In [[1,2],[2,1],[8,9],[12,12]], the first two elements are same. I want to get an output like this: [[8,9],[12,12]].
In:
[1,2,3].product([1,2,3])
# => [[1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1], [3, 2], [3, 3]]
I want to get an output like this:
[[1, 1], [2, 2], [3, 3]]
To get unique output irrespective of individual Array's arrangement, do this:
[1,2,3].product([1,2,3]).map(&:sort).uniq
# => [[1, 1], [1, 2], [1, 3], [2, 2], [2, 3], [3, 3]]
If you require output with only repetitive values, do this:
[1,2,3].product([1,2,3]).map {|k,v| [k,v] if k == v}.compact
# => [[1, 1], [2, 2], [3, 3]]
[[12,12]] instead of [[8,9],[12,12]]).
[[12,12]] will be returned for 2nd case. 1st case returns [[1, 2], [8, 9], [12, 12]]. You may say [1, 2] is extra, but the objective here is to point OP in right direction. he can easily remove that if required.Ruby does have a #uniq method, but that removes any duplicates but KEEPS one of them. So that will not work in your case, since it appears that you're looking to remove ALL elements that have duplicates.
Here's a solution that should do the trick:
def remove_dups(array)
new_array = []
# first, sort each element
start_array = array.map{|e| e.sort}
start_array.each_with_index do |sub_array, idx|
temp_array = start_array.dup
temp_array.delete_at(idx)
# after deleting the element, check to see if the array still includes the element
new_array << sub_array unless temp_array.include?(sub_array)
end
new_array
end
p remove_dups([[1,2],[2,1],[8,9],[12,12]])
#=> [[8, 9], [12, 12]]
p remove_dups([1,2,3].product([1,2,3]))
#=> [[1, 1], [2, 2], [3, 3]]