CODEWITHSUNDEEP
rubyfloating-pointrounding
dotty
dotty

Reputation: 41433

Round a ruby float up or down to the nearest 0.05

I'm getting numbers like 

2.36363636363636
4.567563
1.234566465448465
10.5857447736

How would I get Ruby to round these numbers up (or down) to the nearest 0.05?

Upvotes: 25

Views: 31177

Answers (9)

boulder_ruby
boulder_ruby

Reputation: 39695

To round to the nearest 2 places instead:

(5.65235534).round(2)
#=> 5.65

Upvotes: 12

bimlas
bimlas

Reputation: 2587

I know that the question is old, but I like to share my invention with the world to help others: this is a method for rounding float number with step, rounding decimal to closest given number; it's usefull for rounding product price for example:

def round_with_step(value, rounding)
  decimals = rounding.to_i
  rounded_value = value.round(decimals)

  step_number = (rounding - rounding.to_i) * 10
  if step_number != 0
    step = step_number * 10**(0-decimals)
    rounded_value = ((value / step).round * step)
  end

  return (decimals > 0 ? "%.2f" : "%g") % rounded_value
end

# For example, the value is 234.567
#
# | ROUNDING | RETURN | STEP
# | 1        | 234.60 | 0.1
# | -1       | 230    | 10
# | 1.5      | 234.50 | 5 * 0.1 = 0.5
# | -1.5     | 250    | 5 * 10  = 50
# | 1.3      | 234.60 | 3 * 0.1 = 0.3
# | -1.3     | 240    | 3 * 10  = 30

Upvotes: 0

tokhi
tokhi

Reputation: 21618

Ruby 2 now has a round function:

# Ruby 2.3
(2.5).round
 3

# Ruby 2.4
(2.5).round
 2

There are also options in ruby 2.4 like: :even, :up and :down e.g;

(4.5).round(half: :up)
 5

Upvotes: 5

Rob
Rob

Reputation: 4434

Here's a general function that rounds by any given step value:

place in lib: 

lib/rounding.rb
class Numeric
  # round a given number to the nearest step
  def round_by(increment)
    (self / increment).round * increment
  end
end

and the spec: 

require 'rounding'
describe 'nearest increment by 0.5' do
  {0=>0.0,0.5=>0.5,0.60=>0.5,0.75=>1.0, 1.0=>1.0, 1.25=>1.5, 1.5=>1.5}.each_pair do |val, rounded_val|
    it "#{val}.round_by(0.5) ==#{rounded_val}" do val.round_by(0.5).should == rounded_val end
  end
end

and usage: 

require 'rounding'
2.36363636363636.round_by(0.05)

hth.

Upvotes: 8

Fellow Stranger
Fellow Stranger

Reputation: 34013

To get a rounding result without decimals, use Float's .round

5.44.round
=> 5

5.54.round
=> 6

Upvotes: 3

Smar
Smar

Reputation: 8591

It’s possible to round numbers with String class’s % method.

For example

"%.2f" % 5.555555555

would give "5.56" as result (a string).

Upvotes: 4

Damian
Damian

Reputation: 1553

Check this link out, I think it's what you need. Ruby rounding

class Float
  def round_to(x)
    (self * 10**x).round.to_f / 10**x
  end

  def ceil_to(x)
    (self * 10**x).ceil.to_f / 10**x
  end

  def floor_to(x)
    (self * 10**x).floor.to_f / 10**x
  end
end

Upvotes: 21

Bombe
Bombe

Reputation: 83846

In general the algorithm for “rounding to the nearest x” is:

round(x / precision)) * precision

Sometimes is better to multiply by 1 / precision because it is an integer (and thus it works a bit faster):

round(x * (1 / precision)) / (1 / precision)

In your case that would be:

round(x * (1 / 0.05)) / (1 / 0.05)

which would evaluate to:

round(x * 20) / 20;

I don’t know any Python, though, so the syntax might not be correct but I’m sure you can figure it out.

Upvotes: 20

sepp2k
sepp2k

Reputation: 370102

[2.36363636363636, 4.567563, 1.23456646544846, 10.5857447736].map do |x|
  (x*20).round / 20.0
end
#=> [2.35, 4.55, 1.25, 10.6]

Upvotes: 25

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