count zeros between 1s in same column - sql

Question

I've data like this.
ID IND
1 0
2 0
3 1
4 0
5 1
6 0
7 0
I want to count the zeros before the value 1. So that, the output will be like below.
ID IND OUT
1 0 0
2 0 0
3 1 2
4 0 0
5 1 1
6 0 0
7 0 2
Is it possible without pl/sql? I tried to find the differences between row numbers but couldn't achieve it.

Answer 1

The match_recognize clause, introduced in Oracle 12.1, can do quick work of such "row pattern recognition" problems. The solution is just a bit complex due to the special treatment of a "last row" with ID = 0, but it is straightforward otherwise.
As usual, the with clause is not part of the solution; I include it to test the query. Remove it and use your actual table and column names.
with
inputs (id, ind) as (
select 1, 0 from dual union all
select 2, 0 from dual union all
select 3, 1 from dual union all
select 4, 0 from dual union all
select 5, 1 from dual union all
select 6, 0 from dual union all
select 7, 0 from dual
)
select id, ind, out
from inputs
match_recognize(
order by id
measures case classifier() when 'Z' then 0
when 'O' then count(*) - 1
else count(*) end as out
all rows per match
pattern ( Z* ( O | X ) )
define Z as ind = 0, O as ind != 0
);
ID IND OUT
---------- ---------- ----------
1 0 0
2 0 0
3 1 2
4 0 0
5 1 1
6 0 0
7 0 2

Answer 2

You can treat this as a gaps-and-islands problem. You can define the "islands" by the number of "1"s one or after each row. Then use a window function:
select t.*,
(case when ind = 1 or row_number() over (order by id desc) = 1
then sum(1 - ind) over (partition by grp)
else 0
end) as num_zeros
from (select t.*,
sum(ind) over (order by id desc) as grp
from t
) t;
If id is sequential with no gaps, you can do this without a subquery:
select t.*,
(case when ind = 1 or row_number() over (order by id desc) = 1
then id - coalesce(lag(case when ind = 1 then id end ignore nulls) over (order by id), min(id) over () - 1)
else 0
end)
from t;
I would suggest removing the case conditions and just using the then clause for the expression, so the value is on all rows.