How to elegantly rename all keys in a hash in Ruby?

RubyHashKey

Ruby Problem Overview


I have a Ruby hash:

ages = { "Bruce" => 32,
         "Clark" => 28
       }

Assuming I have another hash of replacement names, is there an elegant way to rename all the keys so that I end up with:

ages = { "Bruce Wayne" => 32,
         "Clark Kent" => 28
       }

Ruby Solutions


Solution 1 - Ruby

ages = { 'Bruce' => 32, 'Clark' => 28 }
mappings = { 'Bruce' => 'Bruce Wayne', 'Clark' => 'Clark Kent' }

ages.transform_keys(&mappings.method(:[]))
#=> { 'Bruce Wayne' => 32, 'Clark Kent' => 28 }

Solution 2 - Ruby

I liked Jörg W Mittag's answer, but if you want to rename the keys of your current Hash and not to create a new Hash with the renamed keys, the following snippet does exactly that:

ages = { "Bruce" => 32, "Clark" => 28 }
mappings = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}

ages.keys.each { |k| ages[ mappings[k] ] = ages.delete(k) if mappings[k] }
ages

There's also the advantage of only renaming the necessary keys.

Performance considerations:

Based on the Tin Man's answer, my answer is about 20% faster than Jörg W Mittag's answer for a Hash with only two keys. It may get even higher performance for Hashes with many keys, specially if there are just a few keys to be renamed.

Solution 3 - Ruby

There's the under-utilized each_with_object method in Ruby as well:

ages = { "Bruce" => 32, "Clark" => 28 }
mappings = { "Bruce" => "Bruce Wayne", "Clark" => "Clark Kent" }

ages.each_with_object({}) { |(k, v), memo| memo[mappings[k]] = v }

Solution 4 - Ruby

Just to see what was faster:

require 'fruity'

AGES = { "Bruce" => 32, "Clark" => 28 }
MAPPINGS = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}

def jörg_w_mittag_test(ages, mappings)
  Hash[ages.map {|k, v| [mappings[k], v] }]
end

require 'facets/hash/rekey'
def tyler_rick_test(ages, mappings)
  ages.rekey(mappings)
end

def barbolo_test(ages, mappings)
  ages.keys.each { |k| ages[ mappings[k] ] = ages.delete(k) if mappings[k] }
  ages
end

class Hash
  def tfr_rekey(h)
    dup.tfr_rekey! h
  end

  def tfr_rekey!(h)
    h.each { |k, newk| store(newk, delete(k)) if has_key? k }
    self
  end
end

def tfr_test(ages, mappings)
  ages.tfr_rekey mappings
end

class Hash
  def rename_keys(mapping)
    result = {}
    self.map do |k,v|
      mapped_key = mapping[k] ? mapping[k] : k
      result[mapped_key] = v.kind_of?(Hash) ? v.rename_keys(mapping) : v
      result[mapped_key] = v.collect{ |obj| obj.rename_keys(mapping) if obj.kind_of?(Hash)} if v.kind_of?(Array)
    end
    result
  end
end

def greg_test(ages, mappings)
  ages.rename_keys(mappings)
end

compare do
  jörg_w_mittag { jörg_w_mittag_test(AGES.dup, MAPPINGS.dup) }
  tyler_rick    { tyler_rick_test(AGES.dup, MAPPINGS.dup)    }
  barbolo       { barbolo_test(AGES.dup, MAPPINGS.dup)       }
  greg          { greg_test(AGES.dup, MAPPINGS.dup)          }
end

Which outputs:

Running each test 1024 times. Test will take about 1 second.
barbolo is faster than jörg_w_mittag by 19.999999999999996% ± 10.0%
jörg_w_mittag is faster than greg by 10.000000000000009% ± 10.0%
greg is faster than tyler_rick by 30.000000000000004% ± 10.0%

Caution: barbell's solution uses if mappings[k], which will cause the resulting hash to be wrong if mappings[k] results in a nil value.

Solution 5 - Ruby

I monkey-patched the class to handle nested Hashes and Arrays:

   #  Netsted Hash:
   # 
   #  str_hash = {
   #                "a"  => "a val", 
   #                "b"  => "b val",
   #                "c" => {
   #                          "c1" => "c1 val",
   #                          "c2" => "c2 val"
   #                        }, 
   #                "d"  => "d val",
   #           }
   #           
   # mappings = {
   #              "a" => "apple",
   #              "b" => "boss",
   #              "c" => "cat",
   #              "c1" => "cat 1"
   #           }
   # => {"apple"=>"a val", "boss"=>"b val", "cat"=>{"cat 1"=>"c1 val", "c2"=>"c2 val"}, "d"=>"d val"}
   #
   class Hash
    def rename_keys(mapping)
      result = {}
      self.map do |k,v|
        mapped_key = mapping[k] ? mapping[k] : k
        result[mapped_key] = v.kind_of?(Hash) ? v.rename_keys(mapping) : v
        result[mapped_key] = v.collect{ |obj| obj.rename_keys(mapping) if obj.kind_of?(Hash)} if v.kind_of?(Array)
      end
    result
   end
  end

Solution 6 - Ruby

If the mapping Hash will be smaller than the data Hash then iterate on mappings instead. This is useful for renaming a few fields in a large Hash:

class Hash
  def rekey(h)
    dup.rekey! h
  end

  def rekey!(h)
    h.each { |k, newk| store(newk, delete(k)) if has_key? k }
    self
  end
end

ages = { "Bruce" => 32, "Clark" => 28, "John" => 36 }
mappings = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}
p ages.rekey! mappings

Solution 7 - Ruby

The [Facets][1] gem provides a rekey method that does exactly what you're wanting.

As long as you're okay with a dependency on the [Facets][1] gem, you can pass a hash of mappings to rekey and it will return a new hash with the new keys:

require 'facets/hash/rekey'
ages = { "Bruce" => 32, "Clark" => 28 }
mappings = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}
ages.rekey(mappings)
=> {"Bruce Wayne"=>32, "Clark Kent"=>28}

If you want to modify ages hash in place, you can use the rekey! version:

ages.rekey!(mappings)
ages
=> {"Bruce Wayne"=>32, "Clark Kent"=>28}

[1]: https://github.com/rubyworks/facets "Facets"

Solution 8 - Ruby

You may wish to use Object#tap to avoid the need to return ages after the keys have been modified:

ages = { "Bruce" => 32, "Clark" => 28 }
mappings = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}

ages.tap {|h| h.keys.each {|k| (h[mappings[k]] = h.delete(k)) if mappings.key?(k)}}
  #=> {"Bruce Wayne"=>32, "Clark Kent"=>28}

Solution 9 - Ruby

ages = { "Bruce" => 32, "Clark" => 28 }
mappings = {"Bruce" => "Bruce Wayne", "Clark" => "Clark Kent"}
ages = mappings.inject({}) {|memo, mapping| memo[mapping[1]] = ages[mapping[0]]; memo}
puts ages.inspect

Solution 10 - Ruby

>> x={ :a => 'qwe', :b => 'asd'}
=> {:a=>"qwe", :b=>"asd"}
>> rename={:a=>:qwe}
=> {:a=>:qwe}
>> rename.each{|old,new| x[new] = x.delete old}
=> {:a=>:qwe}
>> x
=> {:b=>"asd", :qwe=>"qwe"}

This would loop just through renames hash.

Solution 11 - Ruby

I used this to allow "friendly" names in a Cucumber table to be parsed into class attributes such that Factory Girl could create an instance:

Given(/^an organization exists with the following attributes:$/) do |table|
  # Build a mapping from the "friendly" text in the test to the lower_case actual name in the class
  map_to_keys = Hash.new
  table.transpose.hashes.first.keys.each { |x| map_to_keys[x] = x.downcase.gsub(' ', '_') }
  table.transpose.hashes.each do |obj|
    obj.keys.each { |k| obj[map_to_keys[k]] = obj.delete(k) if map_to_keys[k] }
    create(:organization, Rack::Utils.parse_nested_query(obj.to_query))
  end
end

For what it's worth, the Cucumber table looks like this:

  Background:
    And an organization exists with the following attributes:
      | Name            | Example Org                        |
      | Subdomain       | xfdc                               |
      | Phone Number    | 123-123-1234                       |
      | Address         | 123 E Walnut St, Anytown, PA 18999 |
      | Billing Contact | Alexander Hamilton                 |
      | Billing Address | 123 E Walnut St, Anytown, PA 18999 |

And map_to_keys looks like this:

{
               "Name" => "name",
          "Subdomain" => "subdomain",
       "Phone Number" => "phone_number",
            "Address" => "address",
    "Billing Contact" => "billing_contact",
    "Billing Address" => "billing_address"
}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionChanporyView Question on Stackoverflow
Solution 1 - RubyJörg W MittagView Answer on Stackoverflow
Solution 2 - RubybarboloView Answer on Stackoverflow
Solution 3 - RubysteelView Answer on Stackoverflow
Solution 4 - Rubythe Tin ManView Answer on Stackoverflow
Solution 5 - RubyGregView Answer on Stackoverflow
Solution 6 - RubyTFRView Answer on Stackoverflow
Solution 7 - RubyTyler RickView Answer on Stackoverflow
Solution 8 - RubyCary SwovelandView Answer on Stackoverflow
Solution 9 - RubydanView Answer on Stackoverflow
Solution 10 - RubysitesView Answer on Stackoverflow
Solution 11 - RubyJon KernView Answer on Stackoverflow