Merge two count datename queries - sql-server

Question

Query
SELECT
DATENAME(dw, O1.CreationDateTime) AS 'Day of the week',
DATENAME(hh, O1.CreationDateTime) AS hour,
COUNT(*) AS 'Nr. Orders'
FROM
Orders AS O1
WHERE
O1.CreationDateTime >= '2015-10-01'
GROUP BY
DATENAME(dw, O1.CreationDateTime), DATENAME(hh, O1.CreationDateTime)
ORDER BY
[Day of the week], CAST(DATENAME(hh, O1.CreationDateTime) as int)
Gives the total number of orders for each hour of each weekday
Result
Weekday Hour Nr.Orders
Friday 0 10
Friday 1 11
Friday 2 2
The goal is to get the AVERAGE amount of orders for each hour of the day. However some days have no orders those days should not be counted.
To solve this problem I wrote the select distinct query below.
Query
SELECT
DATENAME(dw,Formatteddate.Date) AS "Day of the week",
COUNT(*) AS "Weekdays in period"
FROM
(SELECT DISTINCT
((CONVERT (date,O2.CreationDateTime,112))) AS Date
FROM Orders AS O2
WHERE
O2.CreationDateTime >= '2015-10-01'
) AS Formatteddate
GROUP BY DATENAME(dw,Formatteddate.Dates)
It counts for example the number of fridays with orders like shown below.
Result
Day of the week | Week days in period (with order)
Friday 32
Monday 30
Saturday 21
Basically my questions is how two merge those two query's I used outer
apply as a workaround. However it is very slow due to the fact it
checks each row. I think I need to use the over clause but I have no
idea how with the datename function.
The end result should look like this to make counting the average easy.
Weekday Hour Nr.Orders Weekdays in period
Friday 0 10 32
Friday 1 11 32
Friday 2 2 32
Friday 3 2 32

Answer 1

You can join the two tables together on the day of the week:
SELECT t1.[Day of the week] AS Weekday,
t1.hour,
t1.[Nr. Orders],
COALESCE(t2.[Weekdays in period], 0)
FROM
(
SELECT DATENAME(dw, O1.CreationDateTime) AS 'Day of the week',
DATENAME(hh, O1.CreationDateTime) AS hour,
COUNT(*) AS 'Nr. Orders'
FROM Orders AS O1
WHERE O1.CreationDateTime >= '2015-10-01'
GROUP BY DATENAME(dw, O1.CreationDateTime),
DATENAME(hh, O1.CreationDateTime)
) t1
LEFT JOIN
(
SELECT DATENAME(dw,Formatteddate.Date) AS "Day of the week",
COUNT(*) AS "Weekdays in period"
FROM
(
SELECT DISTINCT CONVERT (date,O2.CreationDateTime, 112) AS Date
FROM Orders AS O2
WHERE O2.CreationDateTime >= '2015-10-01'
) AS Formatteddate
GROUP BY DATENAME(dw,Formatteddate.Dates)
) t2
ON t1.[Day of the week] = t2.[Day of the week]

Answer 2

Try joining on the day:
WITH DayO AS (
SELECT
DATENAME(dw,Formatteddate.Date) AS DayOfWeek,
COUNT(*) AS PeriodDays
FROM
(SELECT DISTINCT
((CONVERT (date,O2.CreationDateTime,112))) AS Date
FROM Orders AS O2
WHERE
O2.CreationDateTime >= '2015-10-01'
) AS Formatteddate
GROUP BY DATENAME(dw,Formatteddate.Dates)
)
, HourO AS (
SELECT
DATENAME(dw, O1.CreationDateTime) AS DayOfWeek,
DATENAME(hh, O1.CreationDateTime) AS OrdHour,
COUNT(*) AS NumOrd
FROM
Orders AS O1
WHERE
O1.CreationDateTime >= '2015-10-01'
GROUP BY
DATENAME(dw, O1.CreationDateTime), DATENAME(hh, O1.CreationDateTime)
ORDER BY
[DayOfWeek], CAST(DATENAME(hh, O1.CreationDateTime) as int)
)
SELECT ho.DayOfWeek as Weekday, ho.OrdHour as Hour, ho.NumOrd as [Nr. Orders], DayO.PeriodDays as [Weekdays in period]
from HourO ho
inner join DayO
on ho.DayOfWeek = DayO.DayOfWeek