Yet another rounding issue

[Manu Sankala]

At school I was taught that five is always rounded up. Later on I've
been told that sometimes five is rounded to closest even number. Neither
of these seem to be the case with ruby:

irb(main):044:0> RUBY_VERSION
=> "1.8.7"
irb(main):045:0> (217.5).round
=> 218
irb(main):046:0> (218.5).round
=> 219
irb(main):047:0> (2.175*100).round
=> 217
irb(main):048:0> (2.185*100).round
=> 219
irb(main):049:0> sprintf('%.2f',2.175)
=> "2.17"
irb(main):050:0> (2.175*100.0).round/100.0
=> 2.17

Can someone please tell me how do I round 2.175 to 2.18, besides
((2.175*100).to_s+'9').to_f.round/100.0 ?
-
Manu S

 

[Jean-Julien Fleck]

Hello,

Can someone please tell me how do I round 2.175 to 2.18, besides
((2.175*100).to_s+'9').to_f.round/100.0 ?

I guess bigdecimal should be the answer
=3D> "0.218E3"

Cheers,

--=20
JJ Fleck
PCSI1 Lyc=E9e Kl=E9ber

 

[Mohit Sindhwani]

At school I was taught that five is always rounded up. Later on I've
been told that sometimes five is rounded to closest even number. Neither
of these seem to be the case with ruby:

irb(main):044:0> RUBY_VERSION
=> "1.8.7"
irb(main):045:0> (217.5).round
=> 218
irb(main):046:0> (218.5).round
=> 219
irb(main):047:0> (2.175*100).round
=> 217
irb(main):048:0> (2.185*100).round
=> 219
irb(main):049:0> sprintf('%.2f',2.175)
=> "2.17"
irb(main):050:0> (2.175*100.0).round/100.0
=> 2.17

Can someone please tell me how do I round 2.175 to 2.18, besides
((2.175*100).to_s+'9').to_f.round/100.0 ?

Actually, this is because of the way floating point variables are
handled in a computer.

You could try:
a = 2.175
b = a + 0.0005
c = ((b * 100).round) / 100.0

irb(main):010:0> a = 2.175
=> 2.175
irb(main):011:0> b = a + 0.0005
=> 2.1755
irb(main):012:0> c = (b * 100).round / 100.0
=> 2.18
irb(main):013:0> a = 2.185
=> 2.185
irb(main):014:0> b = a + 0.0005
=> 2.1855
irb(main):015:0> c = (b * 100).round / 100.0
=> 2.19

Cheers,
Mohit.
22/2/2010 | 1:46 AM.

 

[Manu Sankala]

You could try:
a = 2.175
b = a + 0.0005
c = ((b * 100).round) / 100.0

Well this is basically same as my
((some_float_here*100).to_s+'9').to_f.round/100.0
except that in my version I don't need to know how many decimals the
float originally has. "b = a + 0.0005" would fail if 'a' has mode than 3
decimals.

Since
((float_here*100).to_s+'9').to_f.round/100.0
and
(BigDecimal(float_here.to_s)*100).round.to_s.to_f/100
seem to be about as fast I think I'll go with bigfloat as it seems less
like a hack.

Thank you both for your quick answers

-
Manu S