I have a table like below
id name year
--------------
1 A 2000
2 B 2000
2 B 2000
2 B 2000
5 C 2000
1 D 2001
3 E 2001
as well as you see in the year 2000 we missed id '3' and id '4' and in the year 2001 we missed id '2'. I want to generate my second table which includes missed items.
2nd table :
From-id to-id name year
--------------------------------
3 4 null 2000
2 null null 2001
Which method in a SQL query can solve my problem?
Gaps and Islands in Sequences is the name of this problem. you read this article
Here's something to get you started:
WITH cte AS
(
SELECT *
FROM
(VALUES
(1),(2),(3),(4),(5)
) Tally(number)
), cte2 as
(
SELECT DISTINCT [year]
FROM
(VALUES
(2000),(2000),(2001)
)tbl([year])
), cte3 as
(
SELECT *
FROM cte
CROSS JOIN cte2
)
SELECT *
FROM cte3
LEFT OUTER JOIN YourTable ON cte3.number = YourTable.id AND cte3.[year] = YourTable[year)
A few notes: please avoid using reserved keywords as column names (such as year).
Furthermore, since I didn't know how you'd handle multiple missing ranges I did not format the output to reflect a range. For example: What would be your expected output if only one row with id=3 would be in your table?
I'd probably use ROW_NUMBER for this
This query gives you what the correct ID should be (if I interpreted your question right):
SELECT
ROW_NUMBER() OVER (PARTITION BY yr ORDER BY name, yr) as "Correct ID", *
FROM misorder
It assigns a row number (so a number starting from 1 increasing by 1 every time the year is the same).
And to let you know which ones are missing I think this should be a working solution:
WITH missing AS
(
SELECT
ROW_NUMBER() OVER (PARTITION BY yr ORDER BY name, yr) as "Correct ID", *
FROM misorder
)
SELECT * FROM missing
WHERE "Correct ID" != "id"
It takes the first query as a base to select only those records where the assumed correct ID is not equal to the currently assigned ID. You can turn this into a query to include the ranges you mentioned, but not sure if that is really necessary.
Hope this helps.
Related
Another person asked a similar question found here:
Frequency of all combinations of values for certain column
in it they asked:
*I have a dataset in SQL Server 2012 with a column for id and value, like this:
[id] [value]
--------------
A 15
A 11
A 11
B 13
B 15
B 12
C 12
C 13
D 13
D 12
My goal is to get a frequency count of all combinations of [value], with two caveats:
Order doesn't matter, so [11,12,15] is not counted separately from [12,11,15]
Repeated values are counted separately, so [11,11,12,15] is counted separately from [11,12,15]
I'm interested in all combinations, of any length (not just pairs)
So the outcome would look like:
[combo] [frequency]
---------------------
11,11,15 1
12,13,15 1
12,13 2
I've seen answers here involving recursion that answer similar questions but where order counts, and answers here involving self-joins that yield pair-wise combinations. These come close but I'm not quite sure how to adapt for my specific needs.*
somone then responded with a pretty good response/answer
select vals, count(*) as frequency
from (select string_agg(value, ',') within group (order by value) as vals, id
from t
group by id
) i
group by vals;
The only problem I see is although 12,13,15 occurs once and 12,13,(nothing) happens twice the solution I would like is that 12,13 actually occurs 3 times (while also reporting 12,13,15 count as well)!
#Gordon Linoff
#zealous
select vals, count(*) as frequency
from (select string_agg(value, ',') within group (order by value) as vals, id
from t
group by id
) i
group by vals;
expecting counts of any instance of two or more values in a combination.
It seems that you have an overlapping groups problem here, i.e (12,13) overlaps with (12,13,15) and (11,12,13). You may solve this problem by performing a self left join as the following query.
Note that your version of (SQL Server 2012) does not support the STRING_AGG function so I used the xml path method instead.
WITH CTE AS
(
SELECT T.id
, STUFF((SELECT CONCAT(',', D.[value])
FROM table_name D
WHERE D.id = T.id
ORDER BY D.[value]
FOR XML PATH('')), 1, LEN(','), '') AS combo
FROM table_name T
GROUP BY T.id
)
SELECT A.combo, COUNT(DISTINCT B.id) frequency
FROM CTE A LEFT JOIN CTE B
ON B.combo LIKE CONCAT('%', A.combo, '%')
GROUP BY A.combo
ORDER BY MAX(A.id)
See a demo.
The output according to your input data:
combo frequency
11,11,15 1
12,13,15 1
12,13 3
I looked at other examples, but I don't know enough about SQL to adapt it to my needs. I have a table that looks like this:
ID Month NAME COUNT First LAST TOTAL
------------------------------------------------------
1 JAN2013 fred 4
2 MAR2013 fred 5
3 APR2014 fred 1
4 JAN2013 Tom 6
5 MAR2014 Tom 1
6 APR2014 Tom 1
This could be in separate queries, but I need 'First' to equal the first month that a particular name is used, so every row with fred would have JAN2013 in the first field for example. I need the 'Last" column to equal the month of the last record of each name, and finally I need the 'total' column to be the sum of all the counts for each name, so in each row that had fred the total would be 10 in this sample data. This is over my head. Can one of you assist?
This is crude but should do the trick. I renamed your fields a bit because you are using a bunch of "RESERVED" sql words and that is bad form.
;WITH cte as
(
Select
[NAME]
,[nmCOUNT]
,ROW_NUMBER() over (partition by NAME order by txtMONTH ASC) as 'FirstMonth'
,ROW_NUMBER() over (partition by NAME order by txtMONTH DESC) as 'LastMonth'
,SUM([nmCOUNT]) as 'TotNameCount'
From Table
Group by NAME, [nmCOUNT]
)
,cteFirst as
(
Select
NAME
,[nmCOUNT]
,[TotNameCount]
,[txtMONTH] as 'ansFirst'
From cte
Where FirstMonth = 1
)
,cteLast as
(
Select
NAME
,[txtMONTH] as 'ansLast'
From cte
Where LastMonth = 1
Select c.NAME, c.nmCount, c.ansFirst, l.ansLast, c.TotNameCount
From cteFirst c
LEFT JOIN cteLast l on c.NAME = l.NAME
I have a table that looks like the following but also has more columns that are not needed for this instance.
ID DATE Random
-- -------- ---------
1 4/12/2015 2
2 4/15/2015 2
3 3/12/2015 2
4 9/16/2015 3
5 1/12/2015 3
6 2/12/2015 3
ID is the primary key
Random is a foreign key but i am not actually using table it points to.
I am trying to design a query that groups the results by Random and Date and select the MAX Date within the grouping then gives me the associated ID.
IF i do the following query
select top 100 ID, Random, MAX(Date) from DateBase group by Random, Date, ID
I get duplicate Randoms since ID is the primary key and will always be unique.
The results i need would look something like this
ID DATE Random
-- -------- ---------
2 4/15/2015 2
4 9/16/2015 3
Also another question is there could be times where there are many of the same date. What will MAX do in that case?
You can use NOT EXISTS() :
SELECT * FROM YourTable t
WHERE NOT EXISTS(SELECT 1 FROM YourTable s
WHERE s.random = t.random
AND s.date > t.date)
This will select only those who doesn't have a bigger date for corresponding random value.
Can also be done using IN() :
SELECT * FROM YourTable t
WHERE (t.random,t.date) in (SELECT s.random,max(s.date)
FROM YourTable s
GROUP BY s.random)
Or with a join:
SELECT t.* FROM YourTable t
INNER JOIN (SELECT s.random,max(s.date) as max_date
FROM YourTable s
GROUP BY s.random) tt
ON(t.date = tt.max_date and s.random = t.random)
In SQL Server you could do something like the following,
select a.* from DateBase a inner join
(select Random,
MAX(dt) as dt from DateBase group by Random) as x
on a.dt =x.dt and a.random = x.random
This method will work in all versions of SQL as there are no vendor specifics (you'll need to format the dates using your vendor specific syntax)
You can do this in two stages:
The first step is to work out the max date for each random:
SELECT MAX(DateField) AS MaxDateField, Random
FROM Example
GROUP BY Random
Now you can join back onto your table to get the max ID for each combination:
SELECT MAX(e.ID) AS ID
,e.DateField AS DateField
,e.Random
FROM Example AS e
INNER JOIN (
SELECT MAX(DateField) AS MaxDateField, Random
FROM Example
GROUP BY Random
) data
ON data.MaxDateField = e.DateField
AND data.Random = e.Random
GROUP BY DateField, Random
SQL Fiddle example here: SQL Fiddle
To answer your second question:
If there are multiples of the same date, the MAX(e.ID) will simply choose the highest number. If you want the lowest, you can use MIN(e.ID) instead.
I am planing an SQL Statement right now and would need someone to look over my thougts.
This is my Table:
id stat period
--- ------- --------
1 10 1/1/2008
2 25 2/1/2008
3 5 3/1/2008
4 15 4/1/2008
5 30 5/1/2008
6 9 6/1/2008
7 22 7/1/2008
8 29 8/1/2008
Create Table
CREATE TABLE tbstats
(
id INT IDENTITY(1, 1) PRIMARY KEY,
stat INT NOT NULL,
period DATETIME NOT NULL
)
go
INSERT INTO tbstats
(stat,period)
SELECT 10,CONVERT(DATETIME, '20080101')
UNION ALL
SELECT 25,CONVERT(DATETIME, '20080102')
UNION ALL
SELECT 5,CONVERT(DATETIME, '20080103')
UNION ALL
SELECT 15,CONVERT(DATETIME, '20080104')
UNION ALL
SELECT 30,CONVERT(DATETIME, '20080105')
UNION ALL
SELECT 9,CONVERT(DATETIME, '20080106')
UNION ALL
SELECT 22,CONVERT(DATETIME, '20080107')
UNION ALL
SELECT 29,CONVERT(DATETIME, '20080108')
go
I want to calculate the difference between each statistic and the next, and then calculate the mean value of the 'gaps.'
Thougts:
I need to join each record with it's subsequent row. I can do that using the ever flexible joining syntax, thanks to the fact that I know the id field is an integer sequence with no gaps.
By aliasing the table I could incorporate it into the SQL query twice, then join them together in a staggered fashion by adding 1 to the id of the first aliased table. The first record in the table has an id of 1. 1 + 1 = 2 so it should join on the row with id of 2 in the second aliased table. And so on.
Now I would simply subtract one from the other.
Then I would use the ABS function to ensure that I always get positive integers as a result of the subtraction regardless of which side of the expression is the higher figure.
Is there an easier way to achieve what I want?
The lead analytic function should do the trick:
SELECT period, stat, stat - LEAD(stat) OVER (ORDER BY period) AS gap
FROM tbstats
The average value of the gaps can be done by calculating the difference between the first value and the last value and dividing by one less than the number of elements:
select sum(case when seqnum = num then stat else - stat end) / (max(num) - 1);
from (select period, row_number() over (order by period) as seqnum,
count(*) over () as num
from tbstats
) t
where seqnum = num or seqnum = 1;
Of course, you can also do the calculation using lead(), but this will also work in SQL Server 2005 and 2008.
By using Join also you achieve this
SELECT t1.period,
t1.stat,
t1.stat - t2.stat gap
FROM #tbstats t1
LEFT JOIN #tbstats t2
ON t1.id + 1 = t2.id
To calculate the difference between each statistic and the next, LEAD() and LAG() may be the simplest option. You provide an ORDER BY, and LEAD(something) returns the next something and LAG(something) returns the previous something in the given order.
select
x.id thisStatId,
LAG(x.id) OVER (ORDER BY x.id) lastStatId,
x.stat thisStatValue,
LAG(x.stat) OVER (ORDER BY x.id) lastStatValue,
x.stat - LAG(x.stat) OVER (ORDER BY x.id) diff
from tbStats x
Code Description Whatever
---------------------------------
1 stuff blah
1 something meh
2 yah bong
2 never hammer time
How do I get a results set from this with each Code only present once? (I don't overly care which record for that code it is).
So I want....
1 stuff blah
2 yah bong
SELECT *
FROM (
SELECT * , row_number() over(partition by code order by Description) as id
from yourTable
) temp
WHERE id = 1
I think this is sql server only
You first need to pick a column which determines what counts as 'the first result'. In my example I chose Description:
SELECT * FROM YourTable first
WHERE
(SELECT COUNT(*) FROM YourTable previous
WHERE previous.Code=first.Code AND previous.Description < first.Description) = 0