从深层嵌套的JSON创建Pandas DataFrame

Question

我正在尝试从深层嵌套的JSON字符串创建单个Pandas DataFrame对象.

JSON模式是:

{"intervals": [
{
pivots: "Jane Smith",
"series": [
    {
        "interval_id": 0,
        "p_value": 1
       },
     {
         "interval_id": 1,
         "p_value": 1.1162791357932633e-8
     },
   {
        "interval_id": 2,
        "p_value": 0.0000028675012051504467
     }
    ],
   },
  {

"pivots": "Bob Smith",
  "series": [
       {
            "interval_id": 0,
            "p_value": 1
           },
         {
             "interval_id": 1,
            "p_value": 1.1162791357932633e-8
         },
       {
            "interval_id": 2,
            "p_value": 0.0000028675012051504467
         }
       ]
     }
    ]
 }

期望的结果我需要将其展平以制作表格:

Actor Interval_id Interval_id Interval_id ... 
Jane Smith      1         1.1162        0.00000 ... 
Bob Smith       1         1.1162        0.00000 ... 

第一列是Pivots值,其余列是键的值interval_id并p_value存储在列表中series.

到目前为止我已经有了

import requests as r
import pandas as pd
actor_data = r.get("url/to/data").json['data']['intervals']
df = pd.DataFrame(actor_data)

actor_data是一个长度等于个人数量的列表,即pivots.values().df对象只是返回

<bound method DataFrame.describe of  pivots             Series
0           Jane Smith  [{u'p_value': 1.0, u'interval_id': 0}, {u'p_va...
1           Bob Smith  [{u'p_value': 1.0, u'interval_id': 0}, {u'p_va...
.
.
.

如何迭代该series列表以获取dict值并创建N个不同的列？我应该尝试为series列表创建一个DataFrame ,重新整形,然后使用actor名称进行列绑定吗？

更新:

pvalue_list = [i['p_value'] for i in json_data['series']]

这给了我一份清单.现在我需要弄清楚如何将每个列表作为一行添加到DataFrame中.

value_list = []
for i in pvalue_list:
    pvs = [j['p_value'] for j in i]
    value_list = value_list.append(pvs)
return value_list

这将返回NoneType

解

def get_hypthesis_data():
    raw_data = r.get("/url/to/data").json()['data']
    actor_dict = {}
    for actor_series in raw_data['intervals']:
        actor = actor_series['pivots']
        p_values = []
        for interval in actor_series['series']:
            p_values.append(interval['p_value'])
        actor_dict[actor] = p_values
    return pd.DataFrame(actor_dict).T

这将返回正确的DataFrame.我把它换成了所以个人是行而不是列.

Answer 1

我认为以产生重复列名的方式组织数据只会让您在以后的日子里感到头痛.更好的方法是恕我直言创造每一个列pivots,interval_id和p_value.将数据加载到pandas后,这将非常容易查询数据.

此外,您的JSON中有一些错误.我通过它来查找错误.

jq 帮助到这里

import sh
jq = sh.jq.bake('-M')  # disable colorizing
json_data = "from above"
rule = """[{pivots: .intervals[].pivots, 
            interval_id: .intervals[].series[].interval_id,
            p_value: .intervals[].series[].p_value}]"""
out = jq(rule, _in=json_data).stdout
res = pd.DataFrame(json.loads(out))

这将产生类似的输出

    interval_id       p_value      pivots
32            2  2.867501e-06  Jane Smith
33            2  1.000000e+00  Jane Smith
34            2  1.116279e-08  Jane Smith
35            2  2.867501e-06  Jane Smith
36            0  1.000000e+00   Bob Smith
37            0  1.116279e-08   Bob Smith
38            0  2.867501e-06   Bob Smith
39            0  1.000000e+00   Bob Smith
40            0  1.116279e-08   Bob Smith
41            0  2.867501e-06   Bob Smith
42            1  1.000000e+00   Bob Smith
43            1  1.116279e-08   Bob Smith

改编自此评论

当然,您始终可以调用res.drop_duplicates()删除重复的行.这给了

In [175]: res.drop_duplicates()
Out[175]:
    interval_id       p_value      pivots
0             0  1.000000e+00  Jane Smith
1             0  1.116279e-08  Jane Smith
2             0  2.867501e-06  Jane Smith
6             1  1.000000e+00  Jane Smith
7             1  1.116279e-08  Jane Smith
8             1  2.867501e-06  Jane Smith
12            2  1.000000e+00  Jane Smith
13            2  1.116279e-08  Jane Smith
14            2  2.867501e-06  Jane Smith
36            0  1.000000e+00   Bob Smith
37            0  1.116279e-08   Bob Smith
38            0  2.867501e-06   Bob Smith
42            1  1.000000e+00   Bob Smith
43            1  1.116279e-08   Bob Smith
44            1  2.867501e-06   Bob Smith
48            2  1.000000e+00   Bob Smith
49            2  1.116279e-08   Bob Smith
50            2  2.867501e-06   Bob Smith

[18 rows x 3 columns]