Calculating difference between daily sum and a average for the same day of the week in defined time range. SQL 10g Oracle - sql

Question

Hi I'm working with data depending mostly on the day of the week. Data is formatted in a table
Date - position - count/number.
There are multiple different positions.
I was able to sort my data for a each day of the week using.
select MOD(to_char(time, 'J'),7),
sum(COUNT))
from TABLE
where time > sysdate -x
group by to_char(time, 'J')
order by to_char(time, 'J');
This outputs daily sums according to day of the week.
Now I'm able to get an average for a single day of a week in a year.
This code outputs an average for only Sunday
SELECT AVG(asset_sums)
FROM (
select MOD(to_char(time, 'J'),7),
sum(COUNT)) as asset_sums
from table
where time > sysdate -365
and MOD(TO_CHAR(time, 'J'), 7) + 1 IN (7)
group by to_char(time, 'J')
order by to_char(time, 'J')
);
My goal is to be able to get a table with daily sum compared with yearly average for that particular day of the week.
For example yearly average number for Mondays is 57 , Tuesdays 60.
This week my Monday is 59 and Tuesday is 57. Output of the table is
Monday +2, Tuesday -3.
What is the easiest way / most efficient ?
Thanks for your help.
Edit : Format of my data
Date : yyyy-mm-dd | Place : xxxx | Number( of customers) 0 to 10000
2013-09-16 | AAAA | 1534
2013-09-16 | AAAB | 534
2013-09-17 | AAAA | 1434
2013-09-17 | AAAC | 834
2013-09-18 | AAAA | 134
2013-09-18 | AAAD | 183
Needed output
2013-09-16 | Day of the week | Sum | Average monday this year | Difference Sum-AVG
2013-09-16 | 1 (= Monday) | 2068 | 2015| 53

Answer 1

For clarity I will use subquery factoring. First, select the current weeks data. Next, subquery the sum for the day over the current week. Then, subquery the sum for each day over the past year. Then, average the daily sum of each day for each day of the week. Finally, join the two and display the difference.
with
this_week as (
select
time
from table
where time > x - 7
group by time
),
this_week_dly_sum as (
select
to_char(time, 'd') day,
sum(count) sum
from this_week
group by to_char(time, 'd')
),
this_year_dly_sum as (
select
time,
sum(count) sum
from table
where time > x - 365
group by time
),
this_year_dly_avg as (
select
to_char(day, 'd'),
avg(sum) avg
from this_year_dly_sum
group by to_char(day, 'd')
)
select
this_week.time,
to_char(this_week.time, 'day') day of week,
this_week_dly_sum.sum,
this_year_dly_avg.avg,
this_week_dly_sum.sum - this_year_dly_avg.avg difference
from this_week
inner join this_week_dly_sum
on to_char(this_week.time, 'd') = this_week_dly_sum.day
inner join this_year_dly_avg
on to_char(this_week.time, 'd').day = this_year_dly_avg.
group by time
;

Answer 2

You can use analytic function for this.
select date1, to_char(date1, 'd'),
sum(val) over(partition by to_char(date1, 'd')),
avg(val) over(partition by to_char(date1, 'd')),
sum(val) over(partition by to_char(date1, 'd'))-
avg(val) over(partition by to_char(date1, 'd'))
from table1
time > add_month(sysdate,-12);

Answer 3

This will give you daily counts for the last year:
SELECT TRUNC(time, 'DD') AS date,
SUM(count) AS asset_sum
FROM yourtable
WHERE time > SYSDATE - 365
GROUP BY TRUNC(time, 'DD')
You can modify it to additionally return averages per day of the week for the specified range:
SELECT TRUNC(time, 'DD') AS date,
SUM(count) AS asset_sum,
AVG(SUM(count)) OVER
(PARTITION BY TO_CHAR(TRUNC(time, 'DD'), 'D')) AS asset_sum_avg
FROM yourtable
WHERE time > SYSDATE - 365
GROUP BY TRUNC(time, 'DD')
At this point you have all the initial data you need but probably for more days than necessary. You can use the above query as a derived table to limit the rows to just those where date > SYSDATE - x:
WITH last_year_by_day AS
(
SELECT TRUNC(time, 'DD') AS date,
SUM(count) AS asset_sum,
AVG(SUM(count)) OVER
(PARTITION BY TO_CHAR(TRUNC(time, 'DD'), 'D')) AS asset_sum_avg
FROM yourtable
WHERE time > SYSDATE - 365
GROUP BY TRUNC(time, 'DD')
)
SELECT date,
TO_CHAR(TRUNC(time, 'DD'), 'D') AS day_of_week,
asset_sum,
asset_sum_avg,
asset_sum - asset_sum_avg AS asset_sum_diff
FROM last_year_by_day
WHERE date > SYSDATE - x
;
As some expressions are being repeated multiple times, it can be a good idea to re-factor the query to avoid the repetition. Here's one way:
WITH last_year AS
(
SELECT TRUNC(time, 'DD') AS date,
TO_CHAR(time, 'D') AS day_of_week,
count
FROM yourtable
WHERE time > SYSDATE - 365
),
last_year_by_day AS
(
SELECT date,
day_of_week,
SUM(count) AS asset_sum,
AVG(SUM(count)) OVER (PARTITION BY day_of_week) AS asset_sum_avg
FROM last_year
GROUP BY date, day_of_week
)
SELECT date,
day_of_week,
asset_sum,
asset_sum_avg,
asset_sum - asset_sum_avg AS asset_sum_diff
FROM last_year_by_day
WHERE date > SYSDATE - x
;
One last note is about TO_CHAR('D'), which is used to obtain the day_of_week values. Since you are using a different method for the same results, you may not be aware that the results of TO_CHAR('D') are affected by the NLS_TERRITORY setting. You may want to use an ALTER SESSION statement to set NLS_TERRITORY to the value that would cause TO_CHAR('D') to return 1 for Monday, 2 for Tuesday etc. Here is the list of territories supported.