Adding new column with set of rules in SQL Server - sql

Question

I have a somewhat-complex set of rules that I need to run against a table. The problem is as follows: I have a table that stores medical records and I need to identify the first site that a person goes to after their discharge date. The discharge date is the end_date with a location of 'initial' (this will be the first row for every group). The table is grouped by ID and sorted in the format shown below.
There are 3 rules: (1) within the group of ID, if any of the rows have a begin_date that matches the first rows end_date, return that location as the first site (if there are two rows that meet this condition, either are correct, the first instance is preferred). (2) if the first option does not exist, then if there is an instance that the patient had location 'Health', then return 'Health'. (3) else, if conditions 1 and 2 do not exist, then return 'Home'
table
ID color begin_date end_date location
1 red 2017-01-01 2017-01-07 initial
1 green 2017-01-05 2017-01-07 nursing
1 blue 2017-01-07 2017-01-15 rehab
1 red 2017-01-11 2017-01-22 Health
2 red 2017-02-22 2017-02-26 initial
2 green 2017-02-26 2017-02-28 nursing
2 blue 2017-02-26 2017-02-28 rehab
3 red 2017-03-11 2017-03-22 initial
4 red 2017-04-01 2017-04-07 initial
4 green 2017-04-05 2017-04-07 nursing
4 blue 2017-04-10 2017-04-15 Health
final result:
ID first_site
1 rehab
2 nursing
3 home
4 Health
In sql-server 2008 my attempt: (side note: I considered adding a helper column which would be the end_date of the 'initial' location to each row so its easier to compare within a row. Not sure if this is necessary). I appreciate any guidance!
SELECT
ID,
OVER( PARTITION ID CASE WHEN end_date[0] = begin_date THEN location
WHEN location = 'Health' THEN 'Health'
ELSE 'Home' end) AS [first_site]
FROM table
In python, I was able to get this answer with:
def conditions(x):
#compare each group first
val = x.loc[x['begin_date'] == x['end_date'].iloc[0], 'location']
#if at least one match (not return empty `Series` get first value)
if not val.empty:
return val.iloc[0]
#if value is empty, check if value 'Health' exists within the group
elif (x['location'] == 'Health').any():
return 'Health'
else:
return 'Home'
final = df.groupby('ID').apply(conditions).reset_index(name='first_site')

Answer 1

This can be achieved with a windowed function to rank the visits that start on the same day as the initial end and then a couple simple joins:
declare #t table(ID int,color varchar(20),begin_date date,end_date date,location varchar(20));
insert into #t values(1,'red','20170101','20170107','initial'),(1,'green','20170105','20170107','nursing'),(1,'blue','20170107','20170115','rehab'),(1,'red','20170111','20170122','Health'),(2,'red','20170222','20170226','initial'),(2,'green','20170226','20170228','nursing'),(2,'blue','20170226','20170228','rehab'),(3,'red','20170311','20170322','initial'),(4,'red','20170401','20170407','initial'),(4,'green','20170405','20170407','nursing'),(4,'blue','20170410','20170415','Health');
with d as
(
select ID
,color
,begin_date
,end_date
,location
,row_number() over (partition by ID
,begin_date
order by case when location = 'initial' then '29990101' else begin_date end
) as r
from #t
)
select i.ID
,isnull(d.location,isnull(h.location,'Home')) as first_site
from d as i
left join d
on i.end_date = d.begin_date
and d.r = 1
left join d as h
on i.ID = h.ID
and h.location = 'Health'
where i.location = 'initial'
;
Output:
+----+------------+
| ID | first_site |
+----+------------+
| 1 | rehab |
| 2 | nursing |
| 3 | Home |
| 4 | Health |
+----+------------+

Answer 2

declare #example table (
ExampleID int identity(1,1) not null primary key clustered
, ID int not null
, Color nvarchar(255) not null
, BeginDate date not null
, EndDate date not null
, Loc nvarchar(255) not null
);
insert into #example (ID, color, begindate, enddate, loc)
select 1, 'red' , '2017-01-01', '2017-01-07', 'initial' union all
select 1, 'green' , '2017-01-05', '2017-01-07', 'nursing' union all
select 1, 'blue' , '2017-01-07', '2017-01-15', 'rehab' union all
select 1, 'red' , '2017-01-11', '2017-01-22', 'Health' union all
select 2, 'red' , '2017-02-22', '2017-02-26', 'initial' union all
select 2, 'green' , '2017-02-26', '2017-02-28', 'nursing' union all
select 2, 'blue' , '2017-02-26', '2017-02-28', 'rehab' union all
select 3, 'red' , '2017-03-11', '2017-03-22', 'initial' union all
select 4, 'red' , '2017-04-01', '2017-04-07', 'initial' union all
select 4, 'green' , '2017-04-05', '2017-04-07', 'nursing' union all
select 4, 'blue' , '2017-04-10', '2017-04-15', 'Health';
with cte as (
select a.ID
, a.Color
, a.BeginDate
, a.EndDate
, b.Loc
, rank() over(partition by a.ID order by a.ID, a.begindate, b.enddate desc, b.loc) Ranking
from #example a
left join #example b
on a.EndDate = b.BeginDate
)
, cte2 as (
select id
, Loc
from #example
where loc = 'health'
)
select a.ID
, COALESCE(a.loc, b.loc, 'Home') as Loc
from cte a
left join cte2 b
on a.id = b.id
where Ranking = 1
Output:
ID Loc
1 rehab
2 nursing
3 home
4 Health