Ruby数组比较而不删除重复项

Question

I have seen many questions regarding this, but not one that addresses the issue of duplicate values.

Below are two arrays. I need to verify ary2 is included in ary1 regardless of the extra duplicates. Also needs to work regardless if the arrays are holding numbers or characters.

ary1 = [1, 1, 1, 2, 2, 3, 4, 5, 6]
ary2 = [1, 1, 2, 3, 5]

Should equal [1, 1, 2, 3, 5] , my code equals [1, 1, 1, 2, 2, 3, 5]

Tried many options, including keep_if or keep_at , delete_if or delete_at , slice , map , etc.

Current code:

ary1.keep_if { |x| ary2.include?(x) }

Answer 1

to verify ary2 is included in ary1

(ary2 - ary1).empty?

should equal [1, 1, 2, 3, 5]

ary2.select { |e| ary1.include?(e) }

Answer 2

Reading between the lines, I'm assuming that " ary2 is included in ary1 " means that, for each element of ary2 , there a unique element of ary1 with the same value. By "unique" I mean, for example, that if ary2 contains two 1 's, ary1 must contains two or more 1 's. If this interpretation of the question is incorrect there is no reason to read further.

If my assumption is correct, we could construct a method (with arguments ary1 and ary2 ) that returns true or false , but if there is a match ( true ) it might be more useful to return the elements of ary1 that are "left over" after elements of ary1 have been matched to the elements of ayr2 .

def ary2_included_in_ary1?(ary1, ary2)
  ary1_cpy = ary1.dup
  ary2.all? do |n|
    idx = ary1_cpy.index n
    return false if idx.nil?
    ary1_cpy.delete_at(idx)
  end
  ary1_cpy
end

ary1 = [1, 1, 1, 2, 2, 3, 4, 5, 6]

ary2 = [1, 1, 2, 3, 5]
ary2_included_in_ary1?(ary1, ary2)
  #=> [1, 2, 4, 6]

ary2 = [1, 1, 2, 3, 7]
ary2_included_in_ary1?(ary1, ary2)
#=> false