Mathematical functions

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Mathematical functions

by romsok :: Rate this Message:

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Hi,

I am encountering the following issue:

SELECT
field1, //returns 4
field2, //returns 1
( ( field2 / field1) * 100) //this returns zero - not what I want - I am looking for 25

Am I doing something wrong with type conversion (integer/floating point)?


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Re: Mathematical functions

by Stefan Manegold :: Rate this Message:

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On Wed, Nov 04, 2009 at 06:09:07PM -0500, Roman Sokolyuk wrote:

> Hi,
>
> I am encountering the following issue:
>
> SELECT
> field1, //returns 4
> field2, //returns 1
> ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
> looking for 25
>
> Am I doing something wrong with type conversion (integer/floating point)?

Of what type are your field1 & field2?
If they are integer types, the result you get seems to be as expected with
integer arithmetics --- unless the SQL standard requires implicit convertion
to floating point when processing integer data --- I'm not sure about that.
In case integer arithmetics is indeed what the SQL standard requires, you'll
need to rewrite your query to explicitly convert the integers to decimals,
floats or doubles to force decimal or floating point arithmetics.
Alterntatively, you could simply apply arithmetic rules to rewrite the term
to, say, ((100 * field2) / field1), obviously risking an overflow in case
field2 can be larger than 1/100 of the respective integer type domain.

In case the SQL standard strictly requires decimal/loating point arithmetics
also with integer types, please file a bug report via our bug tracker on
SourceForge.

Stefan

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> what you do best, core application coding. Discover what's new with
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--
| Dr. Stefan Manegold | mailto:Stefan.Manegold@... |
| CWI,  P.O.Box 94079 | http://www.cwi.nl/~manegold/  |
| 1090 GB Amsterdam   | Tel.: +31 (20) 592-4212       |
| The Netherlands     | Fax : +31 (20) 592-4312       |

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Re: Mathematical functions

by romsok :: Rate this Message:

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Could you please tell me how I can use explicit type conversion from say the output of the COUNT() function, to type REAL?

On Wed, Nov 4, 2009 at 6:45 PM, Stefan Manegold <Stefan.Manegold@...> wrote:
On Wed, Nov 04, 2009 at 06:09:07PM -0500, Roman Sokolyuk wrote:
> Hi,
>
> I am encountering the following issue:
>
> SELECT
> field1, //returns 4
> field2, //returns 1
> ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
> looking for 25
>
> Am I doing something wrong with type conversion (integer/floating point)?

Of what type are your field1 & field2?
If they are integer types, the result you get seems to be as expected with
integer arithmetics --- unless the SQL standard requires implicit convertion
to floating point when processing integer data --- I'm not sure about that.
In case integer arithmetics is indeed what the SQL standard requires, you'll
need to rewrite your query to explicitly convert the integers to decimals,
floats or doubles to force decimal or floating point arithmetics.
Alterntatively, you could simply apply arithmetic rules to rewrite the term
to, say, ((100 * field2) / field1), obviously risking an overflow in case
field2 can be larger than 1/100 of the respective integer type domain.

In case the SQL standard strictly requires decimal/loating point arithmetics
also with integer types, please file a bug report via our bug tracker on
SourceForge.

Stefan

> ------------------------------------------------------------------------------
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> trial. Simplify your report design, integration and deployment - and focus on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
> _______________________________________________
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> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users


--
| Dr. Stefan Manegold | mailto:Stefan.Manegold@... |
| CWI,  P.O.Box 94079 | http://www.cwi.nl/~manegold/  |
| 1090 GB Amsterdam   | Tel.: +31 (20) 592-4212       |
| The Netherlands     | Fax : +31 (20) 592-4312       |

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Re: Mathematical functions

by Guillaume Theoret :: Rate this Message:

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I had this problem today as well. What I did to solve it was this:

select cast(field2 as numeric(12,4)) / field1 * 100

This will force monetdb into not simply doing integer division.

On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk <romsok.tech@...> wrote:

> Hi,
>
> I am encountering the following issue:
>
> SELECT
> field1, //returns 4
> field2, //returns 1
> ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
> looking for 25
>
> Am I doing something wrong with type conversion (integer/floating point)?
>
>
> ------------------------------------------------------------------------------
> Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day
> trial. Simplify your report design, integration and deployment - and focus
> on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
> _______________________________________________
> MonetDB-users mailing list
> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>

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Re: Mathematical functions

by romsok :: Rate this Message:

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Thanks a lot - this works - could you tell me please what the 12,4 stand for?

On Wed, Nov 4, 2009 at 10:40 PM, Guillaume Theoret <smokinn@...> wrote:
I had this problem today as well. What I did to solve it was this:

select cast(field2 as numeric(12,4)) / field1 * 100

This will force monetdb into not simply doing integer division.

On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk <romsok.tech@...> wrote:
> Hi,
>
> I am encountering the following issue:
>
> SELECT
> field1, //returns 4
> field2, //returns 1
> ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
> looking for 25
>
> Am I doing something wrong with type conversion (integer/floating point)?
>
>
> ------------------------------------------------------------------------------
> Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day
> trial. Simplify your report design, integration and deployment - and focus
> on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
> _______________________________________________
> MonetDB-users mailing list
> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>

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Re: Mathematical functions

by Guillaume Theoret :: Rate this Message:

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MonetDB's "numeric" basically a decimal type.

Mysql has a good explanation:

http://dev.mysql.com/doc/refman/5.0/en/precision-math-decimal-changes.html

So basically 12,4 means 12 digits, 4 of them to the right of the decimal.

So 12345678.1234 is a representable number but if you try more digits
on either side they will either be rounded in the case of the decimal
side or you will get an overflow error in the case of the left side:

sql>select cast(12345678.1234 as decimal(12,4));
select cast(12345678.1234 as decimal(12,4));
+----------------+
| single_value   |
+================+
| 12345678.1234  |
+----------------+
1 tuple
Timer       0.400 msec 1 rows
sql>select cast(12345678.12349 as decimal(12,4));
select cast(12345678.12349 as decimal(12,4));
+----------------+
| single_value   |
+================+
| 12345678.1235  |
+----------------+
1 tuple
Timer       0.422 msec 1 rows
sql>select cast(123456789.1234 as decimal(12,4));
!SQLException:convert:too many digits (13 > 12)
0 tuples
Timer       0.426 msec 0 rows


On Thu, Nov 5, 2009 at 10:02 AM, Roman Sokolyuk <romsok.tech@...> wrote:

> Thanks a lot - this works - could you tell me please what the 12,4 stand
> for?
>
> On Wed, Nov 4, 2009 at 10:40 PM, Guillaume Theoret <smokinn@...>
> wrote:
>>
>> I had this problem today as well. What I did to solve it was this:
>>
>> select cast(field2 as numeric(12,4)) / field1 * 100
>>
>> This will force monetdb into not simply doing integer division.
>>
>> On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk <romsok.tech@...>
>> wrote:
>> > Hi,
>> >
>> > I am encountering the following issue:
>> >
>> > SELECT
>> > field1, //returns 4
>> > field2, //returns 1
>> > ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
>> > looking for 25
>> >
>> > Am I doing something wrong with type conversion (integer/floating
>> > point)?
>> >
>> >
>> >
>> > ------------------------------------------------------------------------------
>> > Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> > 30-Day
>> > trial. Simplify your report design, integration and deployment - and
>> > focus
>> > on
>> > what you do best, core application coding. Discover what's new with
>> > Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> > _______________________________________________
>> > MonetDB-users mailing list
>> > MonetDB-users@...
>> > https://lists.sourceforge.net/lists/listinfo/monetdb-users
>> >
>> >
>>
>>
>> ------------------------------------------------------------------------------
>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> 30-Day
>> trial. Simplify your report design, integration and deployment - and focus
>> on
>> what you do best, core application coding. Discover what's new with
>> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> _______________________________________________
>> MonetDB-users mailing list
>> MonetDB-users@...
>> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>
> ------------------------------------------------------------------------------
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> trial. Simplify your report design, integration and deployment - and focus
> on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
> _______________________________________________
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> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>

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Re: Mathematical functions

by romsok :: Rate this Message:

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AND for some reason when I apply the cast like this:
(CAST(field1 AS NUMERIC(12,4))/CAST(field2 AS NUMERIC(12,4))) AS ratio

the result is 0.2 when field1=3 and field2=5 - shouldn't the result be 0.6?

On Thu, Nov 5, 2009 at 10:53 AM, Guillaume Theoret <smokinn@...> wrote:
MonetDB's "numeric" basically a decimal type.

Mysql has a good explanation:

http://dev.mysql.com/doc/refman/5.0/en/precision-math-decimal-changes.html

So basically 12,4 means 12 digits, 4 of them to the right of the decimal.

So 12345678.1234 is a representable number but if you try more digits
on either side they will either be rounded in the case of the decimal
side or you will get an overflow error in the case of the left side:

sql>select cast(12345678.1234 as decimal(12,4));
select cast(12345678.1234 as decimal(12,4));
+----------------+
| single_value   |
+================+
| 12345678.1234  |
+----------------+
1 tuple
Timer       0.400 msec 1 rows
sql>select cast(12345678.12349 as decimal(12,4));
select cast(12345678.12349 as decimal(12,4));
+----------------+
| single_value   |
+================+
| 12345678.1235  |
+----------------+
1 tuple
Timer       0.422 msec 1 rows
sql>select cast(123456789.1234 as decimal(12,4));
!SQLException:convert:too many digits (13 > 12)
0 tuples
Timer       0.426 msec 0 rows


On Thu, Nov 5, 2009 at 10:02 AM, Roman Sokolyuk <romsok.tech@...> wrote:
> Thanks a lot - this works - could you tell me please what the 12,4 stand
> for?
>
> On Wed, Nov 4, 2009 at 10:40 PM, Guillaume Theoret <smokinn@...>
> wrote:
>>
>> I had this problem today as well. What I did to solve it was this:
>>
>> select cast(field2 as numeric(12,4)) / field1 * 100
>>
>> This will force monetdb into not simply doing integer division.
>>
>> On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk <romsok.tech@...>
>> wrote:
>> > Hi,
>> >
>> > I am encountering the following issue:
>> >
>> > SELECT
>> > field1, //returns 4
>> > field2, //returns 1
>> > ( ( field2 / field1) * 100) //this returns zero - not what I want - I am
>> > looking for 25
>> >
>> > Am I doing something wrong with type conversion (integer/floating
>> > point)?
>> >
>> >
>> >
>> > ------------------------------------------------------------------------------
>> > Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> > 30-Day
>> > trial. Simplify your report design, integration and deployment - and
>> > focus
>> > on
>> > what you do best, core application coding. Discover what's new with
>> > Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> > _______________________________________________
>> > MonetDB-users mailing list
>> > MonetDB-users@...
>> > https://lists.sourceforge.net/lists/listinfo/monetdb-users
>> >
>> >
>>
>>
>> ------------------------------------------------------------------------------
>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> 30-Day
>> trial. Simplify your report design, integration and deployment - and focus
>> on
>> what you do best, core application coding. Discover what's new with
>> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> _______________________________________________
>> MonetDB-users mailing list
>> MonetDB-users@...
>> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>
> ------------------------------------------------------------------------------
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> trial. Simplify your report design, integration and deployment - and focus
> on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
> _______________________________________________
> MonetDB-users mailing list
> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>

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Re: Mathematical functions

by Guillaume Theoret :: Rate this Message:

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Are you sure your fields are set correctly?

sql>create schema test;
Operation successful
Timer    3410.692 msec 1 rows
sql>create table test.test (field1 int not null, field2 int not null);
Operation successful
Timer    4343.730 msec 1 rows
sql>insert into test.test (field1, field2) values (3,5);
Rows affected 1
Timer    1076.233 msec 0 rows
sql>select (CAST(field1 AS NUMERIC(12,4))/CAST(field2 AS
NUMERIC(12,4))) AS ratio from test.test;
select (CAST(field1 AS NUMERIC(12,4))/CAST(field2 AS NUMERIC(12,4)))
AS ratio from test.test;
+--------------------+
| ratio              |
+====================+
| 0.6000             |
+--------------------+
1 tuple
Timer     671.128 msec 1 rows


On Thu, Nov 5, 2009 at 11:26 AM, Roman Sokolyuk <romsok.tech@...> wrote:

> AND for some reason when I apply the cast like this:
> (CAST(field1 AS NUMERIC(12,4))/CAST(field2 AS NUMERIC(12,4))) AS ratio
>
> the result is 0.2 when field1=3 and field2=5 - shouldn't the result be 0.6?
>
> On Thu, Nov 5, 2009 at 10:53 AM, Guillaume Theoret <smokinn@...>
> wrote:
>>
>> MonetDB's "numeric" basically a decimal type.
>>
>> Mysql has a good explanation:
>>
>> http://dev.mysql.com/doc/refman/5.0/en/precision-math-decimal-changes.html
>>
>> So basically 12,4 means 12 digits, 4 of them to the right of the decimal.
>>
>> So 12345678.1234 is a representable number but if you try more digits
>> on either side they will either be rounded in the case of the decimal
>> side or you will get an overflow error in the case of the left side:
>>
>> sql>select cast(12345678.1234 as decimal(12,4));
>> select cast(12345678.1234 as decimal(12,4));
>> +----------------+
>> | single_value   |
>> +================+
>> | 12345678.1234  |
>> +----------------+
>> 1 tuple
>> Timer       0.400 msec 1 rows
>> sql>select cast(12345678.12349 as decimal(12,4));
>> select cast(12345678.12349 as decimal(12,4));
>> +----------------+
>> | single_value   |
>> +================+
>> | 12345678.1235  |
>> +----------------+
>> 1 tuple
>> Timer       0.422 msec 1 rows
>> sql>select cast(123456789.1234 as decimal(12,4));
>> !SQLException:convert:too many digits (13 > 12)
>> 0 tuples
>> Timer       0.426 msec 0 rows
>>
>>
>> On Thu, Nov 5, 2009 at 10:02 AM, Roman Sokolyuk <romsok.tech@...>
>> wrote:
>> > Thanks a lot - this works - could you tell me please what the 12,4 stand
>> > for?
>> >
>> > On Wed, Nov 4, 2009 at 10:40 PM, Guillaume Theoret <smokinn@...>
>> > wrote:
>> >>
>> >> I had this problem today as well. What I did to solve it was this:
>> >>
>> >> select cast(field2 as numeric(12,4)) / field1 * 100
>> >>
>> >> This will force monetdb into not simply doing integer division.
>> >>
>> >> On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk <romsok.tech@...>
>> >> wrote:
>> >> > Hi,
>> >> >
>> >> > I am encountering the following issue:
>> >> >
>> >> > SELECT
>> >> > field1, //returns 4
>> >> > field2, //returns 1
>> >> > ( ( field2 / field1) * 100) //this returns zero - not what I want - I
>> >> > am
>> >> > looking for 25
>> >> >
>> >> > Am I doing something wrong with type conversion (integer/floating
>> >> > point)?
>> >> >
>> >> >
>> >> >
>> >> >
>> >> > ------------------------------------------------------------------------------
>> >> > Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> >> > 30-Day
>> >> > trial. Simplify your report design, integration and deployment - and
>> >> > focus
>> >> > on
>> >> > what you do best, core application coding. Discover what's new with
>> >> > Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> >> > _______________________________________________
>> >> > MonetDB-users mailing list
>> >> > MonetDB-users@...
>> >> > https://lists.sourceforge.net/lists/listinfo/monetdb-users
>> >> >
>> >> >
>> >>
>> >>
>> >>
>> >> ------------------------------------------------------------------------------
>> >> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> >> 30-Day
>> >> trial. Simplify your report design, integration and deployment - and
>> >> focus
>> >> on
>> >> what you do best, core application coding. Discover what's new with
>> >> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> >> _______________________________________________
>> >> MonetDB-users mailing list
>> >> MonetDB-users@...
>> >> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>> >
>> >
>> >
>> > ------------------------------------------------------------------------------
>> > Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> > 30-Day
>> > trial. Simplify your report design, integration and deployment - and
>> > focus
>> > on
>> > what you do best, core application coding. Discover what's new with
>> > Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> > _______________________________________________
>> > MonetDB-users mailing list
>> > MonetDB-users@...
>> > https://lists.sourceforge.net/lists/listinfo/monetdb-users
>> >
>> >
>>
>>
>> ------------------------------------------------------------------------------
>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008
>> 30-Day
>> trial. Simplify your report design, integration and deployment - and focus
>> on
>> what you do best, core application coding. Discover what's new with
>> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
>> _______________________________________________
>> MonetDB-users mailing list
>> MonetDB-users@...
>> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>
> ------------------------------------------------------------------------------
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> on
> what you do best, core application coding. Discover what's new with
> Crystal Reports now.  http://p.sf.net/sfu/bobj-july
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> MonetDB-users@...
> https://lists.sourceforge.net/lists/listinfo/monetdb-users
>
>

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