[Resolved] Closest to Value

[Sarkilas]

I've recently encountered an obstacle I somehow can't seem to get past. I want to check which of two values that is closest to a defined value.
So I have two variables, which are dynamic, so I am using this:

var1 /= 21 ; var2 /= 21 # splits both on 21
# here I want to call a condition to return a boolean true if var1 is closer to 21 than var2, false if not

So basically, the values of var1 and var2 can range within 10 and 50, so if it's as low as 10, the check value would be ~0,47. So in the case, if var1 and var2 has the check values: 0,47 and 1,1 I want the condition to return false, as var2 had the value 1,1 which proves that it was closest to 21 (1,1 is about 23).

There is probably a ridiculously easy way to do this, but I just couldn't find any ways to do it.

I appreciate it! Thanks in advance.

 

[Sheol]

Like this?

def closerTo21(var1, var2)
	return (21 - var1).abs < (21 - var2).abs
end

 

[Sarkilas]

Like this?

def closerTo21(var1, var2)
	return (21 - var1).abs < (21 - var2).abs
end

Thanks, I just found a method like that, but I had to get which value that was closest to do what I wanted, so I wrote this:

  def closest(a,b,x)
    ary = [x-a, x-b]
    if ary[0].abs < ary[1].abs
      return a
    elsif ary[1].abs < ary[0].abs
      return b
    else
      return nil
    end
  end

 

[Sheol]

You can do it easier ^^

def closest(a, b, x)
  return (a == b) ? nil : ((x - a).abs < (x - b).abs) ? a : b
end

 

[Sarkilas]

You can do it easier ^^

def closest(a, b, x)
  return (a == b) ? nil : ((x - a).abs < (x - b).abs) ? a : b
end

Thanks! That's a much easier and stripped piece. *adds*