Question

我有一张json树

{"reply": 0, "id": 1, "children": [{"reply": 1, "id": 2, "children": [{"reply": 1, "id": 3, "children": [{"reply": 1, "id": 4, "children": []}]}]}, {"reply": 0, "id": 5, "children": []}]}

我必须分配从叶到根的权重：

所有叶节点的重量均为0.1
如果一个节点有一个孩子：那么节点的权重就会变成其单个孩子的0.1 *重量
如果某个节点有两个或更多子节点，则该节点的权重为：

孩子的体重1 *孩子的体重2 * ....孩子* 0.1 * 0.2

例如，输出是节点的最终权重：

4：0.1,5：0.1,3：0.1 * 0.1,2：（0.1 * 0.1 * 0.1）1：0.1 * 0.2（0.1 * 0.1 * 0.1 * 0.1）`

我正在使用python中的代码

我在更新体重时得到了一个keyerror 2：

例外是Traceback（最近一次调用最后一次）：文件“cc1.py”，第42行，in 重量[面值] =重量[面值] *重量[child_of [面值] [I]] KeyError：2

我有三本词典：

import json
from collections import deque
def generate_children(tree):
    queue = deque()
    queue.append((tree, None))
    while queue:
       node, parent = queue.pop()
       children = []
    for child in node['children']:
        queue.append((child, node['id']))
        children.append(child['id'])
    parent_of[node['id']]=parent
    child_of[node['id']]=children
    no_child_of[node['id']]=len(child_of[node['id']])
    yield node['id'], parent, children

f=open('tree_json','r')
for line in f:
    tree=json.loads(line)
    parent_of={}
    child_of={}
    no_child_of={}
    weight={}
    q = list(generate_children(tree))
    #assigning weights to leaf
    for c in no_child_of.keys():
        if no_child_of[c]==0:
            weight[c]=0.1
    # assigning weight to the parent
    for w in weight.keys():
        par=parent_of[w]
        n=len(child_of[par])
        if n > 1:
            weight[par]=0.2
            for i in range(n):
                weight[par]=weight[par]*weight[child_of[par][i]]

Answer 1

我认为在您的代码中，您尝试访问尚未分配的节点的权重。

检查以下伪代码。希望它能解决你的目的。我假设你知道根元素。

for each node in tree
    weight[node] = 0;

current_node = root;
while(weight[root] = 0){
   if(no_child_of(current_node) = 0){
       weight[current_node] = 0.1;
       current_node = parent_of[current_node];
   }else{
      boolean all_child_weighted = true;
      total_weight = 1;
      for each (child[i] of current_node){
         if(weight[child_of[current_node][i]] = 0){
             all_child_weighted = false;
             current_node = child_of[current_node][i];  
             break; 
         }else{
             total_weight = total_weight * weight[child_of[current_node][i]];
         } 
      }   
      if(all_child_weighted){
          if(no_child_of(current_node) = 1){
              weight[current_node] = 0.1*weight[child_of[current_node][1]];
          }else{
              weight[current_node] = total_weight * 0.1 * 0.2;
          }
          current_node = parent_of(current_node);
      }
   }
}

从叶到根更新树中的权重时出错

1 个答案: