4
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I have a nested hash of symbols like this:

{
  a: {
    b: :c,
    d: {
      e: :f
    }
  },
  g: :h
}

I want to build an array that contains all symbols used in the hash, both keys and values, in any order. For the example above:

[:a, :b, :c, :d, :e, :f, :g, :h]

Is there any simple, fast and ruby-friendly way to do this?

I'm doing this with a recursive function that sums hash.keys with hash.values, and finally applies flatten to the result.

def all_keys(hash)
  hash.keys + hash.values.map { |e| e.is_a?(Hash) ? all_keys(e) : e }
end
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1
  • \$\begingroup\$ If h is your hash, a quick and dirty would be h.to_s.scan /:[a-z]\w*/ => [":a", ":b", ":c", ":d", ":e", ":f", ":g", ":h"] . \$\endgroup\$ Feb 22, 2016 at 19:26

1 Answer 1

7
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Some notes:

  • The flattening should be performed where the non-desired nesting is being introduced, not later. Use flat_map instead of map:

  • Why e for the value name instead of v?

I'd write:

def all_keys(hash)
  hash.keys + hash.values.flat_map { |v| v.is_a?(Hash) ? all_keys(v) : [v] }
end

Another way to do it, slightly shorter, is to iterate directly the pairs. Pick the one which is more declarative to you:

def all_keys(hash)
  hash.flat_map { |k, v| [k] + (v.is_a?(Hash) ? all_keys(v) : [v]) }
end
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