From: "Abinoam Jr." <abinoam@...>
Date: 2011-01-17T13:23:23+09:00
Subject: Re: Ruby iterative depth.

Hi Marcello,

Sorry, but I was not able to figure out exactly how this piece of code
is trying to generate your desired outcome. (my fault!)
If you're just willing to represent the data in a kind o hierarchical
tree, there's a lot of "Ruby way" to do that.
It seams (in my humble opinion) that you're trying to do it in "C"
way, or something like that.
But, I can give you some tips. See comments in the code.

# You shouldn't need to define methods outside the object scope (like
the one bellow).

# Métodos busca em profundidade iterativo
def menu_iterative(menu)
  res = []

  # When shifting the first element you're assuming the Array is in a
kind of order,
  # and the first element is the "root". Is this always true?
  el1 = menu.shift
  res << el1
  while not menu.empty?
    res_aux = []
    for i in menu
      if el1.id == i.parent_id
        print el1," == ",i,"* ",res_aux.index(el1),"\n"
        res_aux <<  i
		
		# The same shift/delete problem here
		# This elemente will never be reevaluated after deleted. And...
		# Using delete here you're assuming that the elements are unique.
		# Is this always true. If there's 2 "i" elements in the Array, with
		# this line you'll be deleting the two of them.
		# You can delete by the index instead.
        el1 = menu.delete(i)
		
		# This use of retry should be considered deprecated.
		# It seems removed from Ruby 1.9.
        #       retry		
	  else
        print el1," != ",i," \n"
      end
    end
    res << res_aux unless res_aux.empty?
    el1 = menu.shift
    res << el1
  end
  # You don't need return here, the "return value" is always the last
evaluated expression
  return res
end

Good luck!

On Fri, Jan 14, 2011 at 11:10 PM, Marcello Henrique <faraohh@gmail.com> wrote:
> Hi Abinoam,
>
> Thanks for reply, actually the first number is 'id' and second
> 'parent_id', so correct is:
> ["1=>0",
>  ["2=>1",
>    ["3=>2",
>      ["7=>3"]
>    ],"5=>1","6=>1"
>  ],"4=>0"
> ]
>
> From source flat:
> [ "1=>0","2=>1", "3=>2","4=>0","5=>1","6=>1","7=>3"]
>
> The purpose is to use such as menu like this, based in example above:
> <menu>
>  <ul>
>    <li class="sub">"1=>0"
>      <ul>
>        <li class="sub">"2=>1"
>          <ul>
>            <li class="sub">"3=>2"</li>
>              <ul>
>                <li>"7=>3"</li>
>              </ul>
>            </li>
>          </ul>
>        </li>
>        <li>"5=>1"</li>
>        <li>"6=>1"</li>
>      </ul>
>    </li>
>    <li>"4=>0"</li>
>  </ul>
> </menu>
>
> Thanks in advance!
>
> On Fri, Jan 14, 2011 at 21:29, Abinoam Jr. <abinoam@gmail.com> wrote:
>> Hi Marcello,
>>
>> For helping me understand your desired outcome I manually turned it
>> into something more readable (just to read the nesting [] ).
>>
>> ["1=>0",["2=>1",["3=>2",["7=>3"]],["3=>2"]],["2=>1","5=>1","6=>1"]]
>>
>> arr[0] = "1=>0"
>> arr[1] = ["2=>1", ["3=>2", ["7=>3"]], ["3=>2"]]
>> arr[2] = ["2=>1", "5=>1", "6=>1"]
>>
>> Is this what you want?
>>
>> On Fri, Jan 14, 2011 at 11:54 AM, Marcello Henrique <faraohh@gmail.com> wrote:
>>> Hello,
>>>
>>> Short question:
>>>
>>> How to make an iterative method of stack?
>>>
>>> Long explanation:
>>> I'm trying to create an iterative method for following structure:
>>>
>>> class Obj
>>>   attr_accessor: id,: parent_id
>>>
>>>   def initialize (id, parent_id)
>>>     parent_id = @ parent_id
>>>     @ id = id
>>>   end
>>>
>>>   def to_s
>>>     "id: # {@ id}, parent_id: @ # {parent_id} '
>>>   end
>>> end
>>>
>>> left = [Obj.new (1.0), Obj.new (2.1), Obj.new (3.2), Obj.new (4.0),
>>> Obj.new (5.1), Obj new (6.1), Obj.new (7.3)]
>>>
>>> My challenge is to make an iterative method, see how far I got:
>>>
>>> def menu_iterative (menu)
>>>   res = []
>>>
>>>   el1 = menu.shift
>>>   res << el1
>>>   while not menu.empty?
>>>     res_aux = []
>>>     for i in menu
>>>       if el1.id == i.parent_id
>>>         print el1, "==" i "* ", res_aux.index (el1), "\ n"
>>>         res_aux << i
>>>         el1 = menu.delete (i)
>>>         retry
>>>       else
>>>         print el1, "! =" i, "\ n"
>>>       end
>>>     end
>>>     res << res_aux unless res_aux.empty?
>>>     puts
>>>     el1 = menu.shift
>>>     res << el1
>>>   end
>>>   return res
>>> end
>>>
>>> puts "result:"
>>> pp menu_iterative (left_it)
>>>
>>> I wish the outcome was:
>>>
>>> [*,
>>>  [#<Obj:0x7f35b48cab18 @id=2, @parent_id=1>,
>>>  [#<Obj:0x7f35b48caaf0 @id=3, @parent_id=2>,
>>>   [#<Obj:0x7f35b48caa50 @id=7, @parent_id=3>]],
>>>  [#<Obj:0x7f35b48caaf0 @id=3, @parent_id=2>]],
>>>  [#<Obj:0x7f35b48cab18 @id=2, @parent_id=1>,
>>>  #<Obj:0x7f35b48caaa0 @id=5, @parent_id=1>,
>>>  #<Obj:0x7f35b48caa78 @id=6, @parent_id=1>]]
>>>
>>> It seems that the problem is attribution without stack, correct?
>>> I'm attaching all I did.
>>>
>>> Thanks for any help.
>>> --
>>> Marcello Henrique
>>> Blog - http://faraohh.wordpress.com
>>> Associação Software Livre de Goiás (www.aslgo.org.br)
>>> Cercomp - UFG (www.cercomp.ufg.br)
>>>
>>
>>
>
>
>
> --
> Marcello Henrique
> Blog - http://faraohh.wordpress.com
> Associação Software Livre de Goiás (www.aslgo.org.br)
> Cercomp - UFG (www.cercomp.ufg.br)
>
>