How to compare arrays inside an array with each other in ruby?

Question

[ 1, 1, 3, 5 ] & [ 1, 2, 3 ]                 #=> [ 1, 3 ]
[ 'a', 'b', 'b', 'z' ] & [ 'a', 'b', 'c' ]   #=> [ 'a', 'b' ]

I need the intersection of each array with all other arrays within an array.

So the array could look like ->

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

The result should look like ->

 a = [[3],[3,4,5][4,5]]

Any suggestions?

You should still provide an example for other looking at this issue. I did give a solution, but it was by making an educated guess. — hirolau
– hirolau, Commented Aug 20, 2016 at 12:59
Data ->> a = [[1, 2, 3], [3, 4, 5], [4, 5, 6]] Extraction via intersection (expected result) ->> a = [[3],[3,4,5][4,5]] thats what i need.. — DDD
– DDD, Commented Aug 20, 2016 at 13:02
It's not clear to me what needs to be intersected with what - why is [1,2,3] not intersected with [4,5,6] for example. Add any clarifying details to the question itself rather than just as a comment. — Frederick Cheung
– Frederick Cheung, Commented Aug 21, 2016 at 15:11
@FrederickCheung Question mended. The issue is already solved by the answers. — DDD
– DDD, Commented Aug 21, 2016 at 17:08

hirolau · Accepted Answer · 2016-08-20 16:48:06Z

2

Look into the combination method.

a = [[1, 2, 3], [3, 4, 5], [4, 5, 6],[1,"a","b"]]

p a.combination(2).map{|x,y| x & y } #=> [[3], [], [1], [4, 5], [], []]

And if you do not want the empty arrays in there:

p a.combination(2).map{|x,y| x & y }.reject(&:empty?) #=> [[3], [1], [4, 5]]

Edit: After seeing some examples what OP actually want here is how I would achieve the desired result:

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

def intersect_with_rest(array)
  array.size.times.map do
    first, *rest = array
    array.rotate!
    first & rest.flatten
  end
end

p intersect_with_rest(original) #=> [[3], [3, 4, 5], [4, 5]]
p original #=> [[1, 2, 3], [3, 4, 5], [4, 5, 6]]

Or:

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

result = original.map.with_index do |x,i|
  x & (original[0...i]+original[1+i..-1]).flatten
end

p result #=> [[3], [3, 4, 5], [4, 5]]

edited Aug 20, 2016 at 16:48

answered Aug 20, 2016 at 12:51

hirolau

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DDD · Accepted Answer · 2016-08-20 13:55:21Z

0

Yeah, finally I found a solution. Maybe there is a simpler way, but that works for me now..

c = [[1,2,3],[3,4,5],[4,5,6]]
results = [];c.length.times.each {|e| results.push c.rotate(e).combination(2).map {|x, y| x & y}}
results.map{|x, y| y + x}

=> [[3], [3, 4, 5], [4, 5]]

Thanks to @hirolau for the hint. Best regards

answered Aug 20, 2016 at 13:55

DDD

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