From: Marcello Henrique <faraohh@...>
Date: 2011-01-17T20:10:22+09:00
Subject: Re: Ruby iterative depth.

Hi Abinoam,

You are true, I'm trying doing like C and hierarchical tree, what
better way to do this in ruby? It may be that my data structure is
bad, do you have some suggestion?

The ideia is create a menu hierarchical, where there self nestedly by parent_id.
Thank you for help.

On Mon, Jan 17, 2011 at 02:23, Abinoam Jr. <abinoam@gmail.com> wrote:
> 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)
>>
>>
>
>



-- 
Marcello Henrique
Blog - http://faraohh.wordpress.com
Associação Software Livre de Goiás (www.aslgo.org.br)
Cercomp - UFG (www.cercomp.ufg.br)