Rounding any number (int or float) to 3 significant figures

[Max Williams]

I have a requirement where i need to display (ie convert to a string)
all digits to 3 significant figures. (not the same as 3 decimal places)

eg

23 => "23"
26759 => "26,800"
10.546 => "10.5"
3332332 => "3,330,000"
0.766 => "0.766"
0.00000766 => "0.00000766"

Can anyone show me a neat way to do this? I'd have thought that there'd
be a method for it already but i can't find one.

thanks!
max

 

[Rob Biedenharn]

I have a requirement where i need to display (ie convert to a string)
all digits to 3 significant figures. (not the same as 3 decimal
places)

eg

23 => "23"
26759 => "26,800"
10.546 => "10.5"
3332332 => "3,330,000"
0.766 => "0.766"
0.00000766 => "0.00000766"

Can anyone show me a neat way to do this? I'd have thought that
there'd
be a method for it already but i can't find one.

thanks!
max
--

This should get you started:

irb> h.each do |n,s|
irb> puts( ("%f"%[("%.3g"%n).to_f]).sub(/\.?0*\z/,'') )
irb> end
0.766
23
26800
10.5
3330000
0.000008
=> {0.766=>"0.766", 23=>"23", 26759=>"26,800", 10.546=>"10.5",
3332332=>"3,330,000", 7.66e-06=>"0.00000766"}

It doesn't do so well with the smallest value, but if you know the
ranges you're dealing with, perhaps you can adjust the %f spec.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)

 

[Harry Kakueki]

I have a requirement where i need to display (ie convert to a string)
all digits to 3 significant figures. (not the same as 3 decimal places)

eg

23 => "23"
26759 => "26,800"
10.546 => "10.5"
3332332 => "3,330,000"
0.766 => "0.766"
0.00000766 => "0.00000766"

Can anyone show me a neat way to do this? I'd have thought that there'd
be a method for it already but i can't find one.

thanks!
max

Sorry, but I don't have much time that I can spend to come up with a
better answer right now.
But, I thought I would throw this idea your way.
Check it carefully (it may have bugs) and maybe you can make improvements.


require 'bigdecimal'

arr = [23,26759,10.546,3332332,0.766,0.00000766]
arr.each do |x|
m = 2 - Math.log10(x).floor
p BigDecimal.new(((x*10**m).round*10**(-1*m)).to_s).to_s('F').gsub(/\.0*$/,"")
end

#output
##########
#> "23"
#> "26800"
#> "10.5"
#> "3330000"
#> "0.766"
#> "0.00000766"


Harry

 

[Robert Dober]

I have a requirement where i need to display (ie convert to a string)
all digits to 3 significant figures. =A0(not the same as 3 decimal places= )

eg

=A0 =A023 =3D> "23"
=A0 =A026759 =3D> "26,800"
=A0 =A010.546 =3D> "10.5"
=A0 =A03332332 =3D> "3,330,000"
=A0 =A00.766 =3D> "0.766"
=A0 =A00.00000766 =3D> "0.00000766"

Can anyone show me a neat way to do this? =A0I'd have thought that there'= d
be a method for it already but i can't find one.

thanks!
max

At frirst sight and testing with the data you have given, the
following pretty much seems to do the trick:

x.to_s.sub( /(\.0*\d{0,3}).*/, '\1' )

Did I overlook any edge cases?

HTH
R.

 

[Max Williams]

Thanks rob!

I combined that with the rails helper 'number_with_delimiter, and it
works great:

def number_to_n_significant_digits(number, n = 3)
("%f"%[("%.#{n}g"%number).to_f]).sub(/\.?0*\z/,'')
end

#in my other method
number_with_delimiter(number_to_n_significant_digits(number))

I don't think the 0.0000008 instead of 0.000000766 case will be a
problem.

My next quest is to work out what's going on in your code snippet.
I can see that the sub at the end removes trailing zeros, right?

And the first part looks like some oldschool c-style string
substitution, time to pull out the pickaxe

thanks again!
max

 

[Max Williams]

A great bunch of solutions, thanks a lot guys. More nice tasks for me
trying to reverse engineer them as well (especially Harry's Math
solution)

cheers!
max

 

[Rob Biedenharn]

Thanks rob!

I combined that with the rails helper 'number_with_delimiter, and it
works great:

def number_to_n_significant_digits(number, n = 3)
("%f"%[("%.#{n}g"%number).to_f]).sub(/\.?0*\z/,'')
end

#in my other method
number_with_delimiter(number_to_n_significant_digits(number))

I don't think the 0.0000008 instead of 0.000000766 case will be a
problem.

Okey-dokey! "soft" requirement for the win!

My next quest is to work out what's going on in your code snippet.
I can see that the sub at the end removes trailing zeros, right?

Yes, in English it says:

If there is a series of 0's (0*) at the end of the string (\z)
possibly preceeded by a decimal point (\.?), replace all that with
nothing (i.e., delete it).

And the first part looks like some oldschool c-style string
substitution, time to pull out the pickaxe

thanks again!
max

String#% will direct you to Kernel#sprintf, but you're already on the
right track.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)