Question

我有一个从REST服务返回的JSON，我希望通过组合父键为每个值生成一个唯一的名称。例如。 name+phone+address+city+name，name+phone+address+city+population+skilled+male等等。

{
  "name": "name",
  "phone": "343444444",
  "address": {
    "lat": 23.444,
    "lng": 34.3322,
    "city":{
      "name": "city name",
      "population": {
        "skilled": {
          "male": 2,
          "female": 4
        },
        "uneducated": {
          "male": 20,
          "femail": 4
        }
      }
    }
  },
  "email": "email",
  "education": "phd"
}

我想组合从JSON树的父级开始的所有键名。

这就是我正在做的事情

class TestJson
     def walk_through(self, json_object):
        for k, v in json_object.items():
            self.x_path = self.x_path + k
            if type(v) is dict:
                self.walk_through(v)
            else:
                print(self.x_path)
                self.x_path = ""

此代码是打印密钥，但仅从当前父节点开始。我想将所有密钥组合到json的根目录。

Answer 1

如果忽略name和phone键，因为它们不是city name或skilled male的祖先，并且无法保证键的顺序，您可以递归构建扁平的字典。

def walk_through(json_object):
    d = {}
    for k, v in json_object.items():
        if isinstance(v, dict):
            v = walk_through(v)
            for vk, vv in v.items():
                d["%s+%s" % (k, vk)] = vv
        else:
            d[k] = v
    return d

print(json.dumps(walk_through(json_object), indent=2))

打印：

{
  "address+city+population+skilled+male": 2, 
  "name": "name", 
  "address+lng": 34.3322, 
  "address+city+name": "city name", 
  "address+lat": 23.444, 
  "address+city+population+uneducated+male": 20, 
  "phone": "343444444", 
  "address+city+population+uneducated+femail": 4, 
  "education": "phd", 
  "email": "email", 
  "address+city+population+skilled+female": 4
}

注意：这会忽略列表，但不会在其中找到dicts。

Answer 2

如果要打印python dict的所有键，可以执行以下操作：

def print_keys(d):
    for key, value in d.iteritems():
        print key,
        if isinstance(value, dict):
            print_keys(value)

如何为每个JSON值生成唯一名称

2 个答案: