使用 Python 规范化 JSON

Question

我对 JSON 和 Python 比较陌生，自从过去两天以来我一直在努力压平 JSON。我在 http://pandas.pydata.org/pandas-docs/version/0.19/generated/pandas.io.json.json_normalize.html, but I didn't understand how to unlist some nested elements. I also read a few threads Flatten JSON based on an attribute - python How to normalize complex nested json in python? and https://towardsdatascience.com/flattening-json-objects-in-python-f5343c794b10 阅读了示例。我尝试了所有但没有任何运气。

这是我的 JSON 文件的第一条记录：

d = 
{'city': {'url': 'link',
  'name': ['San Francisco']},
 'rank': 1,
 'resident': [
  {'link': ['bit.ly/0842/'], 'name': ['John A']},
  {'link': ['bit.ly/5835/'], 'name': ['Tedd B']},
  {'link': ['bit.ly/2011/'], 'name': ['Cobb C']},
  {'link': ['bit.ly/0855/'], 'name': ['Jack N']},
  {'link': ['bit.ly/1430/'], 'name': ['Jack K']},
  {'link': ['bit.ly/3081/'], 'name': ['Edward']},
  {'link': ['bit.ly/2001/'], 'name': ['Jack W']},
  {'link': ['bit.ly/0020/'], 'name': ['Henry F']},
  {'link': ['bit.ly/2137/'], 'name': ['Joseph S']},
  {'link': ['bit.ly/3225/'], 'name': ['Ed B']},
  {'link': ['bit.ly/3667/'], 'name': ['George Vvec']},
  {'link': ['bit.ly/6434/'], 'name': ['Robert W']},
  {'link': ['bit.ly/4036/'], 'name': ['Rudy B']},
  {'link': ['bit.ly/6450/'], 'name': ['James K']},
  {'link': ['bit.ly/5180/'], 'name': ['Billy N']},
  {'link': ['bit.ly/7847/'], 'name': ['John S']}]
}

这是预期的输出：

city_url  city_name      rank    resident_link   resident_name  
link      San Francisco   1     'bit.ly/0842/'   'John A'
link      San Francisco   1     'bit.ly/5835/'   'Tedd B'
link      San Francisco   1     'bit.ly/2011/'   'Cobb C'
link      San Francisco   1     'bit.ly/0855/'   'Jack N'
link      San Francisco   1     'bit.ly/1430/'   'Jack K'
link      San Francisco   1     'bit.ly/3081/'   'Edward'
link      San Francisco   1     'bit.ly/2001/'   'Jack W'
link      San Francisco   1     'bit.ly/0020/'   'Henry F'
link      San Francisco   1     'bit.ly/2137/'   'Joseph S'
link      San Francisco   1     'bit.ly/3225/'   'Ed B'
link      San Francisco   1     'bit.ly/3667/'   'George Vvec'
link      San Francisco   1     'bit.ly/6434/'   'Robert W'
link      San Francisco   1     'bit.ly/4036/'   'Rudy B'
link      San Francisco   1     'bit.ly/6450/'   'James K'
link      San Francisco   1     'bit.ly/5180/'   'Billy N'
link      San Francisco   1     'bit.ly/7847/'   'John S'

flatten_json() 函数（来自上面的 Medium.com）破坏了层次结构。这是前几行：

{'city_url': 'link',
 'city_name_0': 'San Francisco',
 'rank': 1,
 'resident_0_link_0': 'bit.ly/0842/',
 'resident_0_name_0': 'John A', ...

有人可以帮我考虑如何转换这些数据集吗？不幸的是，pandas 文档没有为初学者提供指导。这就是我在玩的东西。没有任何效果。

from pandas.io.json import json_normalize
json_normalize(d,['city',['name','rank']])
json_normalize(d,['city','name','rank'])
json_normalize(d,['city','name'])

如果有人指导如何进行这些类型的转换和思考过程，我将不胜感激。

此外，由于原始数据集中的数据量，我正在寻找矢量化操作或 O(N) 操作而不是 O(N2)。因此，任何比 O(N) 慢的都不起作用。

Answer 1

如果你知道 json blob 的结构，这就可以了

resident_link = [k['link'][0] for k in d['resident']]
resident_name = [k['name'][0] for k in d['resident']]
n = len(d['resident'])
city_url = n * [d['city']['url']]
city_name = n * [d['city']['name'][0]]
rank = n * [d['rank']]

df = pandas.DataFrame({
    'resident_name' : resident_name,
    'resident_link' : resident_link,
    'city_url' : city_url,
    'city_name' : city_name,
    'rank' : rank
})

产生

        city_name city_url  rank resident_link resident_name
0   San Francisco     link     1  bit.ly/0842/        John A
1   San Francisco     link     1  bit.ly/5835/        Tedd B
2   San Francisco     link     1  bit.ly/2011/        Cobb C
3   San Francisco     link     1  bit.ly/0855/        Jack N
4   San Francisco     link     1  bit.ly/1430/        Jack K
5   San Francisco     link     1  bit.ly/3081/        Edward
6   San Francisco     link     1  bit.ly/2001/        Jack W
7   San Francisco     link     1  bit.ly/0020/       Henry F
8   San Francisco     link     1  bit.ly/2137/      Joseph S
9   San Francisco     link     1  bit.ly/3225/          Ed B
10  San Francisco     link     1  bit.ly/3667/   George Vvec
11  San Francisco     link     1  bit.ly/6434/      Robert W
12  San Francisco     link     1  bit.ly/4036/        Rudy B
13  San Francisco     link     1  bit.ly/6450/       James K
14  San Francisco     link     1  bit.ly/5180/       Billy N
15  San Francisco     link     1  bit.ly/7847/        John S

编辑

正如 OP 在评论中所说，假设有很多这样的记录，每个记录都具有相同的结构

nrecords = 10
dd = {k : d for k in range(nrecords)}

dd 现在有原始 json blob 的 10 个副本。这就是代码应该如何更新

ff = pandas.DataFrame()

for record in range(nrecords):

    n = len(dd[record]['resident'])

    df = {
        'resident_link' : [k['link'][0] for k in dd[record]['resident']],
        'resident_name' : [k['name'][0] for k in dd[record]['resident']],
        'city_url' : n * [dd[record]['city']['url']],
        'city_name' : n * [dd[record]['city']['name'][0]],
        'rank' : n * [dd[record]['rank']]
        }

    df = pandas.DataFrame(df)
    ff = ff.append(df).reset_index(drop = True)

下面是运行时间的估计值，它是记录数的函数。基于此，完成 150 万条记录大约需要 1 小时

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