Select distinct in week and distinct overall per week - sql-server

Question

I have an example on sql fiddle. What I am trying to do is divide the overall COUNT(DISTINCT ID) by the weekly COUNT(DISTINCT ID). For example if I have the following conceptual setup of what the result should be.
year week id_set overall_distinct week_distinct result
2016 1 A,A,A,B,B,C 0 3 0
2016 2 A,B,C,C,D 1 4 .25
2016 3 A,B,C,E,F 2 5 .4
The table linked to on sql fiddle has the following schema. Also, in reality I do have multiple values for 'year'.
CREATE TABLE all_ids
([year] int, [week] int, [id] varchar(57))
;
INSERT INTO all_ids
([year], [week], [id])
VALUES
(2016, 1, 'A'),
(2016, 1, 'A'),
(2016, 1, 'A'),
(2016, 1, 'B'),
(2016, 1, 'B'),
(2016, 1, 'C'),
(2016, 2, 'A'),
(2016, 2, 'B'),
(2016, 2, 'C'),
(2016, 2, 'C'),
(2016, 2, 'D'),
(2016, 3, 'A'),
(2016, 3, 'B'),
(2016, 3, 'C'),
(2016, 3, 'E'),
(2016, 3, 'F')
;
Edit
I apologize for the confusion. The above table was just a conceptual example of the result. The actual result only needs to look like the following.
year week overall_distinct week_distinct result
2016 1 0 3 0
2016 2 1 4 .25
2016 3 2 5 .4
there is no need to include id_set

Answer 1

I used dense_rank and max() over () to simulate count (distinct ...) with window functions. You could try to do it with another subquery
select
year, week
, id_set = stuff((
select
',' + a.id
from
all_ids a
where
a.year = t.year
and a.week = t.week
order by a.id
for xml path('')
), 1, 1, '')
, overall_distinct = count(case when cnt = 1 then 1 end)
, week_distinct = count(distinct id)
, result = cast(count(case when cnt = 1 then 1 end) * 1.0 / count(distinct id) as decimal(10, 2))
from (
select
year, week, id, cnt = max(dr) over (partition by id)
from (
select
*, dr = dense_rank() over (partition by id order by year, week)
From
all_ids
) t
) t
group by year, week
Output
year week id_set overall_distinct week_distinct result
--------------------------------------------------------------------------
2016 1 A,A,A,B,B,C 0 3 0.00
2016 2 A,B,C,C,D 1 4 0.25
2016 3 A,B,C,E,F 2 5 0.40

Answer 2

This would be one way, probably not the best one:
;with weekly as
(
select year, week, count(distinct id) nr
from all_ids
group by year, week
),
overall as
(
select a.week, count(distinct a.id) nr
from all_ids a
where a.id not in (select id from all_ids where week <> a.week and id = a.id )
group by week
)
select distinct a.year
, a.week
, stuff((select ', ' + id
from all_ids
where year = a.year and week = a.week
for xml path('')), 1, 1, '') ids
, w.Nr weeklyDistinct
, isnull(t.Nr, 0) overallDistinct
from all_ids a join weekly w on a.year = w.year and a.week = w.week
left join overall t on t.week = a.week

Answer 3

One statement count only
declare #t table (y int, w int, id varchar(57));
INSERT #t (y, w, id)
VALUES
(2016, 1, 'A'),
(2016, 1, 'A'),
(2016, 1, 'A'),
(2016, 1, 'B'),
(2016, 1, 'B'),
(2016, 1, 'C'),
(2016, 2, 'A'),
(2016, 2, 'B'),
(2016, 2, 'C'),
(2016, 2, 'C'),
(2016, 2, 'D'),
(2016, 3, 'A'),
(2016, 3, 'B'),
(2016, 3, 'C'),
(2016, 3, 'E'),
(2016, 3, 'F');
select t1.w, count(distinct t1.id) as wk
, (count(distinct t1.id) - count(distinct t2.id)) as [all]
, (cast(1 as smallmoney) - cast(count(distinct t2.id) as smallmoney) / count(distinct t1.id)) as [frac]
from #t t1
left join #t t2
on t2.id = t1.id
and t2.w <> t1.w
group by t1.w
order by t1.w;