ruby array中找出重复元素

#上文的inject现在用上了
module Enumerable
  def dups
    inject({}) {|h,v| h[v]=h[v].to_i+1; h}.reject{|k,v| v==1}.keys
  end
end

arr = %w{foo bar baz bar baz qux foo zub}
puts arr.dups.inspect
# => ["baz", "foo", "bar"]

#替换一
inject(Hash.new(0)) {|h,v| h[v] += 1; h}.reject{|k,v| v==1}.keys

这个是比较好理解的

class Array
  def only_duplicates
    duplicates = []
    self.each {|each| duplicates << each if self.count(each) > 1}
    duplicates.uniq
  end
end

puts [1,2,2,4,5,1].only_duplicates.inspect
==> [1, 2]

require 'benchmark' 

module Enumerable
   def map_with_index
      index = -1
      (block_given? && self.class == Range || self.class == Array)  ?  map { |x| index += 1; yield(x, index) }  :  self
   end
end


class Array

   def find_dups
      inject(Hash.new(0)) { |h,e| h[e] += 1; h }.select { |k,v| v > 1 }.collect { |x| x.first }
   end




   def find_dups2
      uniq.select{ |e| (self-[e]).size < self.size - 1 }
   end

   def find_ndups     # also returns the number of items
      uniq.map { |v| diff = (self.size - (self-[v]).size); (diff > 1) ? [v, diff] : nil}.compact
   end


   # cf. http://www.ruby-forum.com/topic/122008
   def dups_indices   
      (0...self.size).to_a - self.uniq.map{ |x| index(x) }
   end

   def dup_indices(obj)
      i = -1
      ret = map { |x| i += 1; x == obj ? i : nil }.compact
      #ret = map_with_index { |x,i| x == obj ? i : nil }.compact
      ret.shift
      ret
   end

   def delete_dups(obj)
      indices = dup_indices(obj)
      return self if indices.empty?
      indices.reverse.each { |i| self.delete_at(i) }
      self
   end

end  


array = [1,3,5,5,6,7,9,10,14,18,22,22,4,4,4,3,6]

dups = nil


Benchmark.bm(14) do |t| 

 t.report('find_dups:') do
    dups = array.find_dups
 end 

end 

p dups   #=> [5, 22, 6, 3, 4]


p %w(a b a c c d).dups_indices
p %w(a b a c c d).dup_indices('c')
p %w(a b a c c d).delete_dups('a')

class Array

  def find_dups
    uniq.map {|v| (self - [v]).size < (self.size - 1) ? v : nil}.compact
  end

end

 def mostused(ahash)
    original_size = ahash.size
    sizes = {}   
    ahash.uniq.map { |x| sizes[x] = original_size - ((ahash - [x]).size)}
    sizes.sort_by {|k,v| v}.pop      
  end

评论

1 楼 陈金山 2012-10-30  

>> a = [1,1,2,2,3,3,4]
[1, 1, 2, 2, 3, 3, 4]
>> b = a & a
[1, 2, 3, 4]