Asp Forum - Rounding to X digits

Eric Anderson

10/14/2004 1:16:00 PM

This seems like such a basic question yet I can't really find the answer
anywhere in the docs (I'm probably looking in the wrong place).

Anyway I need to compute a percentage and output it. The percentage
should be in the form 39.45% (i.e. round to the nearest two decimal
places). So I have

top = 68
bottom = 271

percentage = (top.to_f / bottom * 100 * 100).round.to_f / 100

The above works but it does not seem very intuitive. What I would rather is:

percentage = (top.to_f / bottom).round(4) * 100

But round does not take any arguments on how many decimal places I want.
It just assumes I don't want any. Perhaps there is another function that
I am not seeing that will round while keeping a certain number of
decimal places. Obviously I could also enhance round to take an optional
argument but I wanted to see if there was an already existing function
in the Ruby std library that will do it for me.

Thanks,

Eric

25 Answers

Eric Anderson

10/14/2004 1:32:00 PM

Eric Anderson wrote:
> Obviously I could also enhance round to take an optional
> argument but I wanted to see if there was an already existing function
> in the Ruby std library that will do it for me.

To follow up my own post. If there isn't a function like what I am
looking for in the standard library, I am using the following to make it
like I want.

class Float
alias :oldround :round
def round( sd=0 )
return (self * (10 ** sd)).oldround.to_f / (10**sd)
end
end

James Gray

10/14/2004 1:33:00 PM

On Oct 14, 2004, at 8:19 AM, Eric Anderson wrote:

> This seems like such a basic question yet I can't really find the
> answer anywhere in the docs (I'm probably looking in the wrong place).
>
> Anyway I need to compute a percentage and output it. The percentage
> should be in the form 39.45% (i.e. round to the nearest two decimal
> places).

You're looking for sprintf():

sprintf "%.2f", 39.456789 # => 39.46

The ri docs for sprintf() will explain the format string options.

It also has a cousin, printf() that can used as the data is sent to
some IO object.

Finally, the String class allows a shortcut:

"%.2f" % 39.456789 # => 39.46

Hope that helps.

James Edward Gray II

Eric Anderson

10/14/2004 2:24:00 PM

James Edward Gray II wrote:
> You're looking for sprintf():

*Slaps hand on head*

I didn't even think about sprintf. I guess I was so fixed on looking for
a way to round numbers in the various Numeric objects that it didn't
occur to me to look in the string functionality.

sprintf() still seems a bit to C-ish for me. I kind of like my round
extension that allows for an optional arg. The question is my
implementation more efficient or using sprintf more efficient? Benchmark
to the rescue:

My implementation through 1 million iterations:

user system total real
No Decimal: 5.410000 0.000000 5.410000 ( 5.413982)
5 Decimal Places: 13.260000 0.000000 13.260000 ( 13.259163)
13 Decimal Places: 21.490000 0.000000 21.490000 ( 21.481821)

Sprintf implementation through 1 million iterations:

user system total real
No Decimal: 12.400000 0.000000 12.400000 ( 12.362547)
5 Decimal Places: 15.310000 0.000000 15.310000 ( 15.549580)
13 Decimal Places: 14.700000 0.020000 14.720000 ( 16.145235)

Looks like my implementation leads for the case where there are less
decimal places remaining (the more common case) and sprintf leads when
there are more. Of course at this point we are really just splitting
hairs. :)

For reference here are the implementations and benchmark script:

# My implementation
class Float
alias :oldround :round
def round( sd=0 )
return (self * (10 ** sd)).oldround.to_f / (10**sd)
end
end

# Sprintf implementation
class Float
alias :oldround :round
def round( sd=0 )
return ("%.#{sd}f" % self).to_f
end
end

Benchmark script:

#!/usr/bin/ruby

require 'benchmark'
include Benchmark

require 'sprintf_round'

n = 1000000
pi = 3.14159265358979
bm(17) do |test|
test.report('No Decimal:') do
n.times do |x|
pi.round
end
end
test.report('5 Decimal Places:') do
n.times do |x|
pi.round(5)
end
end
test.report('13 Decimal Places:') do
n.times do |x|
pi.round(13)
end
end
end

Robert Klemme

10/14/2004 2:49:00 PM

"Eric Anderson" <eric@afaik.us> schrieb im Newsbeitrag
news:GZvbd.111$FV6.46@fe61.usenetserver.com...
> James Edward Gray II wrote:
> > You're looking for sprintf():
>
> *Slaps hand on head*

*Hands over some cooling ice*
:-)

> I didn't even think about sprintf. I guess I was so fixed on looking for
> a way to round numbers in the various Numeric objects that it didn't
> occur to me to look in the string functionality.
>
> sprintf() still seems a bit to C-ish for me. I kind of like my round
> extension that allows for an optional arg. The question is my
> implementation more efficient or using sprintf more efficient? Benchmark
> to the rescue:

The difference in efficiency is one issue. But there is a conceptual
difference: it depends whether you want to align printed output or round a
numeric value in a calculation. For the first task sprintf is probably
better because the amount of places printed is guaranteed (by the format
string), while it's not if you just round the number and print it.

Kind regards

robert

Markus

10/14/2004 5:55:00 PM

Here's mine (with a few extras):

class Numeric
def iff(cond)
cond ? self : self-self
end
def to_the_nearest(n)
n*(self/n).round
end
def aprox(x,n=0.001)
(to_the_nearest n) == (x.to_the_nearest n)
end
def at_least(x); (self >= x) ? self : x; end
def at_most(x); (self <= x) ? self : x; end
end

-- Markus

On Thu, 2004-10-14 at 06:19, Eric Anderson wrote:
> This seems like such a basic question yet I can't really find the answer
> anywhere in the docs (I'm probably looking in the wrong place).
>
> Anyway I need to compute a percentage and output it. The percentage
> should be in the form 39.45% (i.e. round to the nearest two decimal
> places). So I have
>
> top = 68
> bottom = 271
>
> percentage = (top.to_f / bottom * 100 * 100).round.to_f / 100
>
> The above works but it does not seem very intuitive. What I would rather is:
>
> percentage = (top.to_f / bottom).round(4) * 100
>
> But round does not take any arguments on how many decimal places I want.
> It just assumes I don't want any. Perhaps there is another function that
> I am not seeing that will round while keeping a certain number of
> decimal places. Obviously I could also enhance round to take an optional
> argument but I wanted to see if there was an already existing function
> in the Ruby std library that will do it for me.
>
> Thanks,
>
> Eric
>

T. Onoma

10/15/2004 2:28:00 PM

Markus

10/15/2004 2:58:00 PM

Interesting. I hope to heck you used a script to find that edge
case. What made you suspect there would be a difference, given that
they are mathematically equivalent?

I'll definitely come back to this after the coffee hits.

-- Markus

On Fri, 2004-10-15 at 07:28, trans. (T. Onoma) wrote:
> On Thursday 14 October 2004 01:55 pm, Markus wrote:
> | Here's mine (with a few extras):
> |
> | class Numeric
> | def iff(cond)
> | cond ? self : self-self
> | end
> | def to_the_nearest(n)
> | n*(self/n).round
> | end
> | def aprox(x,n=0.001)
> | (to_the_nearest n) == (x.to_the_nearest n)
> | end
> | def at_least(x); (self >= x) ? self : x; end
> | def at_most(x); (self <= x) ? self : x; end
> | end
> |
>
> | def to_the_nearest(n)
> | n*(self/n).round
> | end
> def round_to_nearest( n=0.01 )
> (self * (1/n)).round.to_f / (1/n)
> end
>
> For some reason they don't give the exact same answer. Try an edge case.
>
> irb(main):066:0> 0.1 * (134.45/0.1).round
> => 134.4
> irb(main):067:0> (134.45 * (1/0.1)).round.to_f / (1/0.1)
> => 134.5
>
> T.

Ara.T.Howard

10/15/2004 3:08:00 PM

Markus

10/15/2004 4:03:00 PM

On Fri, 2004-10-15 at 07:28, trans. (T. Onoma) wrote:
> On Thursday 14 October 2004 01:55 pm, Markus wrote:
> | Here's mine (with a few extras):
> |
> | class Numeric
> | def to_the_nearest(n)
> | n*(self/n).round
> | end
> | end
> |
>
> def round_to_nearest( n=0.01 )
> (self * (1/n)).round.to_f / (1/n)
> end
>
> For some reason they don't give the exact same answer. Try an edge case.
>
> irb(main):066:0> 0.1 * (134.45/0.1).round
> => 134.4
> irb(main):067:0> (134.45 * (1/0.1)).round.to_f / (1/0.1)
> => 134.5

Peeling one layer of the onion, they differ because:

irb(main):033:0> printf "%20.15f",(134.45 * (1.0/0.1))
1344.500000000000000=> nil
irb(main):034:0> printf "%20.15f",(134.45/0.1)
1344.499999999999773=> nil

I suppose this is not unexpected (my mama warned me 'bout floats) but it
is a little unexpected--no, I'm wrong, 1/5 is a repeating decimal base
2, so it's perfectly expected. The tricky bit is, which form (if either
of them) will always (or at least, more generally) give the correct
result? I can't see off hand that either will be intrinsically "better"
but I could be missing a point. I suspect (SWAG) that the hybrid:

def to_the_nearest(n)
if self.abs < 1.0
(self/n).round/n
else
m = 1.0/m
(self*m).round/m
end
end

would do better than both, but I haven't tested it.

-- Markus

P.S. Mine has mostly been tested on values < 1.0, thus my suspicion that
it works well in that domain. Typing this though, I realize that it was
tested FOR CONFORMANCE WITH A PRE-EXISTING SYSTEM* which itself may have
been buggy.

* Think 25 year old spaghetti FORTRAN, then sigh in bliss realizing that
your imagination is much nicer than the ugly facts.

T. Onoma

10/15/2004 5:26:00 PM

On Friday 15 October 2004 11:14 am, Ara.T.Howard@noaa.gov wrote:| Â Â jib:~/eg/ruby > cat truncate.rb| Â Â class Float| Â Â Â def truncate sd = 2| Â Â Â Â i = self.to_i| Â Â Â Â prec = 10 ** sd| Â Â Â Â fraction = (i.zero? ? self : (self / i) - 1)| Â Â Â Â i + (fraction * prec).to_i / prec.to_f| Â Â Â end| Â Â end|| Â Â p(0.1 * (134.45/0.1).truncate) Â Â Â Â Â Â Â Â Â #=> 134.4| Â Â p((134.45 * (1/0.1)).truncate.to_f / (1/0.1)) Â Â #=> 134.4|| Â Â jib:~/eg/ruby > ruby truncate.rb| Â Â 134.4| Â Â 134.4
Sorry, I don't follow. Isn't that answer incorrect? The mid-point should round up, not down. At least, that's what I was taught in school.
T.

comp.lang.ruby

Rounding to X digits

Eric Anderson

Eric Anderson

James Gray

Eric Anderson

Robert Klemme

Markus

T. Onoma

Markus

Ara.T.Howard

Markus

T. Onoma

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