将嵌套的json转换为没有嵌套对象的字典格式 [英] Convert the nested json into a dictionary format with no nested objects

问题描述

我具有以下格式的输入数据:

I have the input data in below format:

data = [[u'Richard', u'48', [u'Josh', u'Beth'], {u'city': u'Seattle', u'Disability': u'no', u'enterprenuer': u'yes'}], [u'Bryan', u'32',[], {u'city': u'NY', u'enterprenuer': u'no', u'wfh': u'yes', u'disability': u'no', u'Visa': u'no'}]]

稍后执行json.dumps的内容将变为:

Which later on doing json.dumps becomes:

[["Richard", "48", ["Josh", "Beth"], {"city": "Seattle", "enterprenuer": "yes", "Disability": "no"}], ["Bryan", "32", [], {"Visa": "no", "city": "NY", "wfh": "yes", "enterprenuer": "no", "disability": "no"}]]

此外，我还有另一个列表，其中包含该字典的键:

Also, I have another list which holds the keys for the dict:

key_list = ["Name", "Age", "Children", "details"]

我尝试了以下代码:

list_of_dicts = []
for d in data:
    dict = {}
    for i in range(0, len(key_list)-1):
        dict[key_list[i]] = d[i]
    list_of_dicts.append(dict)

有了这个，我就能得到new_dict:

With this I was able to get new_dict:

[{'Age': u'48', 'Name': u'Richard', 'Children': [u'Josh', u'Beth']}, {'Age': u'32', 'Name': u'Bryan', 'Children': []}]

但是我无法将嵌套的dict从data放入new_dict中，而无需再次在其上运行代码.我不想多次运行操作. 另外，我也在考虑是否还有更好的方法来删除嵌套列表，但是在多次尝试之后，我一直跟踪并弄乱了我的代码.

But I am not able to get the nested dict from data into the new_dict without the need to run code on it again. I don't want to run operation multiple times. Also, I was thinking if there's any better way to remove the nested list as well but after multiple hit and trial I got side tracked and messed up my code.

这是预期的输出:

[{"Name":"Richard","Age":"48","Children":"Josh,Beth","city":"Seattle","enterprenuer":"yes","Disability":"no"},{"Name":"Bryan","Age":"32","Children":"","Visa":"no","city":"NY","wfh":"yes","enterprenuer":"no","disability":"no"}]

推荐答案

您可以尝试:

1)在python3中:

1) in python3:

from pprint import pprint

data = [["Richard", "48", ["Josh", "Beth"], {"city": "Seattle", "enterprenuer": "yes", "Disability": "no"}], ["Bryan", "32", [], {"Visa": "no", "city": "NY", "wfh": "yes", "enterprenuer": "no", "disability": "no"}]]
key_list = ["Name", "Age", "Children", "details"]

pprint([dict(zip(key_list[:2], e[:2]), **{key_list[2]: ','.join(e[2])}, **e[3]) for e in data])

输出:

[{'Name': 'Richard',
  'Age': '48',
  'Children': 'Josh,Beth',
  'city': 'Seattle',
  'Disability': 'no',
  'enterprenuer': 'yes'},
 {'Name': 'Bryan',
  'Age': '32',
  'Children': '',
  'city': 'NY',
  'enterprenuer': 'no',
  'wfh': 'yes',
  'disability': 'no',
  'Visa': 'no'}]

2)在python2中:

2) in python2:

pprint([dict(zip(key_list[:2], e[:2]), **dict([(key_list[2], ','.join(e[2]))], **e[3])) for e in data])

输出:

[{'Age': '48',
  'Children': 'Josh,Beth',
  'Disability': 'no',
  'Name': 'Richard',
  'city': 'Seattle',
  'enterprenuer': 'yes'},
 {'Age': '32',
  'Children': '',
  'Name': 'Bryan',
  'Visa': 'no',
  'city': 'NY',
  'disability': 'no',
  'enterprenuer': 'no',
  'wfh': 'yes'}]

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