堆栈内存溢出
  简体   繁体   English

无法使用python从Json文件中获取特定值

[英]Not able to get specific value from Json file using python

 原文  2019-05-16 07:43:29  4   5    python/ json/ list/ extract

问题描述

I'm having difficulty to get some data from my json file. 我很难从json文件中获取一些数据。 I'm able to get some of the data but when I want to dig more in details I'm getting error and stuck no where. 我能够获得一些数据，但是当我想进一步挖掘细节时，我会出错并且没有卡住。

Running the script below return an error 运行下面的脚本返回错误 

AttributeError: 'list' object has no attribute 'get'

I know its a list but I'm not sure how to get the rest of the object list. 我知道它的列表，但是我不确定如何获取其余的对象列表。

Script 脚本 

ifile=open('source.json', 'r')
ofile=open('extract.json', 'w')
json_decode=json.load(ifile)
myresult=[]
for item in json_decode:
    mydict={}
    mydict['sID']=item.get('Ls id')
    my_dict['dID']=item.get('items').get('Link ID').get('Metric').get('Link Type')
    mydict['type']=item.get('Type')
    myresult.append(mydict)
myjson=json.dumps(myresult, ofile)
ofile.write(myjson)
ofile.close()

source Json file 源Json文件 

[
  {
    "Ls age": "201",
    "items": [
      {
        "Link ID": "1.1.1.2",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.4",
        "Link Type": "P-2-P",
        "Metric": "1",
        "Data": "192.168.100.34"
      },
      {
        "Link ID": "192.168.100.33",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.1",
        "Link Type": "P-2-P",
        "Metric": "1",
        "Data": "192.168.100.53"
      }
    ],
    "Len": "84",
    "Ls id": "1.1.1.2",
    "Adv rtr": "1.1.1.2",
    "Type": "Router",
    "Link count": "5"
  },
  {
    "Ls age": "1699",
    "seq#": "80008d72",
    "items": [
      {
        "Link ID": "1.1.1.1",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.1",
        "Link Type": "StubNet",
        "Metric": "12",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.3",
        "Link Type": "P-2-P",
        "Metric": "10",
        "Data": "192.168.100.26"
      },
      {
        "Link ID": "192.168.100.25",
        "Link Type": "StubNet",
        "Metric": "10",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.2",
        "Link Type": "P-2-P",
        "Metric": "10",
        "Data": "192.168.100.54"
      },
      {
        "Link ID": "192.168.100.53",
        "Link Type": "StubNet",
        "Metric": "10",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      }
    ],
    "Len": "96",
    "Ls id": "1.1.1.1",
    "chksum": "0x16fc",
    "Adv rtr": "1.1.1.1",
    "Type": "Router",
    "Options": "ASBR  E",
    "Link count": "6"
  }
]

Expected to have result as below 预期结果如下 

[
  {
    "type": "Router",
    "sID": "1.1.1.2",
    "dID": "1.1.1.2",
    "LinkType":"StubNet",
    "Metric":"1"
  },
  { 
    "type": "Router",
    "sID": "1.1.1.2", 
    "dID": "1.1.1.4",
    "Link Type": "P-2-P",
    "Metric": "1"
  },
  {
    "type": "Router",
    "sID": "1.1.1.2",
    "dID": "192.168.100.33",
    "LinkType":"StubNet",
    "Metric":"1"   
  },   
  { 
    "type": "Router",     
    "sID": "1.1.1.2",     
    "dID":"1.1.1.1",
    "Link Type": "P-2-P",
    "Metric": "1"
  },
  {
    "type": "Router",
    "sID": "1.1.1.1",
    "dID": "1.1.1.1",     
    "LinkType":"StubNet",     
    "Metric":"1"  
   },   
   {     
    "type": "Router",
    "sID": "1.1.1.1",
    "dID":"1.1.1.1",
    "Link Type": "StubNet",
    "Metric": "12"
  },
  {
    "type": "Router",
    "sID": "1.1.1.1",
    "dID": "1.1.1.3",
    "LinkType":"P-2-P",
    "Metric":"10"   
  } 
]

Appreciate to advise how could proceed further. 感谢您提供进一步的建议。 I have search around and try error and still no able to resolve it. 我到处搜索并尝试错误，仍然无法解决。 Appreciate your advise and support. 感谢您的建议和支持。 Thank you 谢谢

5 个解决方案

解决方案1
 1  2019-05-16 07:56:22

You should be able to access the values like a dictionary, for example: 您应该能够像字典一样访问这些值，例如： 

ifile=open('source.json', 'r')
ofile=open('extract.json', 'w')
json_decode=json.load(ifile)
myresult=[]
for item in json_decode:
    mydict={}
    mydict['sID']=item['Ls id']
    my_dict['dID']=item['items']['Link ID']['Metric']['Link Type']
    mydict['type']=item['Type']
    myresult.append(mydict)
myjson=json.dumps(myresult, ofile)
ofile.write(myjson)
ofile.close()

Does this work for you? 这对您有用吗？ If not, what errors are you running into? 如果没有，您遇到什么错误？

解决方案2
 1 已采纳  2019-05-16 08:05:29

you need to iterate over devices and then over device['items'] 您需要遍历设备，然后遍历设备['items'] 

import json

with open('source.json', 'r') as ifile:
    json_data=json.load(ifile)

my_result=[]

for device in json_data:
    for item in device.get('items', []):
        my_dict={}
        my_dict['type'] = device.get('Type')
        my_dict['sID'] = device.get('Ls id')
        my_dict['dID'] = item.get('Link ID')
        my_dict['Link Type'] = item.get('Link Type')
        my_dict['Metric'] = item.get('Metric')
        my_result.append(my_dict)

with open('extract.json', 'w') as ofile:
    json.dump(my_result, ofile, indent=4)

For more structured code you may want to define a function(s) that takes device/item as argument, parse it and return a list of dicts/dict 对于更结构化的代码，您可能需要定义一个以设备/项目作为参数的函数，对其进行解析并返回字典/字典的列表。

解决方案3
 1  2019-05-16 08:06:10

First, you are getting a list in item['items'] . 首先，您将在item['items']中获得一个列表。 You need to decide whether you need to save all values available inside that list or not. 您需要确定是否需要保存该列表中的所有可用值。

Second, you are trying to access multiple attributes in a dict at once, but you're chaining the get commands. 其次，您尝试一次访问一个字典中的多个属性，但是链接了get命令。 The code however attempts to treat it like a nested dict, and would run into datatype errors. 但是，该代码尝试将其视为嵌套dict，并会遇到数据类型错误。

Third, you don't actually need to type get like that, a cleaner version is to just use the square bracket notation. 第三，您实际上不需要键入get ，更干净的版本是仅使用方括号表示法。

Assuming you need to create a new dict for each of the items in the list of item['items'] a solution would look something like this: 假设您需要为item['items']列表中的每个项目创建一个新的dict，那么解决方案将如下所示： 

import json
s = '''
[
  {
    "Ls age": "201",
    "items": [
      {
        "Link ID": "1.1.1.2",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.4",
        "Link Type": "P-2-P",
        "Metric": "1",
        "Data": "192.168.100.34"
      },
      {
        "Link ID": "192.168.100.33",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.1",
        "Link Type": "P-2-P",
        "Metric": "1",
        "Data": "192.168.100.53"
      }
    ],
    "Len": "84",
    "Ls id": "1.1.1.2",
    "Adv rtr": "1.1.1.2",
    "Type": "Router",
    "Link count": "5"
  },
  {
    "Ls age": "1699",
    "seq#": "80008d72",
    "items": [
      {
        "Link ID": "1.1.1.1",
        "Link Type": "StubNet",
        "Metric": "1",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.1",
        "Link Type": "StubNet",
        "Metric": "12",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.3",
        "Link Type": "P-2-P",
        "Metric": "10",
        "Data": "192.168.100.26"
      },
      {
        "Link ID": "192.168.100.25",
        "Link Type": "StubNet",
        "Metric": "10",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      },
      {
        "Link ID": "1.1.1.2",
        "Link Type": "P-2-P",
        "Metric": "10",
        "Data": "192.168.100.54"
      },
      {
        "Link ID": "192.168.100.53",
        "Link Type": "StubNet",
        "Metric": "10",
        "Data": "255.255.255.255",
        "Priority": "Medium"
      }
    ],
    "Len": "96",
    "Ls id": "1.1.1.1",
    "chksum": "0x16fc",
    "Adv rtr": "1.1.1.1",
    "Type": "Router",
    "Options": "ASBR  E",
    "Link count": "6"
  }
]
'''
input_lst = json.loads(s)
myresult=[]
for item in input_lst:
    mydict={}
    mydict_sID = item['Ls id']
    mydict_type = item['Type']
    temp = []
    for x in item['items']:
        mydict={'Ls id': mydict_sID,
                'Type': mydict_type,
                'Link ID': x['Link ID'],
                'Metric': x['Metric'],
                'Link Type': x['Link Type']
                }
        temp.append(mydict)
    myresult.extend(temp)

Make sure you change the code lines that i used to read the string as necessary. 确保根据需要更改用于读取字符串的代码行。

解决方案4
 1  2019-05-16 08:21:07

The problem is that item['items'] is also a list so you need an inner loop to process all of its elements. 问题在于item['items']也是一个列表，因此您需要一个内部循环来处理其所有元素。 Furthermore you must extract individually each value: 此外，您必须分别提取每个值： 

for item in json_decode:
    for sub in item.get('items'):
        mydict={}
        mydict['type']=item.get('Type')
        mydict['sID']=item.get('Ls id')
        mydict['dID']=sub.get('Link ID')
        mydict['Link Type']=sub.get('Link Type')
        mydict['Metric']=sub.get('Metric')
        myresult.append(mydict)

解决方案5
 1  2019-05-16 08:40:24

to process the elements of a nested object of the list, you need to use an additional loop, for example, 要处理列表的嵌套对象的元素，您需要使用其他循环，例如， 

import json

ifile = 'source.json'
ofile = 'extract.json'

myresult = []
with open(ifile, 'r') as sf:
   json_decode = json.load(sf)
   for item in json_decode:
       for sub_item in item.get('items', []):
           myresult.append(dict(type=item.get('Type'),
                                sID=item.get('Ls id', ''),
                                dID=sub_item.get('Link ID'),
                                LinkType=sub_item.get('Link Type'),
                                Metric=sub_item.get('Metric')
                                )
                           )

with open(ofile, 'w') as of:
    of.write(json.dumps(myresult, indent=4))

or a slightly abbreviated version using list comprehension: 或使用列表理解的略微缩写版本： 

import json

ifile = 'source.json'
ofile = 'extract.json'

with open(ifile, 'r') as sf:
    json_decode = json.load(sf)
    myresult = [dict(type=item.get('Type'), 
                     sID=item.get('Ls id', ''), 
                     dID=sub_item.get('Link ID'),
                     LinkType=sub_item.get('Link Type'), 
                     Metric=sub_item.get('Metric')) for item in json_decode 
                for sub_item in item.get('items', [])]

with open(ofile, 'w') as of:
    of.write(json.dumps(myresult, indent=4))

暂无
暂无

声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 使用 MongoDB python 从 JSON 数据中获取特定值 - Get a specific value from JSON data using MongoDB python 如何使用 python 从 json 文件中获取特定数据 - how to get specific data from json file using python python：从json字符串中获取特定值 - python : get specific value from json string python 3从json字典中获取特定值 - python 3 get specific value from json dictionary 使用 Python 解析此 JSON 文件中的特定值？ - Parse specific value from this JSON file with Python? 使用 Python 从 json 文件中读取特定值 - Read a specific value from a json file with Python 无法使用 python 从 JSON 生成正确的 csv 文件 - Not able to generate a proper csv file from JSON using python python - 在 json 上获取特定值 - python - get specific value on json 在JSON和Python中获取特定值 - Get a specific value in JSON and Python 如何使用python json从JSON文件中提取特定对象 - How to extract a a specific object from JSON file using python json
相关标签
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM