Ruby: split one hash into two based on desired keys

Question

I'm hoping this question has a very simple answer. I can think of ways to do with with boring, annoying looping, but I'm hoping there's a more elegant solution.

If I have the following two variables:

hash = {:a => 1, :b => 2, :c => 3, :d => 4}
keyset = [:a, :c]

How can I get the following two hashes in the simplest way possible?

hash1 = {:a => 1, :c => 3}
hash2 = {:b => 3, :d => 4}

If the example doesn't make my goal clear, in essence, what I want is a hybrid between #delete and #delete_if - #delete returns the deleted value, whereas #delete_if allows me to delete in bulk. I would prefer a way to delete in bulk, and have the deleted values returned - or something equivalent.

Thanks!

Answer 1

Try Active Support with Hash#slice and/or Hash#except. The bang methods also exist:

$ irb
>> require 'rubygems'
=> true
>> require 'activesupport'
=> true

>> hash = {:a => 1, :b => 2, :c => 3, :d => 4}
=> {:a=>1, :d=>4, :b=>2, :c=>3}
>> keyset = [:a, :c]
=> [:a, :c]

>> remainders = hash.slice!(*keyset)
=> {:d=>4, :b=>2}

>> remainders
=> {:d=>4, :b=>2}
>> hash
=> {:a=>1, :c=>3}

Answer 2

new_hash = {}
keyset.each {|i| new_hash[i] = hash.delete(i)}

That seemed to do it for me, without pulling in extra requirements