How do you calculate the number of "levels" of descendants of a Nokogiri node?

Question

You can call Nokogiri::XML::Node#ancestors.size to see how deeply a node is nested. But is there a way to determine how deeply nested the most deeply nested child of a node is?

Alternatively, how can you find all the leaf nodes that descend from a node?

Answer 1

The following code monkey-patches Nokogiri::XML::Node for fun, but of course you can extract them as individual methods taking a node argument if you like. (Only the height method is part of your question, but I thought the deepest_leaves method might be interesting.)

require 'nokogiri'
class Nokogiri::XML::Node
  def depth
    ancestors.size
    # The following is ~10x slower: xpath('count(ancestor::node())').to_i
  end
  def leaves
    xpath('.//*[not(*)]').to_a
  end
  def height
    tallest = leaves.map{ |leaf| leaf.depth }.max
    tallest ? tallest - depth : 0
  end
  def deepest_leaves
    by_height = leaves.group_by{ |leaf| leaf.depth }
    by_height[ by_height.keys.max ]
  end
end

doc = Nokogiri::XML "<root>
  <a1>
    <b1></b1>
    <b2><c1><d1 /><d2><e1 /><e2 /></d2></c1><c2><d3><e3/></d3></c2></b2>
  </a1>
  <a2><b><c><d><e><f /></e></d></c></b></a2>
</root>"

a1 = doc.at_xpath('//a1')
p a1.height                      #=> 4
p a1.deepest_leaves.map(&:name)  #=> ["e1", "e2", "e3"]
p a1.leaves.map(&:name)          #=> ["b1", "d1", "e1", "e2", "e3"]

Edit: To answer just the question asked tersely, without wrapping it in re-usable pieces:

p a1.xpath('.//*[not(*)]').map{ |n| n.ancestors.size }.max - a1.ancestors.size