从相关对象

-1

我需要建立此树建立了一棵树：从这些对象从相关对象

result = [ 
    ['t9'], 
    ['t3', 
    ['t4'], 
    ['t8', 
     ['t6'], 
     ['t1', 
     ['t5'] 
     ] 
    ] 
    ], 
    ['t7', 
    ['t2'] 
    ] 
]

：

{:id => 't1', :tg => 't8', :rank => 2} 
{:id => 't2', :tg => 't7', :rank => 1} 
{:id => 't3', :tg => nil, :rank => 2} 
{:id => 't4', :tg => 't3', :rank => 1} 
{:id => 't5', :tg => 't1', :rank => 1} 
{:id => 't6', :tg => 't8', :rank => 1} 
{:id => 't7', :tg => nil, :rank => 3} 
{:id => 't8', :tg => 't3', :rank => 2} 
{:id => 't9', :tg => nil, :rank => 1}

tg是自我指涉的关联。 rank是阵列中的位置/索引

任何想法（首选红宝石）？

来源

2013-11-01 Aleksander Lopez

你能更好地表达自己吗？ 'result'看起来不像多维数组... –

你声称的“th [e] s [e] objects”不是Ruby对象。什么是't1'，'t2'等？您如何期望来自对象的信息被传递给一个方法？ – sawa

@sawa我简化了你的对象列表。 –

def combine e, a 
    a 
    .inject([]){|a, h| a[h[:rank] - 1] = [h[:id]] if h[:tg] == e; a} 
    .map{|e| e + combine(e.first, a)} 
end 

combine(nil, [ 
    {:id => 't1', :tg => 't8', :rank => 2}, 
    {:id => 't2', :tg => 't7', :rank => 1}, 
    {:id => 't3', :tg => nil, :rank => 2}, 
    {:id => 't4', :tg => 't3', :rank => 1}, 
    {:id => 't5', :tg => 't1', :rank => 1}, 
    {:id => 't6', :tg => 't8', :rank => 1}, 
    {:id => 't7', :tg => nil, :rank => 3}, 
    {:id => 't8', :tg => 't3', :rank => 2}, 
    {:id => 't9', :tg => nil, :rank => 1}, 
]) 
# => [["t9"], ["t3", ["t4"], ["t8", ["t6"], ["t1", ["t5"]]]], ["t7", ["t2"]]]

来源

2013-11-01 11:25:56 sawa

这是他最好的sawa。你明白了，恭喜！ @AlexLopez：不要忘记接受选择的答案！ –

听起来像你正在建立像树一样。该算法可以总结如下：

找到所有节点，其中：TG是零，那么你得到[T3，T7，T9]
查找其所有节点：TG的[T3，T7内， T9]，然后回来[T4，T8]
查找其所有节点：TG内[T4，T8] ...

来源

2013-11-01 10:54:43 nickcen

使用递归（根据@ cenyongh的树的概念）

def tree(arr,parent=nil) 
    arr.select{|n| n[:tg] == parent}. 
      sort_by{|n| n[:rank]}. 
      map{|n| [n[:id]] + tree(arr,n[:id])} 
end 

# call 
tree(array_of_objects)

来源

2013-11-01 11:17:10 tihom

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