase union all
select 'Joe' as _user, 7 as _hour,0 as purchase union all
select 'Joe' as _user, 8 as _hour,1 as purchase union all
select 'Joe' as _user, 9 as _hour,1 as purchase union all
select 'Joe' as _user,10 as _hour,0 as purchase
)
,subset as
(
select 'Joe' as _user, 3 as _hour,0 as purchase union all
select 'Joe' as _user, 9 as _hour,1 as purchase union all
select 'Joe' as _user,10 as _hour,0 as purchase
)
select  a._user,a._hour,any_value(b.purchase) from _data a  join subset b on 
(
    a._user = b._user and 
    (
        a._hour = b._hour 
        or 
        case when b._hour > a._hour then (case when (b._hour - a._hour) <=2 then 1=1 end) end)
)
group by a._user,a._hour

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Mikhail Berlyant
Solution 2 Mr.Batra

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