I am trying to get a total hours from a dataset and because you can have the same asset with the same company (company_B) twice at two different times I have this join issue. I know I want the min for company_B gone and the Max for company_B gone because they represent wrong dates being matched. The negative is easy but what about the Max?
I have:
AssetID------StartDate-------FinishDate-------CompanyName----HoursOnSite
22222-------2016-02-12-------2016-02-20-------Company_A--------192
22222-------2016-02-01-------2016-02-09-------Company_B--------208 (keep)
22222-------2016-02-12-------2016-02-09-------Company_B-------(-56) (remove)
22222-------2016-02-01-------2016-02-21-------Company_B--------480 (remove)
22222-------2016-02-12-------2016-02-21-------Company_B--------216 (keep)
55555-------2016-02-18-------2016-02-22-------Company_C--------96
99584-------2016-02-22-------2016-02-25-------Company_D--------63
I think you can do the query for the records with max and min HoursOnSite for company B, and use (not in) or not equal to exclude those records.
If you still have concern, please paste your query.
I'm assuming that there has to be atleast 3 instances of unique assetid - companyname combination for the Max, Min filters to work. You can change it in the final where statement tO suit your requirement
WITH CTE
AS (
SELECT *
,count(CompanyName) OVER (PARTITION BY AssetID,CompanyName) AS a
FROM <TABLE_NAME>
)
SELECT *
FROM CTE
WHERE HoursOnSite NOT IN (
SELECT MAX(HoursOnSite)
FROM <TABLE_NAME>
)
AND gdp NOT IN (
SELECT min(HoursOnSite)
FROM <TABLE_NAME>
)
AND a > 2 --MODIFY AS PER YOUR REQUIREMENT
Im looking for something like SELECT PRODUCT(table.price) FROM table GROUP BY table.sale similar to how SUM works.
Have I missed something on the documentation, or is there really no PRODUCT function?
If so, why not?
Note: I looked for the function in postgres, mysql and mssql and found none so I assumed all sql does not support it.
For MSSQL you can use this. It can be adopted for other platforms: it's just maths and aggregates on logarithms.
SELECT
GrpID,
CASE
WHEN MinVal = 0 THEN 0
WHEN Neg % 2 = 1 THEN -1 * EXP(ABSMult)
ELSE EXP(ABSMult)
END
FROM
(
SELECT
GrpID,
--log of +ve row values
SUM(LOG(ABS(NULLIF(Value, 0)))) AS ABSMult,
--count of -ve values. Even = +ve result.
SUM(SIGN(CASE WHEN Value < 0 THEN 1 ELSE 0 END)) AS Neg,
--anything * zero = zero
MIN(ABS(Value)) AS MinVal
FROM
Mytable
GROUP BY
GrpID
) foo
Taken from my answer here: SQL Server Query - groupwise multiplication
I don't know why there isn't one, but (take more care over negative numbers) you can use logs and exponents to do:-
select exp (sum (ln (table.price))) from table ...
There is no PRODUCT set function in the SQL Standard. It would appear to be a worthy candidate, though (unlike, say, a CONCATENATE set function: it's not a good fit for SQL e.g. the resulting data type would involve multivalues and pose a problem as regards first normal form).
The SQL Standards aim to consolidate functionality across SQL products circa 1990 and to provide 'thought leadership' on future development. In short, they document what SQL does and what SQL should do. The absence of PRODUCT set function suggests that in 1990 no vendor though it worthy of inclusion and there has been no academic interest in introducing it into the Standard.
Of course, vendors always have sought to add their own functionality, these days usually as extentions to Standards rather than tangentally. I don't recall seeing a PRODUCT set function (or even demand for one) in any of the SQL products I've used.
In any case, the work around is fairly simple using log and exp scalar functions (and logic to handle negatives) with the SUM set function; see #gbn's answer for some sample code. I've never needed to do this in a business application, though.
In conclusion, my best guess is that there is no demand from SQL end users for a PRODUCT set function; further, that anyone with an academic interest would probably find the workaround acceptable (i.e. would not value the syntactic sugar a PRODUCT set function would provide).
Out of interest, there is indeed demand in SQL Server Land for new set functions but for those of the window function variety (and Standard SQL, too). For more details, including how to get involved in further driving demand, see Itzik Ben-Gan's blog.
You can perform a product aggregate function, but you have to do the maths yourself, like this...
SELECT
Exp(Sum(IIf(Abs([Num])=0,0,Log(Abs([Num])))))*IIf(Min(Abs([Num]))=0,0,1)*(1-2*(Sum(IIf([Num]>=0,0,1)) Mod 2)) AS P
FROM
Table1
Source: http://productfunctionsql.codeplex.com/
There is a neat trick in T-SQL (not sure if it's ANSI) that allows to concatenate string values from a set of rows into one variable. It looks like it works for multiplying as well:
declare #Floats as table (value float)
insert into #Floats values (0.9)
insert into #Floats values (0.9)
insert into #Floats values (0.9)
declare #multiplier float = null
select
#multiplier = isnull(#multiplier, '1') * value
from #Floats
select #multiplier
This can potentially be more numerically stable than the log/exp solution.
I think that is because no numbering system is able to accommodate many products. As databases are designed for large number of records, a product of 1000 numbers would be super massive and in case of floating point numbers, the propagated error would be huge.
Also note that using log can be a dangerous solution. Although mathematically log(a*b) = log(a)*log(b), it might not be in computers as we are not dealing with real numbers. If you calculate 2^(log(a)+log(b)) instead of a*b, you may get unexpected results. For example:
SELECT 9999999999*99999999974482, EXP(LOG(9999999999)+LOG(99999999974482))
in Sql Server returns
999999999644820000025518, 9.99999999644812E+23
So my point is when you are trying to do the product do it carefully and test is heavily.
One way to deal with this problem (if you are working in a scripting language) is to use the group_concat function.
For example, SELECT group_concat(table.price) FROM table GROUP BY table.sale
This will return a string with all prices for the same sale value, separated by a comma.
Then with a parser you can get each price, and do a multiplication. (In php you can even use the array_reduce function, in fact in the php.net manual you get a suitable example).
Cheers
Another approach based on fact that the cardinality of cartesian product is product of cardinalities of particular sets ;-)
⚠ WARNING: This example is just for fun and is rather academic, don't use it in production! (apart from the fact it's just for positive and practically small integers)⚠
with recursive t(c) as (
select unnest(array[2,5,7,8])
), p(a) as (
select array_agg(c) from t
union all
select p.a[2:]
from p
cross join generate_series(1, p.a[1])
)
select count(*) from p where cardinality(a) = 0;
The problem can be solved using modern SQL features such as window functions and CTEs. Everything is standard SQL and - unlike logarithm-based solutions - does not require switching from integer world to floating point world nor handling nonpositive numbers. Just number rows and evaluate product in recursive query until no row remain:
with recursive t(c) as (
select unnest(array[2,5,7,8])
), r(c,n) as (
select t.c, row_number() over () from t
), p(c,n) as (
select c, n from r where n = 1
union all
select r.c * p.c, r.n from p join r on p.n + 1 = r.n
)
select c from p where n = (select max(n) from p);
As your question involves grouping by sale column, things got little bit complicated but it's still solvable:
with recursive t(sale,price) as (
select 'multiplication', 2 union
select 'multiplication', 5 union
select 'multiplication', 7 union
select 'multiplication', 8 union
select 'trivial', 1 union
select 'trivial', 8 union
select 'negatives work', -2 union
select 'negatives work', -3 union
select 'negatives work', -5 union
select 'look ma, zero works too!', 1 union
select 'look ma, zero works too!', 0 union
select 'look ma, zero works too!', 2
), r(sale,price,n,maxn) as (
select t.sale, t.price, row_number() over (partition by sale), count(1) over (partition by sale)
from t
), p(sale,price,n,maxn) as (
select sale, price, n, maxn
from r where n = 1
union all
select p.sale, r.price * p.price, r.n, r.maxn
from p
join r on p.sale = r.sale and p.n + 1 = r.n
)
select sale, price
from p
where n = maxn
order by sale;
Result:
sale,price
"look ma, zero works too!",0
multiplication,560
negatives work,-30
trivial,8
Tested on Postgres.
Here is an oracle solution for anyone who needs it
with data(id, val) as(
select 1,1.0 from dual union all
select 2,-2.0 from dual union all
select 3,1.0 from dual union all
select 4,2.0 from dual
),
neg(val , modifier) as(
select exp(sum(ln(abs(val)))), case when mod(count(*),2) = 0 then 1 Else -1 end
from data
where val <0
)
,
pos(val) as (
select exp(sum(ln(val)))
from data
where val >=0
)
select (select val*modifier from neg)*(select val from pos) product from dual
SELECT firstpartno, nOccurrence, nMale, nFemale, COUNT(nMale) / CAST
((SELECT SUM(nOccurrence) AS Expr1
FROM (SELECT COUNT(dbo.vw_Tally1.nMale) AS nOccurrence
FROM dbo.vw_Split4) AS SumTally) AS decimal) AS nMProportion, COUNT(nFemale) / CAST
((SELECT SUM(nOccurrence) AS Expr1
FROM (SELECT COUNT(dbo.vw_Tally1.nFemale) AS nOccurrence
FROM dbo.vw_Split4 AS vw_Split4_1) AS SumTally_1) AS decimal) AS nFProportion
FROM dbo.vw_Tally1
GROUP BY firstpartno, nOccurrence, nMale, nFemale
If i understood your question here's the solution for you :
SELECT
firstpartno
,nOccurrence
,nMale
,nFemale
,CASE WHEN SUM_nOccurrence.SUM_nOccurrenceMale = 0
THEN 0
ELSE COUNT(nMale)/SUM_nOccurrence.SUM_nOccurrenceMale
END AS nMProportion
,CASE WHEN SUM_nOccurrence.nOccurrenceFemale = 0
THEN 0
ELSE COUNT(nFemale)/SUM_nOccurrence.nOccurrenceFemale
END AS nFProportion
FROM
dbo.vw_Tally1
LEFT JOIN
(SELECT
CAST(SUM(nOccurrenceMale)AS decimal) AS SUM_nOccurrenceMale
,CAST(SUM(nOccurrenceFemale)AS decimal) AS SUM_nOccurrenceFemale
FROM (SELECT
COUNT(dbo.vw_Tally1.nMale) AS nOccurrenceMale
,COUNT(dbo.vw_Tally1.nFemale) AS nOccurrenceFemale
FROM dbo.vw_Split4 ) AS SumTally) SUM_nOccurrence
ON 1=1
GROUP BY
firstpartno
,nOccurrence
,nMale
,nFemale
I hope this will help you
Good Luck :)
The query looks dubious to say the least. You select all records of table vw_Tally1. For each of these records you do the following:
select COUNT(vw_Tally1.nMale) from table vw_Split4. This is COUNT(*) of vw_Split4 when vw_Tally1.nMale is not null, otherwise it is null.
Then you sum this value. Which makes no sense, as the sum of a value is the value itself.
You do the same for nFemale.
At last you group by (firstpartno, nOccurrence, nMale, nFemale) and use the values found so strangly to calculate something. As you don't aggregate the found values, you get a random match per group. I.e. the dbms takes one of the matching records. As nMale and nFemale are grouping columns, the values are constant for all records of the group. So no big problem, but a lot of useless work.
So to speed this up, first think of what you want to select actually. This looks like to become a very simple select statement in the end. We can help you, if you tell us what your tables contain, what result set you are after, what does nMale and nFemale stand for, and what are the primary keys or unique columns of the tables involved.
I need to calculate the net total of a column-- sounds simple. The problem is that some of the values should be negative, as are marked in a separate column. For example, the table below would yield a result of (4+3-5+2-2 = 2). I've tried doing this with subqueries in the select clause, but it seems unnecessarily complex and difficult to expand when I start adding in analysis for other parts of my table. Any help is much appreciated!
Sign Value
Pos 4
Pos 3
Neg 5
Pos 2
Neg 2
Using a CASE statement should work in most versions of sql:
SELECT SUM( CASE
WHEN t.Sign = 'Pos' THEN t.Value
ELSE t.Value * -1
END
) AS Total
FROM YourTable AS t
Try this:
SELECT SUM(IF(sign = 'Pos', Value, Value * (-1))) as total FROM table
I am adding rows from a single field in a table based on values from another field in the same table using oracle 11g as database and sql developer as user interface.
This works:
SELECT COUNTRY_ID, SUM(
CASE
WHEN ACCOUNT IN 'PTBI' THEN AMOUNT
WHEN ACCOUNT IN 'MLS_ENT' THEN AMOUNT
WHEN ACCOUNT IN 'VAL_ALLOW' THEN AMOUNT
WHEN ACCOUNT IN 'RSC_DEV' THEN AMOUNT * -1
END) AS TI
FROM SAMP_TAX_F4
GROUP BY COUNTRY_ID;
select a= sum(Value) where Sign like 'pos'
select b = sum(Value) where Signe like 'neg'
select total = a-b
this is abit sql-agnostic, since you didnt say which db you are using, but it should be easy to adapat it to any db out there.
I have a sql query in which i have a calculated field which calculates the Contribution Margin. I get it to display and the math works fine. The problem i'm having is that i want to only display the records in which the Contribution Margin is lower than 0.25. I know you cant use column alias in the where clause. I was wondering what the best way to go about doing this would be. I'm also using Visual Studio for this.
SELECT *
FROM (
SELECT m.*,
compute_margin(field1, field2) AS margin
FROM mytable m
) q
WHERE margin < 0.25
You can't use the column alias (unless you use your original query as a subquery), but you can use the expression that you're using to define the calculated value.
For example, if your query is this now:
select
contribution_amount,
total_amount,
contribution_amount / total_amount as contribution_margin
from records
You could do this:
select
contribution_amount,
total_amount,
contribution_amount / total_amount as contribution_margin
from records
where contribution_amount / total_amount < 0.25
Or this:
select * from
(
select
contribution_amount,
total_amount,
contribution_amount / total_amount as contribution_margin
from records
)
where contribution_margin < 0.25
(Personally I find the first version to be preferable, but both will likely perform the same)
You can either
repeat the calculation in the where clause
wrap the query in a table expression (CTE or derived table) and use the alias in the where clause
assign the alias in a cross apply.
To give an example of the last approach
select doubled_schema_id,*
from sys.objects
cross apply (select schema_id*2 as doubled_schema_id) c
where doubled_schema_id= 2
two ways, either the solution that Quassnoi posted(you can also use a CTE which is similar)
or WHERE compute_margin(field1, field2) < 0.25