Ruby: Get All Keys in a Hash (Including Sub Keys)

Ruby: Get all keys in a hash (including sub keys)

This will give you an array of all the keys for any level of nesting.

def get_em(h)
  h.each_with_object([]) do |(k,v),keys|      
    keys << k
    keys.concat(get_em(v)) if v.is_a? Hash
  end
end

hash = {"a" => 1, "b" => {"c" => {"d" => 3}}}
get_em(hash) #  => ["a", "b", "c", "d"]

How to retrieve all (nested) keys of a Hash?

Simple recursive solution:

def recursive_keys(data)
  data.keys + data.values.map{|value| recursive_keys(value) if value.is_a?(Hash) }
end

def all_keys(data)
  recursive_keys(data).flatten.compact.uniq
end 

Usage:

all_keys({ a: 3, b: { b: 2, b: { c: 5 } } })
=> [:a, :b, :c]

Get all keys from ruby hash

You can call .keys on a Hash to get an array of keys back.

See: Hash#keys

Find key/value pairs deep inside a hash containing an arbitrary number of nested hashes and arrays

Here's a simple recursive solution:

def nested_hash_value(obj,key)
  if obj.respond_to?(:key?) && obj.key?(key)
    obj[key]
  elsif obj.respond_to?(:each)
    r = nil
    obj.find{ |*a| r=nested_hash_value(a.last,key) }
    r
  end
end

h = { foo:[1,2,[3,4],{a:{bar:42}}] }
p nested_hash_value(h,:bar)
#=> 42

How do I extract a sub-hash from a hash?

If you specifically want the method to return the extracted elements but h1 to remain the same:

h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h2 = h1.select {|key, value| [:b, :d, :e, :f].include?(key) } # => {:b=>:B, :d=>:D} 
h1 = Hash[h1.to_a - h2.to_a] # => {:a=>:A, :c=>:C} 

And if you want to patch that into the Hash class:

class Hash
  def extract_subhash(*extract)
    h2 = self.select{|key, value| extract.include?(key) }
    self.delete_if {|key, value| extract.include?(key) }
    h2
  end
end

If you just want to remove the specified elements from the hash, that is much easier using delete_if.

h1 = {:a => :A, :b => :B, :c => :C, :d => :D}
h1.delete_if {|key, value| [:b, :d, :e, :f].include?(key) } # => {:a=>:A, :c=>:C} 
h1  # => {:a=>:A, :c=>:C} 

Find a value in a nested hash

Found a way to do it with nested maps, a compact and a flatten.first:

fnd="theFoo"
data.map{|t,th|th.map{|s,sh|sh.map{|f,fh|fh["id"]if fh["name"]==fnd}.compact}}.flatten.first

==> 1

Accessing elements of nested hashes in ruby

The way I usually do this these days is:

h = Hash.new { |h,k| h[k] = {} }

This will give you a hash that creates a new hash as the entry for a missing key, but returns nil for the second level of key:

h['foo'] -> {}
h['foo']['bar'] -> nil

You can nest this to add multiple layers that can be addressed this way:

h = Hash.new { |h, k| h[k] = Hash.new { |hh, kk| hh[kk] = {} } }

h['bar'] -> {}
h['tar']['zar'] -> {}
h['scar']['far']['mar'] -> nil

You can also chain indefinitely by using the default_proc method:

h = Hash.new { |h, k| h[k] = Hash.new(&h.default_proc) }

h['bar'] -> {}
h['tar']['star']['par'] -> {}

The above code creates a hash whose default proc creates a new Hash with the same default proc. So, a hash created as a default value when a lookup for an unseen key occurs will have the same default behavior.

EDIT: More details

Ruby hashes allow you to control how default values are created when a lookup occurs for a new key. When specified, this behavior is encapsulated as a Proc object and is reachable via the default_proc and default_proc= methods. The default proc can also be specified by passing a block to Hash.new.

Let's break this code down a little. This is not idiomatic ruby, but it's easier to break it out into multiple lines:

1. recursive_hash = Hash.new do |h, k|
2.   h[k] = Hash.new(&h.default_proc)
3. end

Line 1 declares a variable recursive_hash to be a new Hash and begins a block to be recursive_hash's default_proc. The block is passed two objects: h, which is the Hash instance the key lookup is being performed on, and k, the key being looked up.

Line 2 sets the default value in the hash to a new Hash instance. The default behavior for this hash is supplied by passing a Proc created from the default_proc of the hash the lookup is occurring in; ie, the default proc the block itself is defining.

Here's an example from an IRB session:

irb(main):011:0> recursive_hash = Hash.new do |h,k|
irb(main):012:1* h[k] = Hash.new(&h.default_proc)
irb(main):013:1> end
=> {}
irb(main):014:0> recursive_hash[:foo]
=> {}
irb(main):015:0> recursive_hash
=> {:foo=>{}}

When the hash at recursive_hash[:foo] was created, its default_proc was supplied by recursive_hash's default_proc. This has two effects:

1. The default behavior for recursive_hash[:foo] is the same as recursive_hash.
2. The default behavior for hashes created by recursive_hash[:foo]'s default_proc will be the same as recursive_hash.

So, continuing in IRB, we get the following:

irb(main):016:0> recursive_hash[:foo][:bar]
=> {}
irb(main):017:0> recursive_hash
=> {:foo=>{:bar=>{}}}
irb(main):018:0> recursive_hash[:foo][:bar][:zap]
=> {}
irb(main):019:0> recursive_hash
=> {:foo=>{:bar=>{:zap=>{}}}}

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