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

preferably something like

hash.includes_keys? [ :a, :b, :c, :d ]

I came up with

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

but I dont like having to add the array twice though

Awesome answer here which does not use any loops: http://stackoverflow.com/a/4743729/238880 — Pratik Khadloya, Jun 27 '13 at 15:08
@PratikKhadloya - I'm not sure that applies here: this is about ensuring _all_ keys are present rather than any. — SRack, Jun 19 '19 at 10:58

Mori · Accepted Answer · 2018-08-03T18:54:53.473

123

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

edited Aug 03 '18 at 18:54

answered Jul 19 '12 at 00:58

Mori

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3

If you want to make it more complicated, you could use `[:a, :b, :c, :d].all?(&hash.method(:key?))`. – Patrick Brinich-Langlois Aug 23 '16 at 21:11

score 21 · Answer 2 · answered Jul 19 '12 at 01:04

21

@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

answered Jul 19 '12 at 01:04

John Douthat

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score 15 · Answer 3 · answered Jul 19 '12 at 01:16

15

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

answered Jul 19 '12 at 01:16

Mark Thomas

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score 9 · Answer 4 · answered Jul 25 '16 at 15:24

9

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?

answered Jul 25 '16 at 15:24

Chris Hanson

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score 8 · Answer 5 · answered May 07 '15 at 11:43

8

I like this way to solve this:

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

It is fast and clear.

answered May 07 '15 at 11:43

Seb Wilgosz

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2

Works unless one needs a strict match. [:a, :b, :c, :d] & [:a, :b, :c, :d, :e] => [:a, :b, :c, :d] – Vadym Tyemirov May 21 '15 at 22:49
4

Yeah, however the question wasn't about that. – Seb Wilgosz May 23 '15 at 08:51

score 1 · Answer 6 · edited Aug 28 '20 at 04:55

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

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

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