Create a counter in a hash to return first 10 results only

ruby-talk
1

New to ruby..

I created a hash array and everything works fine except I can't figure
out how to include a counter to return the first 10 results only.

Thanks John

Here is the code

h = Hash.new
good_words.each { |w|
    if h.has_key?(w)
      h[w] = h[w] + 1
    else
      h[w] = 1
    end
  }

# sort the hash by value, and then print it in this sorted order

h.sort{|a,b| b[1]<=>a[1]}.each { |elem|

   puts "\"#{elem[0]}\" has #{elem[1]} occurrences"

}

···

--
Posted via http://www.ruby-forum.com/.

2

John Dale wrote:

New to ruby..

I created a hash array and everything works fine except I can't figure
out how to include a counter to return the first 10 results only.

Thanks John

Here is the code

h = Hash.new
good_words.each { |w|
    if h.has_key?(w)
      h[w] = h[w] + 1
    else
      h[w] = 1
    end
  }

# sort the hash by value, and then print it in this sorted order

h.sort{|a,b| b[1]<=>a[1]}.each { |elem|

   puts "\"#{elem[0]}\" has #{elem[1]} occurrences"

}

referred to an earlier post and figured this out.

Thanks

John

···

--
Posted via http://www.ruby-forum.com/\.

3

Like?

h = Hash.new 0
...
h[w] += 1
...

puts h.sort_by {|w,c| -c}[0,10]

Kind regards

  robert

···

On 24.02.2010 20:52, John Dale wrote:

John Dale wrote:

New to ruby..

I created a hash array and everything works fine except I can't figure
out how to include a counter to return the first 10 results only.

Thanks John

Here is the code

h = Hash.new
good_words.each { |w|
     if h.has_key?(w)
       h[w] = h[w] + 1
     else
       h[w] = 1
     end
   }

# sort the hash by value, and then print it in this sorted order

h.sort{|a,b| b[1]<=>a[1]}.each { |elem|

    puts "\"#{elem[0]}\" has #{elem[1]} occurrences"

}

referred to an earlier post and figured this out.

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