[EMAIL PROTECTED] graced perl with these words of wisdom:
> Srt f lk wht wer dng 2 Nglsh
>.
*We*? Speak for yourself.
I find it intensely irritating to read posts on an email list like this
where people write "u" and "ur" when they're not talking about Burmese
honorifics or ancient Mesopotam
> Remember when one state (Kansas, Nebraska, or Arkansas, I think it
was)
> wanted to declare pi = 3, which would've made all the wheels in the
> state turn square overnight?
This isn't quite true, but the truth is equally weird. It was a bill
with a method to square a circle, which is to find a s
>But at some point it has to happen, whether u do it urself
>in some code or the arithmetic engine does it on its floats.
Understood. But the difference is that computers round beyond where we
humans generally care about. Not that it's actually rounding--as you
pointed out in another post--but
At 08:39 AM 3/19/2007 -0500, [EMAIL PROTECTED] wrote:
>I have a recollection from a computer math class many years ago that you
>should never, ever, round intermediate numbers in a calculation. The
>"rounding" itself then becomes part of the next step, then the next, and
>so on. While that may n
At 04:34 AM 3/18/2007 +, [EMAIL PROTECTED] wrote:
>Do 'human rounding principles' say that you would expect 10 / 3 rounded
>to 2 DP would produce 3.34 or 3.33? How about -10 / 3?
10/3 rounded to two places is 3.33. 3.34 is simply wrong. To say that 3.34
is 10/3 rouned to two places means tha
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On 16 Mar 2007 at 21:52, Bill Luebkert wrote:
> That can be resolved by using a smaller value than .05. Actually, I
> use something like this to do my rounding when CPU isn't important :
>
> sub round ($;$) {# $ret = round ($num, [$digi
>From wagnerc:
>That's all good for rounding intermediate numbers in a calculation...
I have a recollection from a computer math class many years ago that you
should never, ever, round intermediate numbers in a calculation. The
"rounding" itself then becomes part of the next step, then the next
I decided to write my own rounding algorithm that takes it out of Perl's
hands and does all the math itself using strings. Since it uses strings it
operates on the same literal value that a human would be looking at rather
than an opaque object. Floats are better thought of as opaque objects than
Chris Wagner wrote:
> OP was trying to round for display. In this case whatever rounding function
> u use has to output a string consistent with human rounding principles.
> Floating point numbers that a machine uses internally have no real relation
> to the real numbers that humans use. And the
At 05:32 PM 3/17/2007 +, Mark Dootson wrote:
>If you are using floating point arithmetic, you want unbiased rounding,
>'even' if you don't realise that this is good for you.
That's all good for rounding intermediate numbers in a calculation but the
OP was trying to round for display. In this
I'd disagree that there is any problem with sprintf and 'rounding'.
Basically, you can round towards zero, away from zero, or use unbiased
rounding. Towards even is the accepted method of unbiased rounding.
Unbiased rounding is the method least likely to introduce bias into your
end results.
If y
I think the safest way to round is to add 0.5, then use the
POSIX::floor function. The Perl docs warn against using int() for
just about anything (I don't have the exact quote handy):
use strict;
use POSIX qw(floor);
MAIN: {
foreach my $x (
0.0, 0.1, 0.2, 0.3, 0.4, 0.5,
At 12:48 AM 3/17/2007 -0500, [EMAIL PROTECTED] wrote:
>Incidentally, perl is a typeless language, compounding the problem even
more. That is, perl
>converts a variable or literal to an integer, a floating point double, or
text string on-the-fly as
>needed. A single line of perl code may hide seve
At 08:04 PM 3/16/2007 -0400, Su, Yu \(Eugene\) wrote:
>I expect
>print sprintf("%.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f", 0.5,
1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5);
>will give me
>1 2 3 4 5 6 7 8 9 10
>but instead I get
>0 2 2 4 4 6 6 8 8 10
&g
Casteele/ShadowLord wrote:
>
> Actually, Bill's answer has an error.. Adding .05 just introduces a subtle
> error because some
> decimal floating point numbers cannot be exactly represented in binary form.
> For example:
>
> my ($i, $j);
> foreach $i (0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8
planation why sprintf treats 3.5 and 4.5
> differently in rounding?
>
> Thanks Bill.
>
> -Eugene
>
>
> -Original Message-
> From: Bill Luebkert [mailto:[EMAIL PROTECTED]
> Sent: Friday, March 16, 2007 5:33 PM
> To: Su, Yu (Eugene)
> Cc: perl-win32-users@lists
get
>> 0 2 2 4 4 6 6 8 8 10
>>
>> How to round a floating-point value to an integer value using
>> (.5) up rule?
>
> I'd add .5 in the next place down :
>
> foreach (0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0) {
> pr
On Fri, 16 Mar 2007, Jan Dubois wrote:
> On Fri, 16 Mar 2007, Su, Yu (Eugene) wrote:
> > Interesting. Any explanation why sprintf treats 3.5 and 4.5
> > differently in rounding?
>
> It is called "unbiased rounding", which has the advantage of not
> adding an upward trend to your numbers:
>
> ht
> I expect
> > print sprintf("%.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f", 0.5,
> > 1.5, 2.5, 3.5, 4.5, 5.5,
> 6.5, 7.5, 8.5, 9.5);
> > will give me
> > 1 2 3 4 5 6 7 8 9 10
> > but instead I get
> > 0 2 2 4 4 6 6 8 8 10
> >
> >
Su, Yu (Eugene) wrote:
> Interesting. Any explanation why sprintf treats 3.5 and 4.5 differently in
> rounding?
Try adding this to the loop:
printf "%.20f", $_;
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To uns
: how to round a floating-point value to an integer value
Su, Yu (Eugene) wrote:
> I expect
> print sprintf("%.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f", 0.5, 1.5,
> 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5);
> will give me
> 1 2 3 4 5 6 7 8 9 10
> but instead I
Su, Yu (Eugene) wrote:
> I expect
> print sprintf("%.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f", 0.5, 1.5,
> 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5);
> will give me
> 1 2 3 4 5 6 7 8 9 10
> but instead I get
> 0 2 2 4 4 6 6 8 8 10
>
> How to round
I expect
print sprintf("%.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f %.0f", 0.5, 1.5,
2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5);
will give me
1 2 3 4 5 6 7 8 9 10
but instead I get
0 2 2 4 4 6 6 8 8 10
How to round a floating-point value to an integer value using (.5) up rule?
Thanks
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