Ruby:在树表示中转换平面数组
我试图编写一个函数,将带有路径信息的平面数组转换为该数组的树表示形式 我们的目标是使阵列如下所示:Ruby:在树表示中转换平面数组,ruby,tree,Ruby,Tree,我试图编写一个函数,将带有路径信息的平面数组转换为该数组的树表示形式 我们的目标是使阵列如下所示: [ { :name => "a", :path => [ 'a' ] }, { :name => "b", :path => [ 'a', 'b' ] }, { :name => "c", :path => [ 'a', 'b', 'c' ] }, { :name => "d", :path => [ 'a', 'd' ] }, { :name =&
[
{ :name => "a", :path => [ 'a' ] },
{ :name => "b", :path => [ 'a', 'b' ] },
{ :name => "c", :path => [ 'a', 'b', 'c' ] },
{ :name => "d", :path => [ 'a', 'd' ] },
{ :name => "e", :path => [ 'e' ] }
]
变成这样的一个:
[{:node=>{:name=>"a", :path=>["a"]},
:children=>
[{:node=>{:name=>"b", :path=>["a", "b"]},
:children=>
[{:node=>{:name=>"c", :path=>["a", "b", "c"]}, :children=>[]}]},
{:node=>{:name=>"d", :path=>["a", "d"]}, :children=>[]}]},
{:node=>{:name=>"e", :path=>["e"]}, :children=>[]}]
我得到的最接近的结果是以下代码:
class Tree
def initialize
@root = { :node => nil, :children => [ ] }
end
def from_array( array )
array.inject(self) { |tree, node| tree.add(node) }
@root[:children]
end
def add(node)
recursive_add(@root, node[:path].dup, node)
self
end
private
def recursive_add(parent, path, node)
if(path.empty?)
parent[:node] = node
return
end
current_path = path.shift
children_nodes = parent[:children].find { |child| child[:node][:path].last == current_path }
unless children_nodes
children_nodes = { :node => nil, :children => [ ] }
parent[:children].push children_nodes
end
recursive_add(children_nodes, path, node)
end
end
flat = [
{ :name => "a", :path => [ 'a' ] },
{ :name => "b", :path => [ 'a', 'b' ] },
{ :name => "c", :path => [ 'a', 'b', 'c' ] },
{ :name => "d", :path => [ 'a', 'd' ] },
{ :name => "e", :path => [ 'e' ] }
]
require 'pp'
pp Tree.new.from_array( flat )
但这是相当冗长的,我有一种感觉,它可能不是非常有效的非常大的集
在ruby中实现这一点最干净、最有效的方法是什么?这是我的尝试
array = [
{ :name => "a", :path => [ 'a' ] },
{ :name => "b", :path => [ 'a', 'b' ] },
{ :name => "c", :path => [ 'a', 'b', 'c' ] },
{ :name => "d", :path => [ 'a', 'd' ] },
{ :name => "e", :path => [ 'e' ] }
]
array
.sort_by{|h| -h[:path].length}
.map{|h| {node: h, children: []}}
.tap{|array|
while array.first[:node][:path].length > 1
child = array.shift
array
.find{|h| h[:node][:name] == child[:node][:path][-2]}[:children]
.push(child)
end
}
# => [
{:node=>{:name=>"e", :path=>["e"]}, :children=>[]},
{:node=>{:name=>"a", :path=>["a"]}, :children=>[
{:node=>{:name=>"d", :path=>["a", "d"]}, :children=>[]},
{:node=>{:name=>"b", :path=>["a", "b"]}, :children=>[
{:node=>{:name=>"c", :path=>["a", "b", "c"]}, :children=>[]}
]}
]}
]
张贴你所拥有的,即使你认为它不好。请不要在问题中使用无法解释的方法或变量。什么是路径?@sawa抱歉,这是一个打字错误。路径是一个符号。我认为你应该试着简化你的树。例如,如果名称是唯一的,那么可以使用名称作为键,以便您可以轻松搜索。@AndrewMarshall谢谢您的指点,我添加了我当前的源代码。Ruby有时感觉像魔术。干净有效。谢谢!