How to loop through a hash in ruby and determine a final value

Question

I have a variable that prints like this:

{:type1=>:poor, :type2=>:avg, :type3=>:best}

I want to iterate through this hash and print out a final value based on following rules

if value for any of the types is :best then final value is best
if value for any of the types is :avg and there are no best then final value is avg
otherwise final value is poor

basically the strongest ones wins.

I've tries the following

def final_value(values)
  val = "poor"
  values.each do |key, val|

  end
  val
end

You also have symbols :poor etc, but you wrote as if there are variable/methods poor etc. Then, you are even mentioning strings "poor" etc. — sawa
– sawa, Commented Apr 8, 2013 at 17:54
I'm simply treating them as strings in my loop even though they are symbols — birdy
– birdy, Commented Apr 8, 2013 at 17:54

Fred · Accepted Answer · 2013-04-08 17:58:03Z

5

Use Hash's has_value? method.

val = :poor;
val = :avg  if values.has_value?(:avg);
val = :best if values.has_value?(:best);
val

answered Apr 8, 2013 at 17:58

Fred

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4 Comments

Nir Alfasi Over a year ago

+1 good answer - indeed, iteration is unnecessary! you can slightly improve it by looking for best first, cause then you might not need to look for avg

Alex Wayne Over a year ago

@alfasin The implementation of the has_value? method does iteration. Iteration is necessary, it's just a matter of explicit or implicit iteration.

Nir Alfasi Over a year ago

@AlexWayne I think that you understood what I meant didn't you ? :) In case not: I was comparing it to cainy393's answer - I meant that it is unnecessary to implement iteration since Ruby does it for you using has_value?. The "implicit" way is more elegant - that's what I meant.

Nir Alfasi Over a year ago

BTW, on a second thought, iteration is actually NOT needed, if the implementation would change to hold an "opposite" Hash (one in which values point to the boolean "true") -it's could be done in O(1). I think that the reason it's not implemented is in order to not waste memory-space, which is a good reason. Just a thought.

cainy393 · Accepted Answer · 2013-04-08 17:56:04Z

0

Some thing like this?

final_val = :poor
values.each() do |val|
    if val = :avg && final_val = :poor
        final_val = :avg
    end
    if val = :best && final_val = :avg
        final_val = :best
    end
end
return final_val

answered Apr 8, 2013 at 17:56

cainy393

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Comments

Roney Michael · Accepted Answer · 2013-04-08 18:05:52Z

0

You can achieve what you need as follows:

def final_value(values)
    val = :poor;
    values.each do|k, v|
        if(v==:best)
            val=:best;
            break;
        elsif(val==:poor && v==:avg)
            val=:avg;
        end
    end
    puts "Final value is: "+val.to_s;
end

values = {:type1=>:poor, :type2=>:avg, :type3=>:best};
final_value(values)

answered Apr 8, 2013 at 18:05

Roney Michael

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Comments

Arup Rakshit · Accepted Answer · 2013-04-08 19:04:11Z

0

def compare(h,val)
 m = ((h.has_value?(val) and val != :poor) ? ((val == :best or h.has_value?(:best))? :best : :avg ) : :poor)
end
h = {:type1=>:poor, :type2=>:avg, :type3=>:best}

p compare(h,:avg) #:best
p compare(h,:trr) #:poor
p compare(h,:best) #:best

edited Apr 8, 2013 at 19:04

answered Apr 8, 2013 at 18:22

Arup Rakshit

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