递归修改字典

Question

我创建了一个应该将嵌套列表转换为字典的类。以下是我输入：

['function:and', 
    ['variable:X', 'function:>=', 'value:13'], 
    ['variable:Y', 'function:==', 'variable:W']] 

和输出应该是以下形式的字典：

{ 
    "function": "and", 
    "args": [ 
    { 
     "function": ">=", 
     "args": [ 
     { 
      "variable": "X" 
     }, 
     { 
      "value": 13 
     } 
     ] 
    }, 
    { 
     "function": "==", 
     "args": [ 
     { 
      "variable": "Y" 
     }, 
     { 
      "variable": "W" 
     } 
     ] 
    } 
    ] 
} 

这是接收输入列表，并应返回所需的字典类。

class Tokenizer(object): 
    def __init__(self, tree): 
     self.tree = tree 
     self.filter = {} 

    def to_dict(self, triple): 
     my_dict = {} 
     try: 
      first = triple[0] 
      second = triple[1] 
      third = triple[2] 
     except KeyError: 
      return 
     if type(second) == str and type(third) == str: 
      my_dict['function'] = second.split(':')[-1] 
      my_dict['args'] = [ 
       {first.split(':')[0]: first.split(':')[1]}, 
       {third.split(':')[0]: third.split(':')[1]}] 
     # case recursive 
     if type(second) == list: 
      my_dict['function'] = first.split(':')[-1] 
      my_dict['args'] = [second, third] 
     return my_dict 

    def walk(self, args): 
     left = self.to_dict(args[0]) 
     right = self.to_dict(args[1]) 
     if isinstance(left, dict): 
      if 'args' in left.keys(): 
       left = self.walk(left['args']) 
     if isinstance(right, dict): 
      if 'args' in right.keys(): 
       right = self.walk(right['args']) 
     args = [left, right] 
     return args 

    def run(self): 
     self.filter.update(self.to_dict(self.tree)) 
     if 'args' in self.filter.keys(): 
      self.filter['args'] = self.walk(self.filter['args']) 


tree = [ 
    'function:and', 
     ['variable:X', 'function:>=', 'value:13'], 
     ['variable:Y', 'function:==', 'variable:W'] 
    ] 

import pprint 
pp = pprint.PrettyPrinter(indent=4) 
t = Tokenizer(tree) 
t.run() 
pp.pprint(t.filter) 

我的递归方法walk没有做它应该是什么，我在递归共吸盘，所以我不能图什么我做错了。

我得到的输出是：

{ 'args': [[None, None], [None, None]], 'function': 'and'} 

Answer 1

对于不需要进入递归在所有特定的测试用例。您可以发表您的电话评论：

def walk(self, args): 
    left = self.to_dict(args[0]) 
    right = self.to_dict(args[1]) 
    #if isinstance(left, dict): 
    # if 'args' in left.keys(): 
    #  left = self.walk(left['args']) 
    #if isinstance(right, dict): 
    # if 'args' in right.keys(): 
    #  right = self.walk(right['args']) 
    args = [left, right] 
    return args 

并获得所需的输出。 你只需要进入递归，如果你允许你输入嵌套函数：

['function:and', 
     ['variable:X', 'function:>=', 'value:13'], 
     ['function:==', 
      ['variable:R', 'function:>=', 'value:1'], 
      ['variable:Z', 'function:==', 'variable:K'] 
     ] 
] 

，那么你必须检查基本情况，让你进入递归只有当你的args键的值包含未处理值：

def walk(self, args): 
    left = self.to_dict(args[0]) 
    right = self.to_dict(args[1]) 
    if isinstance(left, dict): 
     if 'args' in left.keys() and isinstance(left['args'][0], list): 
      left = self.walk(left['args']) 
    if isinstance(right, dict): 
     if 'args' in right.keys() and isinstance(right['args'][0], list): 
      right = self.walk(right['args']) 
    args = [left, right] 
    return args 

，然后你会得到这样的：

{ 'args': [ { 'args': [{ 'variable': 'X'}, { 'value': '13'}], 
       'function': '>='}, 
      { 'args': [ { 'args': [ { 'variable': 'R'}, 
              { 'value': '1'}], 
           'function': '>='}, 
          { 'args': [ { 'variable': 'Z'}, 
              { 'variable': 'K'}], 
           'function': '=='}], 
       'function': '=='}], 
'function': 'and'} 

而且它会更容易，如果你的输入列表是一个普通结构在函数名字段后面始终有参数字段。然后，您可以大大简化您的to_dict方法。