I have a table:
spent totalsent totalused
----------------------------
4 1234 123
6 12 4
7 45 32
I need to group totalused/totalsent in groups of 0 <= spent <= 5, 6 <= spent <= 10 and so on.
How can this be done?
If you are just wanting groups of 5, you can probably just set up a column to group on which divides by 5.
with my_source_data as (
select 4 as spent, 1234 as totalsent, 123 as totalused from dual union all
select 6 as spent, 12 as totalsent, 4 as totalused from dual union all
select 7 as spent, 45 as totalsent, 32 as totalused from dual
)
select
(spent_group -1) * 5 + 1 as lower_bound,
spent_group * 5 as upper_bound, totalsent, totalused
from (
select
greatest(ceil(spent/5),1) as spent_group,
sum(totalsent) as totalsent, sum(totalused) totalused
from my_source_data
group by greatest(ceil(spent/5),1)
)
This code doesn't quite handle 0 or anything below 0 correctly since it puts everything in the lower group and labels it 1-5, but your requirements are a little vague in that respect.
Also, this is a sparse grouping so there will only be a row for 11-15 if there is source data which allows it to be produced.
Related
My database is an amazon redshift.
I have a table that looks like this -
id
nested_id
date
value
1
10
'2021-01-01'
5
1
20
'2021-01-01'
10
1
10
'2021-01-02'
6
1
20
'2021-01-02'
11
1
10
'2021-01-03'
7
1
20
'2021-01-03'
12
2
30
'2021-01-01'
5
2
40
'2021-01-01'
10
2
30
'2021-01-02'
6
2
40
'2021-01-02'
11
2
30
'2021-01-03'
7
2
40
'2021-01-03'
12
So this is basically a table that tracks values by id over time, except for every id there can be a nested_id. And the dates and values are primarily connected to the nested_id.
However, let's say I'm starting with the id field, but for each id I want to only return the points over time for the nested_id that has the greater sum of points.
So right now I'm just grabbing it like this...
select *
from mytable
where id in (1, 2)
except I only want it to return nested_id rows where the maximum value of that nested_id is the greatest.
So here's how I would do this manually.
For id of 1, the maximum value is 12, and the nested_id of that value is 20
For id of 2, the maximum value is 12, and the nested_id of that value is 40
So my return table should be
id
nested_id
date
value
1
20
'2021-01-01'
10
1
20
'2021-01-02'
11
1
20
'2021-01-03'
12
2
40
'2021-01-01'
10
2
40
'2021-01-02'
11
2
40
'2021-01-03'
12
Is there an easy way of performing this query? I'm assuming you have to partition somehow?
You can solve this with row_number window functions
with maxs as (
select id,
nested_id,
value,
row_number() over (partition by id order by value desc) rn
from mytable
)
select mt.*
from mytable mt
left join maxs on mt.id = maxs.id and mt.nested_id = maxs.nested_id
where maxs.rn = 1
I have the following data in the table
Period Total_amount R_total
01/01/20 2 2
01/02/20 5 null
01/03/20 3 null
01/04/20 8 null
01/05/20 31 null
Based on the above data I would like to have the following situation.
Period Total_amount R_total
01/01/20 2 2
01/02/20 5 3
01/03/20 3 0
01/04/20 8 8
01/05/20 31 23
Additional data
01/06/20 21 0 (previously it would be -2)
01/07/20 25 25
01/08/20 29 4
Pattern to the additional data is:
if total_amount < previous(r_total) then 0
Based on the filled data, we can spot the pattern is:
R_total = total_amount - previous(R_total)
Could you please help me out with this issue?
As Gordon Linoff suspected, it is possible to solve this problem with analytic functions. The benefit is that the query will likely be much faster. The price to pay for that benefit is that you need to do a bit of math beforehand (before ever thinking about "programming" and "computers").
A bit of elementary arithmetic shows that R_TOTAL is an alternating sum of TOTAL_AMOUNT. This can be arranged easily by using ROW_NUMBER() (to get the signs) and then an analytic SUM(), as shown below.
Table setup:
create table sample_data (period, total_amount) as
select to_date('01/01/20', 'mm/dd/rr'), 2 from dual union all
select to_date('01/02/20', 'mm/dd/rr'), 5 from dual union all
select to_date('01/03/20', 'mm/dd/rr'), 3 from dual union all
select to_date('01/04/20', 'mm/dd/rr'), 8 from dual union all
select to_date('01/05/20', 'mm/dd/rr'), 31 from dual
;
Query and result:
with
prep (period, total_amount, sgn) as (
select period, total_amount,
case mod(row_number() over (order by period), 2) when 0 then 1 else -1 end
from sample_data
)
select period, total_amount,
sgn * sum(sgn * total_amount) over (order by period) as r_total
from prep
;
PERIOD TOTAL_AMOUNT R_TOTAL
-------- ------------ ----------
01/01/20 2 2
01/02/20 5 3
01/03/20 3 0
01/04/20 8 8
01/05/20 31 23
This may be possible with window functions, but the simplest method is probably a recursive CTE:
with t as (
select t.*, row_number() over (order by period) as seqnum
from yourtable t
),
cte(period, total_amount, r_amount, seqnum) as (
select period, total_amount, r_amount, seqnum
from t
where seqnum = 1
union all
select t.period, t.total_amount, t.total_amount - cte.r_amount, t.seqnum
from cte join
t
on t.seqnum = cte.seqnum + 1
)
select *
from cte;
This question explicitly talks about "recursively" adding values. If you want to solve this using another mechanism, you might explain the logic in detail and ask if there is a non-recursive CTE solution.
Suppose I have my table like:
uid day_used_app
--- -------------
1 2012-04-28
1 2012-04-29
1 2012-04-30
2 2012-04-29
2 2012-04-30
2 2012-05-01
2 2012-05-21
2 2012-05-22
Suppose I want the number of unique users who returned to the app at least 2 different days in the last 7 days (from 2012-05-03).
So as an example to retrieve the number of users who have used the application on at least 2 different days in the past 7 days:
select count(distinct case when num_different_days_on_app >= 2
then uid else null end) as users_return_2_or_more_days
from (
select uid,
count(distinct day_used_app) as num_different_days_on_app
from table
where day_used_app between current_date() - 7 and current_date()
group by 1
)
This gives me:
users_return_2_or_more_days
---------------------------
2
The question I have is:
What if I want to do this for every day up to now so that my table looks like this, where the second field equals the number of unique users who returned 2 or more different days within a week prior to the date in the first field.
date users_return_2_or_more_days
-------- ---------------------------
2012-04-28 2
2012-04-29 2
2012-04-30 3
2012-05-01 4
2012-05-02 4
2012-05-03 3
Would this help?
WITH
-- your original input, don't use in "real" query ...
input(uid,day_used_app) AS (
SELECT 1,DATE '2012-04-28'
UNION ALL SELECT 1,DATE '2012-04-29'
UNION ALL SELECT 1,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-04-29'
UNION ALL SELECT 2,DATE '2012-04-30'
UNION ALL SELECT 2,DATE '2012-05-01'
UNION ALL SELECT 2,DATE '2012-05-21'
UNION ALL SELECT 2,DATE '2012-05-22'
)
-- end of input, start "real" query here, replace ',' with 'WITH'
,
one_week_b4 AS (
SELECT
uid
, day_used_app
, day_used_app -7 AS day_used_1week_b4
FROM input
)
SELECT
one_week_b4.uid
, one_week_b4.day_used_app
, count(*) AS users_return_2_or_more_days
FROM one_week_b4
JOIN input
ON input.day_used_app BETWEEN one_week_b4.day_used_1week_b4 AND one_week_b4.day_used_app
GROUP BY
one_week_b4.uid
, one_week_b4.day_used_app
HAVING count(*) >= 2
ORDER BY 1;
Output is:
uid|day_used_app|users_return_2_or_more_days
1|2012-04-29 | 3
1|2012-04-30 | 5
2|2012-04-29 | 3
2|2012-04-30 | 5
2|2012-05-01 | 6
2|2012-05-22 | 2
Does that help your needs?
Marco the Sane ...
SELECT DISTINCT
t1.day_used_app,
(
SELECT SUM(CASE WHEN t.num_visits >= 2 THEN 1 ELSE 0 END)
FROM
(
SELECT uid,
COUNT(DISTINCT day_used_app) AS num_visits
FROM table
WHERE day_used_app BETWEEN t1.day_used_app - 7 AND t1.day_used_app
GROUP BY uid
) t
) AS users_return_2_or_more_days
FROM table t1
I have an SQL Table in which I keep project information coming from primavera.
Suppose that i have columns for Start Date,End Date,Duration, and Total Qty as shown below .
How can i distribute Total Qty over Months using these information. What kind of additional columns, sql queries i need in order to get correct monthly distribution?
Thanks in Advance.
Columns in order:
itemname,quantity,startdate,duration,enddate
item1 -- 108 -- 2013-03-25 -- 720 -- 2013-07-26
item2 -- 640 -- 2013-03-25 -- 720 -- 2013-07-26
.
.
I think the key is to break the records apart by month. Here is an example of how to do it:
with months as (
select 1 as mon union all select 2 union all select 3 union all
select 4 as mon union all select 5 union all select 6 union all
select 7 as mon union all select 8 union all select 9 union all
select 10 as mon union all select 11 union all select 12
)
select item, m.mon, quantity / nummonths
from (select t.*, (month(enddate) - month(startdate) + 1) as nummonths
from t
) t join
months m
on month(t.startDate) <= m.mon and
months(t.endDate) >= m.mon;
This works because all the months are within the same year -- as in your example. You are quite vague on how the split should be calculated. So, I assumed that every month from the start to the end gets an equal amount.
I have two sets of pricing data (A and B). Set A consists of all of my pricing data per order over a month. Set B consists of all of my competitor's pricing data over the same month. I want to compare my competitor's lowest price to each of my prices per day.
Graphically, the data appears like this:
Date:-- Set A: -- Set B:
1---------25---------31
1---------54---------47
1---------23---------56
1---------12---------23
1---------76---------40
1---------42
I want pass only the lowest price to a case statement which evaluates which prices are better. I would like to process an entire month's worth of data all at one time, so in my example, Dates 1 thru 30(1) would be included and crunched all at once, and for each day, there would only be one value from set B included: the lowest price in the set.
Important notes: Set B does not have a datapoint for each point in Set A
Hopefully this makes sense. Thanks in advance for any help you may be able to render.
That's a strange example you have - do you really have prices ranging from 12 to 76 within a single day?
Anyway, left joining your (grouped) data with their (grouped) data should work (untested):
with
my_prices as (
select price_date, min(price_value) min_price from my_prices group by price_date),
their_prices as (
select price_date, min(price_value) min_price from their_prices group by price_date)
select
mine.price_date,
(case
when theirs.min_price is null then mine.min_price
when theirs.min_price >= mine.min_price then mine.min_price
else theirs.min_price
end) min_price
from
my_min_prices mine
left join their_prices theirs on mine.price_date = theirs.price_date
I'm still not sure that I understand your requirements. My best guess is that you want something like
SQL> ed
Wrote file afiedt.buf
1 with your_data as (
2 select 1 date_id, 25 price_a,31 price_b from dual
3 union all
4 select 1, 54, 47 from dual union all
5 select 1, 23, 56 from dual union all
6 select 1, 12, 23 from dual union all
7 select 1, 76, 40 from dual union all
8 select 1, 42, null from dual)
9 select date_id,
10 sum( case when price_a < min_price_b
11 then 1
12 else 0
13 end) better,
14 sum( case when price_a = min_price_b
15 then 1
16 else 0
17 end) tie,
18 sum( case when price_a > min_price_b
19 then 1
20 else 0
21 end) worse
22 from( select date_id,
23 price_a,
24 min(price_b) over (partition by date_id) min_price_b
25 from your_data )
26* group by date_id
SQL> /
DATE_ID BETTER TIE WORSE
---------- ---------- ---------- ----------
1 1 1 4