Find most common string in an array

Question

I have this array, for example (the size is variable):

   x = [\"1.111\", \"1.122\", \"1.250\", \"1.111\"]

and I need to find the most comm

Answer 1

One pass through the hash to accumulate the counts. Use .max() to find the hash entry with the largest value.

#!/usr/bin/ruby

a = Hash.new(0)
["1.111", "1.122", "1.250", "1.111"].each { |num|
  a[num] += 1
}

a.max{ |a,b| a[1] <=> b[1] } # => ["1.111", 2]

or, roll it all into one line:

ary.inject(Hash.new(0)){ |h,i| h[i] += 1; h }.max{ |a,b| a[1] <=> b[1] } # => ["1.111", 2]

If you only want the item back add .first():

ary.inject(Hash.new(0)){ |h,i| h[i] += 1; h }.max{ |a,b| a[1] <=> b[1] }.first # => "1.111"

---

The first sample I used is how it would be done in Perl usually. The second is more Ruby-ish. Both work with older versions of Ruby. I wanted to compare them, plus see how Wayne's solution would speed things up so I tested with benchmark:

#!/usr/bin/env ruby

require 'benchmark'

ary = ["1.111", "1.122", "1.250", "1.111"] * 1000 

def most_common_value(a)
  a.group_by { |e| e }.values.max_by { |values| values.size }.first
end

n = 1000
Benchmark.bm(20) do |x|
  x.report("Hash.new(0)") do
    n.times do 
      a = Hash.new(0)
      ary.each { |num| a[num] += 1 }
      a.max{ |a,b| a[1] <=> b[1] }.first
    end 
  end

  x.report("inject:") do
    n.times do
      ary.inject(Hash.new(0)){ |h,i| h[i] += 1; h }.max{ |a,b| a[1] <=> b[1] }.first
    end
  end

  x.report("most_common_value():") do
    n.times do
      most_common_value(ary)
    end
  end
end

Here's the results:

                          user     system      total        real
Hash.new(0)           2.150000   0.000000   2.150000 (  2.164180)
inject:               2.440000   0.010000   2.450000 (  2.451466)
most_common_value():  1.080000   0.000000   1.080000 (  1.089784)