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Python-动态构建嵌套树

tree recursion algorithm json python

AutoTester213 · 技术社区 · 7 年前

我正在尝试用python构建一个树层次结构,假设我有他的结构,我需要能够动态地添加更多的子元素,例如 Bannana .

更新: Apple(Children) 对Lemon和其他孩子也一样。苹果节点也是根节点。

Apple
   Bannana
   Lemon
      Juice
        Drink
   Watermelon
       Red
         Round

Apple
   Bannana
   Lemon
      Juice
        Drink
          Watermelon
       Red
          Round

如果是这样的话

{
    'Apple': 'Fruit',
    'children': [{
        'Bannana': 'fruit',
        'children': None
    },  {
        'Lemon': 'Fruit',
        'children': [{
            'Juice': 'Food',
            'children': [{
                'Drink': 'Action',
                'children': None
             And so on...

如何使层次结构动态化?例如,特定父项下的行数?

我从我找到的一个例子中尝试过类似的东西

import collections
def add_element(root, path, data):
    if len(path) == 1:
        root[path[0]] = data
    else:
        add_element(root[path[0]], path[1:], data)

count = 1


tree = lambda: collections.defaultdict(tree)
root = tree()
n= 10
for i in range(1,n):
    path_list= ['Apple', 'Lemon', 'Juice' + str(count)]
    print (path_list)
    count += 1
    add_element(root,path_list, 1 )
print (root)

编辑1

args = {'Apple': 'Apple', 'Lemon': 'Lemon', 'Juice': 'Juice', 'Drink': 'Drink'}

s = """
{Apple}
   {Lemon}
      {Juice}
         {Drink}
""".format(**args)

def group_data(vals):
  if len(vals) == 1:
     return {vals[0]:'Fruit', 'Children':None}
  new_data = [list(b) for _, b in itertools.groupby(vals, key=lambda x:bool(re.findall('^\s', x)))]
  new_group = [[new_data[i], new_data[i+1]] for i in range(0, len(new_data), 2)]
  result = []
  for a, b in new_group:
     result.extend([{i:'Fruit', 'Children':None} for i in a[:-1]])
     result.append({a[-1]:'Fruit', 'Children':group_data([re.sub('^\s{3}', '', c) for c in b])})
  return result

_new_data = [i.strip('\n') for i in filter(None, s.split('\n'))]


print(json.dumps(group_data(_new_data), indent=4))

这工作很好,但它仍然硬编码,这不是我要找的。

1 回复 | 直到 7 年前

Ajax1234 7 年前

您可以分析每个水果前的空格:

s = """
Apple
   Bannana
   Lemon
      Juice
         Drink
            Watermelon
      Red
         Round
"""

import itertools, re
def group_data(vals):
  if len(vals) == 1:
     return {vals[0]:'Fruit', 'Children':None}
  new_data = [list(b) for _, b in itertools.groupby(vals, key=lambda x:bool(re.findall('^\s', x)))]
  new_group = [[new_data[i], new_data[i+1]] for i in range(0, len(new_data), 2)]
  result = []
  for a, b in new_group:
     result.extend([{i:'Fruit', 'Children':None} for i in a[:-1]])
     result.append({a[-1]:'Fruit', 'Children':group_data([re.sub('^\s{3}', '', c) for c in b])})
  return result

_new_data = [i.strip('\n') for i in filter(None, s.split('\n'))]

import json
print(json.dumps(group_data(_new_data), indent=4))

输出:

[
     {
       "Apple": "Fruit",
       "Children": [
         {
            "Bannana": "Fruit",
            "Children": null
           },
           {
               "Lemon": "Fruit",
               "Children": [
                  {
                      "Juice": "Fruit",
                       "Children": [
                            {
                             "Drink": "Fruit",
                             "Children": {
                                 "Watermelon": "Fruit",
                                 "Children": null
                              }
                          }
                      ]
                  },
                  {
                       "Red": "Fruit",
                       "Children": {
                          "Round": "Fruit",
                          "Children": null
                     }
                  }
              ]
          }
       ]
    }
]