Reputation: 107
Lets say I have the following arrays:
arr1 = [
['a', 'b'],
['c', 'd'],
['e', 'f']
]
arr2 = [
['g', 'h'],
['i', 'k'],
['a', 'b']
]
I want to find the elements in arr1
that do not exist in arr2
and the index of the elements in arr1
I can do this with the following but this isn't very efficient and is not a pretty solution. There can also be many elements in the arrays so this does not scale. Is there a better way to do this?
diff = arr1 - arr2
diff_with_index = diff.map { |x| { index: arr1.index(x), values: x } }
print diff_with_index
# [{:index=>1, :values=>["c", "d"]}, {:index=>2, :values=>["e", "f"]}]
Upvotes: 0
Views: 309
Reputation: 26768
When you have to do multiple include?
checks, the most efficient way is to turn one of the lists into a set or hash beforehand, so you can have O(1) lookup time, so something like this:
require 'set'
arr2_set = Set.new(arr2)
arr1.each_index.select { |idx| !arr2_set.include?(arr1[idx]) }
Upvotes: 2
Reputation: 110675
Here is one way.
i1 = (0..arr1.size-1).to_a
#=> [0, 1, 2]
h1 = arr1.zip(i1).to_h
#=> {["a", "b"]=>0, ["c", "d"]=>1, ["e", "f"]=>2}
i1 - arr2.map { |a| h1[a] }
#=> [1, 2]
Note that
arr2.map { |a| h1[a] }
#=> [nil, nil, 0]
Upvotes: 0