Python 打印嵌套JSON的父数据和子数据_Python_Json

Python 打印嵌套JSON的父数据和子数据

python json

Python 打印嵌套JSON的父数据和子数据,python,json,Python,Json,我有以下类型的JSON。在另一个表中，我有jsonpath，它告诉我获取某些值。如果要获取子元素的值，则需要获取所有相应的父属性值，并将所有这些值作为一行存储在数据帧中 { "Parent": { "Name": "Bob", "Age": "80", "Children": [ { "Name": "Michael", "Gender":"M", "Children": [ {

我有以下类型的JSON。在另一个表中，我有jsonpath，它告诉我获取某些值。如果要获取子元素的值，则需要获取所有相应的父属性值，并将所有这些值作为一行存储在数据帧中

{
  "Parent": {
    "Name": "Bob",
    "Age": "80",
    "Children": [
      {
        "Name": "Michael",
        "Gender":"M",
        "Children": [
          {
            "Name": "Ezee",
            "Gender": "M",
            "Age": 20
          },
          {
            "Name": "Ezee",
            "Gender": "M",
            "Age": 28,
            "Children": [
              {
                "Name": "Dre",
                "Age": 1
              },
              {
                "Name": "George",
                "Age": 2
              }
            ]
          }
        ],
        "Age": 50,
        "MiddleName": "Jay"
      },
      {
        "Name": "Justin",
        "Gender": "M",
        "Children": [
          {
            "Name": "Emily",
            "Age": 18,
            "Gender": "F"
          }
        ],
        "Age": 45
      }
    ]
  }
}

假设我需要获取JSON路径的值：

Parent/Children/0/Children/0/Name

，我需要获取对应父级（Parent/Children/0/[Name或age或etc.]）的Name、age等属性值，并将上述所有值存储为一行

{
  "Parent": {
    "Name": "Bob",
    "Age": "80",
    "Children": [
      {
        "Name": "Michael",
        "Gender":"M",
        "Children": [
          {
            "Name": "Ezee",
            "Gender": "M",
            "Age": 20
          },
          {
            "Name": "Ezee",
            "Gender": "M",
            "Age": 28,
            "Children": [
              {
                "Name": "Dre",
                "Age": 1
              },
              {
                "Name": "George",
                "Age": 2
              }
            ]
          }
        ],
        "Age": 50,
        "MiddleName": "Jay"
      },
      {
        "Name": "Justin",
        "Gender": "M",
        "Children": [
          {
            "Name": "Emily",
            "Age": 18,
            "Gender": "F"
          }
        ],
        "Age": 45
      }
    ]
  }
}

目前，我可以通过传递相关路径分别获取父值，通过传递该路径分别获取子值

def findValue(path, json_data):
    paths = path.split("/")    
    data = json_data
    for i in range(0,len(paths)):
        if isinstance(data, list):
            paths[i]=int(paths[i])
            data = data[paths[i]]
        else:
            data = data.get(paths[i])
    return data

我怎样才能做到这一点呢？

如果我理解得很好，您所需要的只是一个路径，如

Parent/Children/0/Children/0/Name

获取其父属性的相同路径。在这种情况下，它将是

Parent/Children/0/Name

下面是我对python解释器的尝试，希望它能帮助您：

>>> path = "Parent/Children/0/Children/0/Name"
>>> path_l = path.split('/')
>>> rev = path_l[::-1]
>>> rev
['Name', '0', 'Children', '0', 'Children', 'Parent']
>>> rev.index('Children')
2
>>> rev = rev[rev.index('Children')+1:]
>>> rev
['0', 'Children', 'Parent']
>>> final = rev[::-1] + [path_l[-1]]
>>> final
['Parent', 'Children', '0', 'Name']
>>> parent_path = '/'.join(final)
>>> parent_path
'Parent/Children/0/Name'

然后使用您的函数，您可以将两个值添加到数据帧中

>>> df = pandas.DataFrame({'Parent': [], 'Children':[]})
>>> df.append([parent], [children])

如果我理解得很好，那么您所需要的只是一个路径，如

Parent/Children/0/Children/0/Name

get其父属性的相同路径。在这种情况下，它将是

Parent/Children/0/Name

下面是我对python解释器的尝试，希望它能帮助您：

>>> path = "Parent/Children/0/Children/0/Name"
>>> path_l = path.split('/')
>>> rev = path_l[::-1]
>>> rev
['Name', '0', 'Children', '0', 'Children', 'Parent']
>>> rev.index('Children')
2
>>> rev = rev[rev.index('Children')+1:]
>>> rev
['0', 'Children', 'Parent']
>>> final = rev[::-1] + [path_l[-1]]
>>> final
['Parent', 'Children', '0', 'Name']
>>> parent_path = '/'.join(final)
>>> parent_path
'Parent/Children/0/Name'

然后使用您的函数，您可以将两个值添加到数据帧中

>>> df = pandas.DataFrame({'Parent': [], 'Children':[]})
>>> df.append([parent], [children])

要获取最后一个父级，您需要遍历路径，直到最后一次跟随

子级

列表

例如，给定路径：

'Parent/Children/0/Children/0/Name'

您希望在

'Parent/Children/0'

处返回父级的数据

这在Python中很容易做到，只需将

路径

字符串切分到子字符串的最后一个匹配项

/Children

：

path[:path.rfind('/Children')]

然后，您可以使用与当前获取父级数据类似的代码：

parent = json_data
path = path[:path.rfind('/Children')]
for attr in path.split('/'):
    parent = parent[int(attr) if isinstance(parent, list) else attr]

在本例中，这将为我们提供

parent

，如下所示：

{
  "Name": "Michael",
  "Gender": "M",
  "Children": [
    {
      "Name": "Ezee",
      "Gender": "M",
      "Age": 20
    },
    {
      "Name": "Ezee",
      "Gender": "M",
      "Age": 28,
      "Children": [
        {
          "Name": "Dre",
          "Age": 1
        },
        {
          "Name": "George",
          "Age": 2
        }
      ]
    }
  ],
  "Age": 50,
  "MiddleName": "Jay"
}

为了完整地完成您的问题，如果您希望此人的属性（没有其

子项列表

）作为一行，您必须决定以一种固定的方式（如字母顺序）存储这些属性，然后您可以使用dict的

.items（）

方法将其提取为正确的格式：

[v for k,v in sorted(t for t in parent.items() if t[0] != 'Children')]

举个例子：

[50, 'M', 'Jay', 'Michael']
#Age, Gender, Middle Name, Name

哦，如果您愿意，第一段代码可以压缩成一行：

__import__('functools').reduce(lambda d,a:d[int(a) if isintance(d,list) else a], path[:path.rfind('/Children')].split('/'), json_data)