Question

假设我们在下面给出了嵌套字典。最终，我们希望列出每个字典中出现的所有嵌套命名字段。使用句点'。'连接嵌套字段名称。为嵌套记录定义命名字段。按字母顺序显示列表。例如，如果我们的数据文件包含以下内容那么有序的字段列表将是：

    {
     "name":{"firstName": "ff", "lastName": "ll"},
     "phone": "555",
      "address": {
                  "personal": {"city":"cc", "street": "ss"},
                  "work": "ww"
                  }
     }


["address.personal.city", "address.personal.street", "email", 
 "name.firstName", "name.lastName", "phone"]

怎么做？我真的想对每个字典做这样的事情

到目前为止我做了什么：

let X = {"name":{"first":"FFF","last":"LLL"},
     "address":{"personal":{"city": "cc", "home":"hh"}, "work":"ww"}, 
     "phone":"555"}

def find_deep(dictionary):
    key_list = list(dictionary)
    for key in key_list:
        if type(dictionary[key])==dict:
            key_list[key_list.index(key)] = [key, find_deep(dictionary[key])]
            # key_list[key_list.index(key)] = appen_word_to_list(key, list(dictionary[key]))
    return key_list

以上功能产生：

    y = find_deep(X)
    y = ['phone', ['name', ['last', 'first']], ['address', 
         [['personal', ['city', 'home']], 'work']]]

展开这个，我有错误！

Answer 1

这是使用递归解决的经典问题，递归方法一次添加一个键级别：

def find_deep(dictionary, parent):
    ans = []
    for key in list(dictionary):
        # To make sure the first level doesn't get a preceding dot
        initial_dot = "" if parent == "" else "."

        if type(dictionary[key]) == dict:
            # The recursion progresses
            ans.extend(find_deep(dictionary[key], initial_dot.join((parent, key))))
        else:
            # The recursion terminates
            ans.extend([initial_dot.join((parent, key))])
    return ans


x = {"name":{"first":"FFF","last":"LLL"},
     "address":{"personal":{"city": "cc", "home":"hh"}, "work":"ww"}, 
     "phone":"555"}
print(find_deep(x, ""))

输出正如预期的那样：

'name.first', 'name.last', 'address.personal.city', 'address.personal.home', 'address.work', 'phone'

Answer 2

您可以使用递归：

s = {
 "name":{"firstName": "ff", "lastName": "ll"},
 "phone": "555",
  "address": {
              "personal": {"city":"cc", "street": "ss"},
              "work": "ww"
              }
 }
def trim(f):
   return ['.'.join(i.split('.')[:-1]) for i in f(s)]

@trim
def object_notation(s, last = ''):
   if not isinstance(s, dict):
      yield s
   else:
      for a, b in s.items():
          yield from ['{}.{}'.format(a, i) for i in object_notation(b, a)]
print(object_notation)

输出：

['name.firstName', 'name.lastName', 'phone', 'address.personal.city', 'address.personal.street', 'address.work']

列出嵌套字典

2 个答案: