Check if a hash's keys include all of a set of keys

Question

I\'m looking for a better way to do

if hash.key? :a &&
   hash.key? :b &&
   hash.key? :c &&
   hash.key? :d

prefer

Answer 1

Just in the spirit of TIMTOWTDI, here's another way. If you require 'set' (in the std lib) then you can do this:

Set[:a,:b,:c,:d].subset? hash.keys.to_set

Answer 2

%i[a b c d].all? {|s| hash.key? s}

Answer 3

I like this way to solve this:

subset = [:a, :b, :c, :d]
subset & hash.keys == subset

It is fast and clear.

Answer 4

@Mori's way is best, but here's another way:

([:a, :b, :c, :d] - hash.keys).empty?

or

hash.slice(:a, :b, :c, :d).size == 4

Answer 5

Here is my solution:

(also given as answer at)

class Hash
    # doesn't check recursively
    def same_keys?(compare)
      return unless compare.class == Hash
        
      self.size == compare.size && self.keys.all? { |s| compare.key?(s) }
    end
end

a = c = {  a: nil,    b: "whatever1",  c: 1.14,     d: false  }
b     = {  a: "foo",  b: "whatever2",  c: 2.14,   "d": false  }
d     = {  a: "bar",  b: "whatever3",  c: 3.14,               }

puts a.same_keys?(b)                    # => true
puts a.same_keys?(c)                    # => true
puts a.same_keys?(d)                    # => false   
puts a.same_keys?(false).inspect        # => nil
puts a.same_keys?("jack").inspect       # => nil
puts a.same_keys?({}).inspect           # => false

Answer 6

You can get a list of missing keys this way:

expected_keys = [:a, :b, :c, :d]
missing_keys = expected_keys - hash.keys

If you just want to see if there are any missing keys:

(expected_keys - hash.keys).empty?