[ 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?
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]]
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
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