Question

I want to put in a space at every third character when formatting a number. According to this spec:

  it "should format an amount" do
    spaces_on( 1202003 ).should == "1 202 003"
  end

and I came up with this piece of code that does the job

  def spaces_on amount
    thousands = amount / 1000
    remainder = amount % 1000
    if thousands == 0
      "#{remainder}"
    else
      zero_padded_remainder = '%03.f' % remainder
      "#{spaces_on thousands} #{zero_padded_remainder}"
    end
  end

So my question is if this was the best way to do it. I suspect that there may be a regex way around it but I am not sure I will like the readability of that. (On the other hand – the %03.f magic is not very readable either….)

Answer 1

>> def spaces_on number
>>   number.to_s.gsub(/\D/, '').reverse.gsub(/.{3}/, '\0 ').reverse
>> end
=> nil
>> spaces_on 12345678
=> "12 345 678"

Maybe there’s an argument that regular expressions aren’t amazingly readable, but personally I think they’re simpler to understand than having to think in recursion.