Converting a hash into a nested hash

Question

This question is the inverse of this question.

Given a hash that has an array for each key like

{
    [:a, :b, :c] => 1,
    [:a, :b, :d] => 2,         


        

Answer 1

For a mixed hash/array nested structure you can use this. (Modified for arrays as well)

def unflatten(h={})
  ret = {}
  h.each do |k,v|
    node = ret
    keys = k.split('.').collect { |x| x.to_i.to_s == x ? x.to_i : x }
    keys.each_cons(2) do |x, next_d|
      if(next_d.is_a? Fixnum)
        node[x] ||= []
        node=node[x]
      else
        node[x] ||={}
        node=node[x]
      end
    end                                                                                                                                                                                                                                                                         
    node[keys[-1]] = v
  end
  ret
end

provided you used the below for flattening. ( dot separate string for key instead of array [split on . if you need] )

def flatten_hash(hash)
  hash.each_with_object({}) do |(k, v), h|
    if v.is_a? Hash
      flatten_hash(v).map do |h_k, h_v|
        h["#{k}.#{h_k}"] = h_v 
      end 
    elsif v.is_a? Array
      flatten_array(v).map do |h_k,h_v|
        h["#{k}.#{h_k}"] = h_v 
      end 
    else
      h[k] = v 
    end 
  end 
end


def flatten_array(array)
  array.each_with_object({}).with_index do |(v,h),i|
    pp v,h,i
    if v.is_a? Hash
      flatten_hash(v).map do |h_k, h_v|
        h["#{i}.#{h_k}"] = h_v 
      end 
    elsif v.is_a? Array
      flatten_array(v).map do |h_k,h_v|
        h["#{i}.#{h_k}"] = h_v 
      end 
    end 
  end 
end  

Answer 2

Another way:

def convert(h)
  h.each_with_object({}) { |(a,n),f| f.update({ a.first=>(a.size==1 ? n :
    convert({ a[1..-1]=>n })) }) { |_,ov,nv| ov.merge(nv) } }
end

Try it:

h = {
    [:a, :b, :c] => 1,
    [:a, :b, :d] => 2,
    [:a, :e] => 3,
    [:f] => 4,
}

convert(h) #=> {:a=>{:b=>{:d=>2}, :e=>3},
           #    :f=>4}

Answer 3

Functional recursive algorithm:

require 'facets'

class Hash
  def nestify
    map_by { |ks, v| [ks.first, [ks.drop(1), v]] }.mash do |key, pairs|
      [key, pairs.first[0].empty? ? pairs.first[1] : Hash[pairs].nestify]
    end
  end
end

p {[:a, :b, :c]=>1, [:a, :b, :d]=>2, [:a, :e]=>3, [:f]=>4}.nestify
# {:a=>{:b=>{:c=>1, :d=>2}, :e=>3}, :f=>4}

Answer 4

Here's an iterative solution, a recursive one is left as an exercise to the reader:

def convert(h={})
  ret = {}
  h.each do |k,v|
    node = ret
    k[0..-2].each {|x| node[x]||={}; node=node[x]}
    node[k[-1]] = v
  end
  ret
end

convert(your_hash) # => {:f=>4, :a=>{:b=>{:c=>1, :d=>2}, :e=>3}}

Answer 5

There is already a good answer, but I worked on this recursive solution, so here it is:

def to_nest(hash)
  {}.tap do |nest|
    hash.each_pair do |key, value|
      nodes = key.dup
      node  = nodes.shift
      if nodes.empty?
        nest[node] = value
      else
        nest[node] ||= {}
        nest[node].merge!({nodes => value})
      end
    end
    nest.each_pair do |key, value|
      nest[key] = to_nest(value) if value.kind_of?(Hash)
    end
  end
end

Answer 6

Using DeepEnumerable:

require DeepEnumerable

h = {[:a, :b, :c]=>1, [:a, :b, :d]=>2, [:a, :e]=>3, [:f]=>4}

h.inject({}){|hash, kv| hash.deep_set(*kv)}