Round to the nearest .5 value

2009-11-25 Thread Che Vilnonis

I'm trying to create a ratings system that shows 1/2 values. Quick
question, what would be the easiest way to round a value to the nearest .5
of a number? I'm trying to avoid multiple if/elses.

i.e. 3.21 would round to 3
i.e. 3.31 would round to 3.5
i.e. 3.61 would round to 3.5
i.e. 3.91 would round to 4

Thanks, Che



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Re: Round to the nearest .5 value

2009-11-25 Thread Barney Boisvert

Assuming 'n' is your number, use #round(n * 2) / 2#

For example:

3.21 * 2 == 6.42
round(6.42) == 6
6 / 2 == 3

3.31 * 2 = 6.62
round(6.62) == 7
7 / 2 == 3.35

cheers,
barneyb

On Wed, Nov 25, 2009 at 11:13 AM, Che Vilnonis ch...@asitv.com wrote:

 I'm trying to create a ratings system that shows 1/2 values. Quick
 question, what would be the easiest way to round a value to the nearest .5
 of a number? I'm trying to avoid multiple if/elses.

 i.e. 3.21 would round to 3
 i.e. 3.31 would round to 3.5
 i.e. 3.61 would round to 3.5
 i.e. 3.91 would round to 4

 Thanks, Che



 

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Re: Round to the nearest .5 value

2009-11-25 Thread Ian Skinner

Che Vilnonis wrote:

Multiply number by 2, round it, divide it by 2.



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Re: Round to the nearest .5 value

2009-11-25 Thread Ian Skinner

Barney Boisvert wrote:
 7 / 2 == 3.35

No it doesn't!  I suspect your typing thumbs, or is it your ring finger, 
is making you look like a bad mathematician! ;-)

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RE: Round to the nearest .5 value

2009-11-25 Thread Che Vilnonis

Sweet! Many years ago, in high school Algebra class, I would have been able
to figure that out. It's amazing what we forget.

-Original Message-
From: Barney Boisvert [mailto:bboisv...@gmail.com] 
Sent: Wednesday, November 25, 2009 2:23 PM
To: cf-talk
Subject: Re: Round to the nearest .5 value


Assuming 'n' is your number, use #round(n * 2) / 2#

For example:

3.21 * 2 == 6.42
round(6.42) == 6
6 / 2 == 3

3.31 * 2 = 6.62
round(6.62) == 7
7 / 2 == 3.35

cheers,
barneyb



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Re: Round to the nearest .5 value

2009-11-25 Thread Allen Souliere

cfoutput
#Round(3.21/0.5) *0.5#
#Round(3.31/0.5) *0.5#
#Round(3.61/0.5) *0.5#
#Round(3.91/0.5) *0.5#
#Round(4.24/0.5) *0.5#
#Round(4.49/0.5) *0.5#
#Round(4.50/0.5) *0.5#
/cfoutput

produces:

3
3.5
3.5
4
4
4.5
4.5

Cheers,
Allen

Che Vilnonis wrote:
 I'm trying to create a ratings system that shows 1/2 values. Quick
 question, what would be the easiest way to round a value to the nearest .5
 of a number? I'm trying to avoid multiple if/elses.

 i.e. 3.21 would round to 3
 i.e. 3.31 would round to 3.5
 i.e. 3.61 would round to 3.5
 i.e. 3.91 would round to 4

 Thanks, Che



 

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Re: Round to the nearest .5 value

2009-11-25 Thread Barney Boisvert

Oops, you're right, Ian.  That should be 7 / 2 == 3.5 (remove the
second three), of course.  I wonder who had coffee this morning and
therefore can't control his fine motor skills.  At least with a
computer your assertions are actually checked, so typos get caught.
:)

cheers,
barneyb

On Wed, Nov 25, 2009 at 11:32 AM, Ian Skinner h...@ilsweb.com wrote:

 Barney Boisvert wrote:
 7 / 2 == 3.35

 No it doesn't!  I suspect your typing thumbs, or is it your ring finger,
 is making you look like a bad mathematician! ;-)



-- 
Barney Boisvert
bboisv...@gmail.com
http://www.barneyb.com

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Re: Round to the nearest .5 value

2009-11-25 Thread Phillip Vector

That is beautiful in form and simplicity. Bravo.

On Wed, Nov 25, 2009 at 11:27 AM, Ian Skinner h...@ilsweb.com wrote:

 Multiply number by 2, round it, divide it by 2.

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