How to change all the keys of a hash by a new set of given keys

Question

How do I change all the keys of a hash by a new set of given keys?

Is there a way to do that elegantly?

Answer 1

Another way to do it is:

hash = {
  'foo' => 1,
  'bar' => 2
}

new_keys = {
  'foo' => 'foozle',
  'bar' => 'barzle'
}

new_keys.values.zip(hash.values_at(*new_keys.keys)).to_h 
# => {"foozle"=>1, "barzle"=>2}

Breaking it down:

new_keys
.values # => ["foozle", "barzle"]
.zip(
  hash.values_at(*new_keys.keys) # => [1, 2]
) # => [["foozle", 1], ["barzle", 2]]
.to_h 
# => {"foozle"=>1, "barzle"=>2}

---

It's benchmark time...

While I like the simplicity of Jörn's answer, I'm wasn't sure it was as fast as it should be, then I saw selvamani's comment:

require 'fruity'

HASH = {
  'foo' => 1,
  'bar' => 2
}

NEW_KEYS = {
  'foo' => 'foozle',
  'bar' => 'barzle'
}

compare do
  mittag    { HASH.dup.map {|k, v| [NEW_KEYS[k], v] }.to_h }
  ttm       { h = HASH.dup; NEW_KEYS.values.zip(h.values_at(*NEW_KEYS.keys)).to_h }
  selvamani { h = HASH.dup; h.keys.each { |key| h[NEW_KEYS[key]] = h.delete(key)}; h }
end

# >> Running each test 2048 times. Test will take about 1 second.
# >> selvamani is faster than ttm by 39.99999999999999% ± 10.0%
# >> ttm is faster than mittag by 10.000000000000009% ± 10.0%

These are running very close together speed wise, so any will do, but 39% pays off over time so consider that. A couple answers were not included because there are potential flaws where they'd return bad results.