```On Thu, Oct 11, 2012 at 8:14 PM, Anna Baas <lists / ruby-forum.com> wrote:
> Hi,
>
> I've tried searching for this on the forum and couldn't find an answer.
> Apologies if it's already out there.
>
> I've converted the digits of an integer into elements of an array, like
> so:
>
> x = some_integer
>
> y = "#{x}".scan(/./)

Better

y = x.to_s.scan /./

You can also do this numerically

irb(main):008:0> x = 113453
=> 113453
irb(main):009:0> y = []
=> []
irb(main):010:0> while x > 0; x, b = x.divmod 10; y.unshift b; end
=> nil
irb(main):011:0> y
=> [1, 1, 3, 4, 5, 3]

respectively

irb(main):015:0> x = 113453
=> 113453
irb(main):016:0> y = []
=> []
irb(main):017:0> while x > 0; x, b = x.divmod 10; y.unshift b.to_s; end
=> nil
irb(main):018:0> y
=> ["1", "1", "3", "4", "5", "3"]

or even

irb(main):030:0> x = 113453
=> 113453
irb(main):031:0> y = []
=> []
irb(main):032:0> while x > 0; y.unshift((x, b = x.divmod 10).last.to_s); end
=> nil
irb(main):033:0> y
=> ["1", "1", "3", "4", "5", "3"]

> Now, after using y for my nefarious purposes, I would like to re-convert
> the elements (still containing one digit each) into an integer.

You make me curious...

> I don't mind writing a routine for it, but was wondering if there's some
> magical way to do it quicker :)

No magic needed

z = y.join.to_i

Kind regards

robert

--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

```