From: Markus <markus@...>
Date: 2004-10-15T02:55:19+09:00
Subject: Re: Rounding to X digits

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
>