Iwant to round of the value upto 2 decimal point when third decimal digit is greater then 5:
39.956 should be round off to 39.96,
35.665 should be round off to 35.66 ,
39.997 should be round off to 40.00 ,
56.684 should be round off to 56.68.
I am trying to do below
SELECT CAST(FLOOR(VALUE) AS VARCHAR2(30))
+ CASE
WHEN CAST(SUBSTR(SUBSTR(VALUE, INSTR(VALUE, '.')), 4) AS INT) > 5
THEN
CONCAT(
'.',
( SUBSTR(
SUBSTR(VALUE, INSTR(VALUE, '.')),
2,
2
)
+ 1)
)
ELSE
CONCAT(
'.',
SUBSTR(
SUBSTR(VALUE, INSTR(VALUE, '.')),
2,
2
)
)
END
FROM DUAL;
but for the border cases, for example 39.897 and 39.997 it is not working.
Maybe you simply need this:
SQL> with test(num) as (
2 select 39.956 from dual union all
3 select 35.665 from dual union all
4 select 39.997 from dual union all
5 select 56.684 from dual
6 )
7 select num, round(num -0.001, 2)
8 from test;
NUM ROUND(NUM-0.001,2)
---------- ------------------
39,956 39,96
35,665 35,66
39,997 40
56,684 56,68
Aleksej's solution will work fine and is probably the most efficient if it is known beforehand that the input numbers have at most three decimal places.
The problem can be generalized though, like so: round 38.445 down to 38.44; however, round 38.44503 to 38.45. (That is, if there are non-zero digits after the "5" in the third decimal position, then round up.)
Something like the query below can be used in the general case. The only time the result is different from "usual" rounding is when the input number has exactly three non-zero decimal places, and the third decimal place is 5. This is exactly how the solution reads.
with inp (n) as (select 38.445 from dual union all select 38.44503 from dual)
select n,
round(n,2) - case when n = round(n, 3) and mod(1000*n, 10) = 5
then 0.01
else 0 end as custom_rounded
from inp;
N CUSTOM_ROUNDED
---------- --------------
38.445 38.44
38.44503 38.45
Related
I have this
RowNum
Value
1
X
2
X
3
Y
4
Z
5
Z
6
Z
7
V
and I want something like this
RowNum
Value
1
X
1
X
2
Y
3
Z
3
Z
3
Z
4
V
How can I do that in Oracle?
Thanks
Here is one way - using the match_recognize clause, available since Oracle 12.1. Note that rownum is a reserved keyword, so it can't be a column name; I changed it to rnum in the input and rn in the output, adapt as needed.
The with clause is not part of the query - it's there just to include your sample data. Remove it before using the query on your actual data.
with
inputs (rnum, value) as (
select 1, 'X' from dual union all
select 2, 'X' from dual union all
select 3, 'Y' from dual union all
select 4, 'Z' from dual union all
select 5, 'Z' from dual union all
select 6, 'Z' from dual union all
select 7, 'V' from dual
)
select rn, value
from inputs
match_recognize (
order by rnum
measures match_number() as rn
all rows per match
pattern ( a+ )
define a as value = first(value)
);
RN VALUE
-- -----
1 X
1 X
2 Y
3 Z
3 Z
3 Z
4 V
In Oracle 11.2 and earlier, you can use the start-of-group method (the flags created in the subquery and counted in the outer query):
select count(flag) over (order by rnum) as rn, value
from (
select rnum, value,
case lag(value) over (order by rnum)
when value then null else 1 end as flag
from inputs
)
;
Consider following table where I am doing row data multiplication:
with v1 (member_id, the_number) as
(
select 1, 3 from dual union all
select 1, 5 from dual union all
select 2, 2 from dual union all
select 2, 3 from dual union all
select 2, 4 from dual union all
select 3, 9 from dual union all
select 3, 3 from dual union all
select 3, 2 from dual
)
select member_id, EXP(SUM(LN(the_number))) from v1
GROUP BY member_id;
It gives the correct result as:
MEMBER_ID EXP(SUM(LN(THE_NUMBER)))
1 15
2 24
3 54
The moment I put a negative value in the the_number column, I get the following Oracle error: ORA-01428: argument 'x' is out of range This is because the range for LN () argument is > 0.
How can I modify the query so that I can have negative values as well in the_number column? I am using Oracle 11g.
Get the product of the absolute values of the numbers and finally multiply by -1 or 1 depending on whether there is an odd or even number of negative numbers:
select
member_id,
CASE WHEN MOD(SUM(CASE WHEN the_number < 0 THEN 1 ELSE 0 END), 2) = 1 THEN -1 ELSE 1 END *
EXP(SUM(LN(ABS(the_number)))) from v1
GROUP BY member_id;
See the demo.
I have a table that have date grouped.
I need to select only groups that have positive and negative value inside.
For example:
id value1
2 7
2 8
2 -1
3 3
3 4
4 -1
4 -2
5 7
5 -5
the result should be
id value1
2 7
2 8
2 -1
5 7
5 -5
because the group with id 3 just have positive number and the group with id 4 just have negative number.
any idea how can I do it using case (when then) in a select or using if else inside a function. Or any other idea?
Try this.
select id,value1 FROM
(
select t.*,
count( DISTINCT SIGN (value1 ) ) OVER (PARTITION BY id ) n
from yourtable t
) WHERE n = 2
;
The Sign() function gives 1 for positive and -1 for negative numbers.
DEMO
If you group by the ID, you can use the aggregate functions MIN and MAX to find out if there are both positive and negative values. You need to decide how to treat 0 though... I have treated it as positive below :)
with your_table as(
-- Your example data here, this is not really part of the solution
select 2 as id, 7 as value1 from dual union all
select 2 as id, 8 as value1 from dual union all
select 2 as id, -1 as value1 from dual union all
select 3 as id, 3 as value1 from dual union all
select 3 as id, 4 as value1 from dual union all
select 4 as id, -1 as value1 from dual union all
select 4 as id, -2 as value1 from dual union all
select 5 as id, 7 as value1 from dual union all
select 5 as id, -5 as value1 from dual
)
select *
from your_table
where id in(select id
from your_table
group by id
having min(value1) < 0
and max(value1) >= 0);
For example if you take the following example into consideration.
100.00 - Original Number
33.33 - 1st divided by 3
33.33 - 2nd divided by 3
33.33 - 3rd divided by 3
99.99 - Is the sum of the 3 division outcomes
But i want it to match the original 100.00
One way that i saw it could be done was by taking the original number minus the first two divisions and the result would be my third number. Now if i take those 3 numbers i get my original number.
100.00 - Original Number
33.33 - 1st divided by 3
33.33 - 2nd divided by 3
33.34 - 3rd number
100.00 - Which gives me my original number correctly. (33.33+33.33+33.34 = 100.00)
Is there a formula for this either in Oracle PL/SQL or a function or something that could be implemented?
Thanks in advance!
This version takes precision as a parameter as well:
with q as (select 100 as val, 3 as parts, 2 as prec from dual)
select rownum as no
,case when rownum = parts
then val - round(val / parts, prec) * (parts - 1)
else round(val / parts, prec)
end v
from q
connect by level <= parts
no v
=== =====
1 33.33
2 33.33
3 33.34
For example, if you want to split the value among the number of days in the current month, you can do this:
with q as (select 100 as val
,extract(day from last_day(sysdate)) as parts
,2 as prec from dual)
select rownum as no
,case when rownum = parts
then val - round(val / parts, prec) * (parts - 1)
else round(val / parts, prec)
end v
from q
connect by level <= parts;
1 3.33
2 3.33
3 3.33
4 3.33
...
27 3.33
28 3.33
29 3.33
30 3.43
To apportion the value amongst each month, weighted by the number of days in each month, you could do this instead (change the level <= 3 to change the number of months it is calculated for):
with q as (
select add_months(date '2013-07-01', rownum-1) the_month
,extract(day from last_day(add_months(date '2013-07-01', rownum-1)))
as days_in_month
,100 as val
,2 as prec
from dual
connect by level <= 3)
,q2 as (
select the_month, val, prec
,round(val * days_in_month
/ sum(days_in_month) over (), prec)
as apportioned
,row_number() over (order by the_month desc)
as reverse_rn
from q)
select the_month
,case when reverse_rn = 1
then val - sum(apportioned) over (order by the_month
rows between unbounded preceding and 1 preceding)
else apportioned
end as portion
from q2;
01/JUL/13 33.7
01/AUG/13 33.7
01/SEP/13 32.6
Use rational numbers. You could store the numbers as fractions rather than simple values. That's the only way to assure that the quantity is truly split in 3, and that it adds up to the original number. Sure you can do something hacky with rounding and remainders, as long as you don't care that the portions are not exactly split in 3.
The "algorithm" is simply that
100/3 + 100/3 + 100/3 == 300/3 == 100
Store both the numerator and the denominator in separate fields, then add the numerators. You can always convert to floating point when you display the values.
The Oracle docs even have a nice example of how to implement it:
CREATE TYPE rational_type AS OBJECT
( numerator INTEGER,
denominator INTEGER,
MAP MEMBER FUNCTION rat_to_real RETURN REAL,
MEMBER PROCEDURE normalize,
MEMBER FUNCTION plus (x rational_type)
RETURN rational_type);
Here is a parameterized SQL version
SELECT COUNT (*), grp
FROM (WITH input AS (SELECT 100 p_number, 3 p_buckets FROM DUAL),
data
AS ( SELECT LEVEL id, (p_number / p_buckets) group_size
FROM input
CONNECT BY LEVEL <= p_number)
SELECT id, CEIL (ROW_NUMBER () OVER (ORDER BY id) / group_size) grp
FROM data)
GROUP BY grp
output:
COUNT(*) GRP
33 1
33 2
34 3
If you edit the input parameters (p_number and p_buckets) the SQL essentially distributes p_number as evenly as possible among the # of buckets requested (p_buckets).
I've solved this problem yesterday by subtracting 2 of 3 parts from the starting number, e.g. 100 - 33.33 - 33.33 = 33.34 and the result of summing it up is still 100.
I have a table which has a column call numbers
Numbers
------
3
5
I am trying to get the factorial of those. I am using the below logic but not with proper result
Select
Numbers
,EXP(SUM(LN(Numbers)) OVER (ORDER BY Numbers)) Factorial
FROM testTbl
*Output
*
Numbers Factorial
------ ---------
3 3.00000000000000000000000000000000000001
5 15.0000000000000000000000000000000000002
What is wrong? Please help
Expected
--------
Numbers Factorial
------ ---------
3 6
5 120
Thanks in advance
I've had a go at this from another angle, trying to do it all in a SQL statement (using your table testTbl and the column numbers).
This is what I've come up with, see if it suits you:
SELECT testtbl.numbers,
ROUND( EXP( SUM( LN( t1.n ) ) ) ) AS factorial
FROM ( SELECT UNIQUE LEVEL n
FROM testtbl
CONNECT BY LEVEL <= numbers) t1,
( SELECT UNIQUE LEVEL n
FROM testtbl
CONNECT BY LEVEL <= numbers) t2,
testTbl
WHERE t1.n <= t2.n
AND t2.n = testTbl.numbers
GROUP BY testtbl.numbers
ORDER BY testtbl.numbers;
Gives the output:
Numbers Factorial
3 6
5 120
Hope it helps...
Were it me, I'd create a factorial function and call that user-defined function in my query. Something like
SQL> create function factorial( p_n in number )
2 return number
3 is
4 begin
5 if( p_n = 1 )
6 then
7 return p_n;
8 else
9 return p_n * factorial( p_n - 1 );
10 end if;
11 end;
12 /
Function created.
SQL> with t as (
2 select 3 num from dual
3 union all
4 select 5 from dual
5 )
6 select num,
7 factorial(num)
8 from t;
NUM FACTORIAL(NUM)
---------- --------------
3 6
5 120
If for some reason you cannot define a new function and you really want to do it in SQL, you'll can generate all the numbers less than the number in your table and then aggregate those generated numbers.
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select 3 num from dual
3 union all
4 select 5 from dual
5 )
6 select t.num,
7 exp( sum(ln(gen.num))) factorial
8 from (select level num
9 from dual
10 connect by level <= (select max(t.num) from t)) gen,
11 t
12 where gen.num <= t.num
13* group by t.num
SQL> /
NUM FACTORIAL
---------- ----------
5 120
3 6