Number format

ruby-talk
1

You can use the number.round method, but this doesn't take a number of decimal places.

To round 1.234 to 1.23 use this code:
x = 1.234
x = x * 100 # -> 123.4
x = x.round -> 123
x = x / 100.0 => 1.23

···

At least, that's the best method I can find from the Ruby docs... Dan Li Chen wrote:
Hi all,

I have a number, for example, 1.123456789. What is the Ruby way to
change it into whatever number of floating points such as 1.12,
1.123,1.1234568 or 1.12345679.

Thanks,

Li

class Float
  def places(places)
    (self * 10 ** places).truncate / 10.0 ** places
  end
end

1.123456789.places(5) => 1.12345

you can use round instead of truncate if you want to round off

alternate method
class Float
  def place(places)
    sprintf("%0.#{places}f", self).to_f
  end
end

not sure which method is most efficient

2

Bernard Kenik wrote:

You can use the number.round method, but this doesn't take a number of
decimal places.

To round 1.234 to 1.23 use this code:
x = 1.234
x = x * 100 # -> 123.4
x = x.round -> 123
x = x / 100.0 => 1.23

At least, that's the best method I can find from the Ruby docs...

Second reply to the second copy of this identical post. This is not a good
idea. The problem is that Ruby's numbers are stored internally in binary
form, and this multiply -- truncate -- divide process will not work as
intended for a lot of numbers.

It is better to retain the full resolution of binary numbers internally, and
simply print the number with the desired number of decimal places:

···

--------------------------------

#!/usr/bin/ruby -w

0.upto(10) do
  n = rand() * 100
  printf("%5.02f\n",n)
end
--------------------------------

Output:

18.64
61.11
58.18
61.80
93.98
50.86
55.28
44.93
87.23
1.38
11.63

--
Paul Lutus
http://www.arachnoid.com