How to `sum( DISTINCT <column> ) OVER ()` using window function? - postgresql

Question

I have next data:
Here I already calculated total for conf_id. But want also calculate total for whole partition. eg:
Calculate total suma by agreement for each its order (not goods at order which are with slightly different rounding)
How to sum 737.38 and 1238.3? eg. take only one number among group
(I can not sum( item_suma ), because it will return 1975.67. Notice round for conf_suma as intermediate step)
UPD
Full query. Here I want to calculate rounded suma for each group. Then I need to calculate total suma for those groups
SELECT app_period( '2021-02-01', '2021-03-01' );
WITH
target_date AS ( SELECT '2021-02-01'::timestamptz ),
target_order as (
SELECT
tstzrange( '2021-01-01', '2021-02-01') as bill_range,
o.*
FROM ( SELECT * FROM "order_bt" WHERE sys_period #> sys_time() ) o
WHERE FALSE
OR o.agreement_id = 3385 and o.period_id = 10
),
USAGE AS ( SELECT
ocd.*,
o.agreement_id as agreement_id,
o.id AS order_id,
(dense_rank() over (PARTITION BY o.agreement_id ORDER BY o.id )) as zzzz_id,
(dense_rank() over (PARTITION BY o.agreement_id, o.id ORDER BY (ocd.ic).consumed_period )) as conf_id,
sum( ocd.item_suma ) OVER( PARTITION BY (ocd.o).agreement_id ) AS agreement_suma2,
(sum( ocd.item_suma ) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period )) AS x_suma,
(sum( ocd.item_cost ) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period )) AS x_cost,
(sum( ocd.item_suma ) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period ))::numeric( 10, 2) AS conf_suma,
(sum( ocd.item_cost ) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period ))::numeric( 10, 2) AS conf_cost,
max((ocd.ic).consumed) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period ) AS consumed,
(sum( ocd.item_suma ) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id )) AS order_suma2
FROM target_order o
LEFT JOIN order_cost_details( o.bill_range ) ocd
ON (ocd.o).id = o.id AND (ocd.ic).consumed_period && o.app_period
)
SELECT
*,
(conf_suma/6) ::numeric( 10, 2 ) as group_nds,
(SELECT sum(x) from (SELECT sum( DISTINCT conf_suma ) AS x FROM usage sub_u WHERE sub_u.agreement_id = usage.agreement_id GROUP BY agreement_id, order_id) t) as total_suma,
(SELECT sum(x) from (SELECT (sum( DISTINCT conf_suma ) /6)::numeric( 10, 2 ) AS x FROM usage sub_u WHERE sub_u.agreement_id = usage.agreement_id GROUP BY agreement_id, order_id) t) as total_nds
FROM USAGE
WINDOW w AS ( PARTITION BY usage.agreement_id ROWS CURRENT ROW EXCLUDE TIES)
ORDER BY
order_id,
conf_id
My old question

Answer 1

I found solution. See dbfiddle.
To run window function for distinct values I should get first value from each peer. To complete this I
aggregate IDs of rows for this peer
lag this aggregation by one
Mark rows that are not aggregated yet (this is first row at peer) as _distinct
sum( ) FILTER ( WHERE _distinct ) over ( ... )
Voila. You get sum over DISTINCT values at target PARTITION
which are not implemented yet by PostgreSQL
with data as (
select * from (values
( 1, 1, 1, 1.0049 ), (2, 1,1,1.0049), ( 3, 1,1,1.0049 ) ,
( 4, 1, 2, 1.0049 ), (5, 1,2,1.0057),
( 6, 2, 1, 1.53 ), ( 7,2,1,2.18), ( 8,2,2,3.48 )
) t (id, agreement_id, order_id, suma)
),
intermediate as (select
*,
sum( suma ) over ( partition by agreement_id, order_id ) as fract_order_suma,
sum( suma ) over ( partition by agreement_id ) as fract_agreement_total,
(sum( suma::numeric(10,2) ) over ( partition by agreement_id, order_id )) as wrong_order_suma,
(sum( suma ) over ( partition by agreement_id, order_id ))::numeric( 10, 2) as order_suma,
(sum( suma ) over ( partition by agreement_id ))::numeric( 10, 2) as wrong_agreement_total,
id as xid,
array_agg( id ) over ( partition by agreement_id, order_id ) as agg
from data),
distinc as (select *,
lag( agg ) over ( partition by agreement_id ) as prev,
id = any (lag( agg ) over ()) is not true as _distinct, -- allow to match first ID from next peer
order_suma as xorder_suma, -- repeat column to easily visually compare with _distinct
(SELECT sum(x) from (SELECT sum( DISTINCT order_suma ) AS x FROM intermediate sub_q WHERE sub_q.agreement_id = intermediate.agreement_id GROUP BY agreement_id, order_id) t) as correct_total_suma
from intermediate
)
select
*,
sum( order_suma ) filter ( where _distinct ) over ( partition by agreement_id ) as also_correct_total_suma
from distinc

Answer 2

better approach dbfiddle:
Assign row_number at each order: row_number() over (partition by agreement_id, order_id ) as nrow
Take only first suma: filter nrow = 1
with data as (
select * from (values
( 1, 1, 1, 1.0049 ), (2, 1,1,1.0049), ( 3, 1,1,1.0049 ) ,
( 4, 1, 2, 1.0049 ), (5, 1,2,1.0057),
( 6, 2, 1, 1.53 ), ( 7,2,1,2.18), ( 8,2,2,3.48 )
) t (id, agreement_id, order_id, suma)
),
intermediate as (select
*,
row_number() over (partition by agreement_id, order_id ) as nrow,
(sum( suma ) over ( partition by agreement_id, order_id ))::numeric( 10, 2) as order_suma,
from data)
select
*,
sum( order_suma ) filter (where nrow = 1) over (partition by agreement_id)
from intermediate```