From: Harry Kakueki <list.push@...>
Date: 2009-08-06T06:51:06+09:00
Subject: Re: zero placeholder exponential sprintf?

On Wed, Aug 5, 2009 at 10:57 PM, Harry Kakueki<list.push@gmail.com> wrote:
>>
>
> x = 1.12345e2
>
> sig,y = 9, Math.log10(x).floor + 1
> p (x*10**(y*-1)).to_s[0..sig+1].ljust(sig+2,"0") + "e+" +
> y.to_s.rjust(2,"0") if y >= 0
> p (x*10**(y*-1)).to_s[0..sig+1].ljust(sig+2,"0") + "e-" +
> y.abs.to_s.rjust(2,"0") if y < 0
>
>
> Harry
>

This is the same code I posted earlier, just a little DRYer.

x = 1.12345e2

s,y,h = 7, Math.log10(x).floor+1,{-1=>"-",0=>"+",1=>"+"}
p (x*10**(y*-1)).to_s[0..s+1].ljust(s+2,"0")+"e"+h[(y<=>0)]+y.abs.to_s.rjust(2,"0")


Harry


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