Re: zero placeholder exponential sprintf?


ruby -e "puts sprintf('%.6e',1.123450e+2)" #=> 1.123450e+02

but I want '0.112345e+03', i.e., a leading zero placeholder.

Hmmm, now just need the '+0' part in the exponent and control
of the number of decimal places?


I am still not sure if you require sprintf.
I am not familiar with that but I will learn about it starting tomorrow. Thanks.

This code looks a bit strange to me and it is not very DRY.
Maybe someone will have a better solution soon. There must be a better way.
It may have some problems, so check it carefully. It is the best I can
offer you this late at night.

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

--
A Look into Japanese Ruby List in English
http://www.kakueki.com/ruby/list.html

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