SQL - cumulative sum in postgres
0
I have my data like this:
item - initial_value - amount - dateofpurchase
A 100 -3 2018-11-22
A 100 -2 2018-11-22
B 200 -5 2018-11-22
B 200 6 2018-11-22
B 200 -1 2018-11-22
(everything is ordered by date and time)
I want to calculate this column, that shows how much stock do you have after each step and taking in count the last amount
item - initial_value - amount - dateofpurchase - cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
I've been trying a sum function with unbounded preceding and current row with no luck
sql postgresql
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
add a comment |
0
I have my data like this:
item - initial_value - amount - dateofpurchase
A 100 -3 2018-11-22
A 100 -2 2018-11-22
B 200 -5 2018-11-22
B 200 6 2018-11-22
B 200 -1 2018-11-22
(everything is ordered by date and time)
I want to calculate this column, that shows how much stock do you have after each step and taking in count the last amount
item - initial_value - amount - dateofpurchase - cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
I've been trying a sum function with unbounded preceding and current row with no luck
sql postgresql
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
add a comment |
0
0
0
2
I have my data like this:
item - initial_value - amount - dateofpurchase
A 100 -3 2018-11-22
A 100 -2 2018-11-22
B 200 -5 2018-11-22
B 200 6 2018-11-22
B 200 -1 2018-11-22
(everything is ordered by date and time)
I want to calculate this column, that shows how much stock do you have after each step and taking in count the last amount
item - initial_value - amount - dateofpurchase - cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
I've been trying a sum function with unbounded preceding and current row with no luck
sql postgresql
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
I have my data like this:
item - initial_value - amount - dateofpurchase
A 100 -3 2018-11-22
A 100 -2 2018-11-22
B 200 -5 2018-11-22
B 200 6 2018-11-22
B 200 -1 2018-11-22
(everything is ordered by date and time)
I want to calculate this column, that shows how much stock do you have after each step and taking in count the last amount
item - initial_value - amount - dateofpurchase - cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
I've been trying a sum function with unbounded preceding and current row with no luck
sql postgresql
sql postgresql
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
asked Nov 23 '18 at 12:49
Naty BizzNaty Bizz
80951841
80951841
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
1
You can use window functions and subtraction:
select t.*,
( initial_amount +
sum(amount) over (partition by item order by date_of_purchase)
) as cumulative
from t;
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
add a comment |
0
use window function
with cte as
(
select 'A' item, 100 as initial_value, -3 amount, '2018-11-22'::date as dateofpurchase
union all
select 'A' ,100, -2, '2018-11-22'
union all
select 'B',200, -5,'2018-11-22'
union all
select 'B',200, 6,'2018-11-22'
union all
select 'B',200, -1,'2018-11-22'
)
, t1 as
(select t.*, row_number() over(partition by item order by dateofpurchase) rn
from cte t
)
, t3 as
(select *, case when rn=1 then initial_value else 0 end as val from t1
) select item,initial_value,amount,dateofpurchase, sum(val+amount) over(partition by item order by rn) as cumulative from t3
Sample output
item initial_value amount dateofpurchase cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
demo link
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You can use window functions and subtraction:
select t.*,
( initial_amount +
sum(amount) over (partition by item order by date_of_purchase)
) as cumulative
from t;
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
add a comment |
1
You can use window functions and subtraction:
select t.*,
( initial_amount +
sum(amount) over (partition by item order by date_of_purchase)
) as cumulative
from t;
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
add a comment |
1
1
1
You can use window functions and subtraction:
select t.*,
( initial_amount +
sum(amount) over (partition by item order by date_of_purchase)
) as cumulative
from t;
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
You can use window functions and subtraction:
select t.*,
( initial_amount +
sum(amount) over (partition by item order by date_of_purchase)
) as cumulative
from t;
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
answered Nov 23 '18 at 12:51
Gordon LinoffGordon Linoff
771k35304404
771k35304404
add a comment |
add a comment |
0
use window function
with cte as
(
select 'A' item, 100 as initial_value, -3 amount, '2018-11-22'::date as dateofpurchase
union all
select 'A' ,100, -2, '2018-11-22'
union all
select 'B',200, -5,'2018-11-22'
union all
select 'B',200, 6,'2018-11-22'
union all
select 'B',200, -1,'2018-11-22'
)
, t1 as
(select t.*, row_number() over(partition by item order by dateofpurchase) rn
from cte t
)
, t3 as
(select *, case when rn=1 then initial_value else 0 end as val from t1
) select item,initial_value,amount,dateofpurchase, sum(val+amount) over(partition by item order by rn) as cumulative from t3
Sample output
item initial_value amount dateofpurchase cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
demo link
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
add a comment |
0
use window function
with cte as
(
select 'A' item, 100 as initial_value, -3 amount, '2018-11-22'::date as dateofpurchase
union all
select 'A' ,100, -2, '2018-11-22'
union all
select 'B',200, -5,'2018-11-22'
union all
select 'B',200, 6,'2018-11-22'
union all
select 'B',200, -1,'2018-11-22'
)
, t1 as
(select t.*, row_number() over(partition by item order by dateofpurchase) rn
from cte t
)
, t3 as
(select *, case when rn=1 then initial_value else 0 end as val from t1
) select item,initial_value,amount,dateofpurchase, sum(val+amount) over(partition by item order by rn) as cumulative from t3
Sample output
item initial_value amount dateofpurchase cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
demo link
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
add a comment |
0
0
0
use window function
with cte as
(
select 'A' item, 100 as initial_value, -3 amount, '2018-11-22'::date as dateofpurchase
union all
select 'A' ,100, -2, '2018-11-22'
union all
select 'B',200, -5,'2018-11-22'
union all
select 'B',200, 6,'2018-11-22'
union all
select 'B',200, -1,'2018-11-22'
)
, t1 as
(select t.*, row_number() over(partition by item order by dateofpurchase) rn
from cte t
)
, t3 as
(select *, case when rn=1 then initial_value else 0 end as val from t1
) select item,initial_value,amount,dateofpurchase, sum(val+amount) over(partition by item order by rn) as cumulative from t3
Sample output
item initial_value amount dateofpurchase cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
demo link
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
use window function
with cte as
(
select 'A' item, 100 as initial_value, -3 amount, '2018-11-22'::date as dateofpurchase
union all
select 'A' ,100, -2, '2018-11-22'
union all
select 'B',200, -5,'2018-11-22'
union all
select 'B',200, 6,'2018-11-22'
union all
select 'B',200, -1,'2018-11-22'
)
, t1 as
(select t.*, row_number() over(partition by item order by dateofpurchase) rn
from cte t
)
, t3 as
(select *, case when rn=1 then initial_value else 0 end as val from t1
) select item,initial_value,amount,dateofpurchase, sum(val+amount) over(partition by item order by rn) as cumulative from t3
Sample output
item initial_value amount dateofpurchase cumulative
A 100 -3 2018-11-22 97
A 100 -2 2018-11-22 95
B 200 -5 2018-11-22 195
B 200 6 2018-11-22 201
B 200 -1 2018-11-22 200
demo link
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
edited Nov 23 '18 at 13:15
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
answered Nov 23 '18 at 12:51
Zaynul Abadin TuhinZaynul Abadin Tuhin
12.9k2932
12.9k2932
add a comment |
add a comment |
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