Creating a dynamic hash

ruby-talk
1

Hi
     I am trying to draw a graph My question is not related to that
In the below @tic_num_array is an array contains values like
["IN1","IN2","IN3".......] and @tic_elapsed_time_array contains values
like [10,2,12,.......] They are dynamic arrays

chart = Ziya::Charts::Mixed.new( 'licence', "status_chart" )
    chart.add :chart_types, %w[column line line]
    chart.add :axis_category_text, @tic_num_array
    chart.add :series, "Ticket" , report(
"ticket",@tic_elapsed_time_array )
    chart.add :series, "Warning" , report(
"warning",@tic_elapsed_time_array )
    chart.add :theme , "them1"

def report( charttype,elapsedtime )
      case charttype
        when "ticket"
          elapsedtime.each { |etime| Hash.new("shadow" => 'high',:bevel
=> 'bevel1', :value => etime.to_i) }
       when "warning"
         [
            { :bevel => 'bevel1', :value => 5 },
            { :bevel => 'bevel1', :value => 5 },
            { :bevel => 'bevel1', :value => 5 }
         ]
     end
end

       The above works fine ..But it is clear the case when "warning" is
dynamic So suppose some 20 values theer the above approch is wrong So
for that I tried the "warning " like
when "warning"
    (elapsedtime.length-1).times { Hash.new(:bevel => 'bevel1', :value
=> 5)}

         Here my I thought the above code for a value elapsedtime.length
= 4 generate
                [
            { :bevel => 'bevel1', :value => 5 },
            { :bevel => 'bevel1', :value => 5 },
            { :bevel => 'bevel1', :value => 5 }
         ]
        But this does not happen and also the graph now is not working?
Please help me to solve this

Thanks in advance
sijo

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2

Hi
    I could solve the problem like
hashes = []
          for i in 0..elapsedtime.length-1
             hashes[i] = {:bevel => 'bevel1',:value => 5}
          end
return hashes

sijo

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3

You can do it like this

hashes = [{:bevel => 'bevel1',:value => 5}] * (elapsedtime.length-1)

HTH,

Park Heesob

···

2008/11/27 Sijo Kg <sijo@maxxion.com>:

Hi
   I could solve the problem like
hashes =
         for i in 0..elapsedtime.length-1
            hashes[i] = {:bevel => 'bevel1',:value => 5}
         end
return hashes

4

Hi --

Hi
   I could solve the problem like
hashes =
         for i in 0..elapsedtime.length-1
            hashes[i] = {:bevel => 'bevel1',:value => 5}
         end
return hashes

Let Ruby do the work :slight_smile:

   hashes = Array.new(elapsedtime.length-1) { {:bevel => 'bevel1',
                                               :value => 5 } }

That gives you an array of the right length, where each element is
initialized by running the code block, which gives you a new hash each
time.

David

···

On Thu, 27 Nov 2008, Sijo Kg wrote:

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   Intro to Ruby on Rails January 12-15 Fort Lauderdale, FL
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   * Co-taught with Patrick Ewing!
See http://www.rubypal.com for details and updates!

5

Hi --

···

On Thu, 27 Nov 2008, Heesob Park wrote:

2008/11/27 Sijo Kg <sijo@maxxion.com>:

Hi
   I could solve the problem like
hashes =
         for i in 0..elapsedtime.length-1
            hashes[i] = {:bevel => 'bevel1',:value => 5}
         end
return hashes

You can do it like this

hashes = [{:bevel => 'bevel1',:value => 5}] * (elapsedtime.length-1)

That's not the same, because you get the same hash n times, instead of
n hashes once each.

hashes = [{:x => 1, :y => 2}] * 3
hashes[0].delete(:x)
p hashes # [{:y=>2}, {:y=>2}, {:y=>2}]

David

--
Rails training from David A. Black and Ruby Power and Light:
   Intro to Ruby on Rails January 12-15 Fort Lauderdale, FL
   Advancing with Rails January 19-22 Fort Lauderdale, FL *
   * Co-taught with Patrick Ewing!
See http://www.rubypal.com for details and updates!

6

Hi
  Thnaks for all the reply..So could you please tell this differnece
also..I tried both ways
First

hashes = Array.new(elapsedtime.length) { {:bevel => 'bevel1',
                                               :value => 5 } }
   Here I did not want the statement
   return hashes

In the second way
          hashes = []
          for i in 0..elapsedtime.length-1
             hashes[i] = {:bevel => 'bevel1',:value => 5}
          end
          return hashes
     Here when I commented return hashes it did not work

Sijo

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