如何将嵌套的dict转换为数据

Question

假设我有一个API响应，如：

{
    "fact": {
        "UP": [{
            "SCODE": "CNB",
            "SNAME": "Kanpur Central"
        }, {
            "SCODE": "JHS",
            "SNAME": "Jhansi Junction"
        }],
        "MP": [{
            "SCODE": "BPL",
            "SNAME": "Bhopal Junction"
        }, {
            "SCODE": "JBP",
            "SNAME": "Jabalpur Junction"
        }]
    }
}

我必须将其转换为如下所示的数据格式(预期输出)：

fact    SCODE   SNAME
UP      CNB     Kanpur Central
UP      JHS     Jhansi Junction
MP      BPL     Bhopal Junction
MP      JBP     Jabalpur Junction

我的努力:我尝试使用json_normalize()，但没有达到预期的输出：

pd.json_normalize(response).apply(pd.Series.explode)

Answer 1

一种选择是使用python重新构建：

df = pd.DataFrame([{'fact': k, **item}
                   for k, lst in response['fact'].items()
                   for item in lst])

  fact SCODE              SNAME
0   UP   CNB     Kanpur Central
1   UP   JHS    Jhansi Junction
2   MP   BPL    Bhopal Junction
3   MP   JBP  Jabalpur Junction

pandas通过explode + apply pd.Series的选项

df = (
    pd.DataFrame(response)['fact']
        .explode()
        .apply(pd.Series)
        .rename_axis('fact')
        .reset_index()
)

  fact SCODE              SNAME
0   MP   BPL    Bhopal Junction
1   MP   JBP  Jabalpur Junction
2   UP   CNB     Kanpur Central
3   UP   JHS    Jhansi Junction

Answer 2

• 使用OP.
• 中的response，您必须创建另一个结构，因为json_normalize使用字典列表，而fact必须包含在那里：

。

new_response = [{"fact": rfact, **r} for rfact in response["fact"] for r in response["fact"][rfact]]

最后，只需应用该函数：

final_result = pd.json_normalize(new_response)

  fact SCODE              SNAME
0   UP   CNB     Kanpur Central
1   UP   JHS    Jhansi Junction
2   MP   BPL    Bhopal Junction
3   MP   JBP  Jabalpur Junction

Answer 3

不像直接在字典中工作那样有效(所选的答案做得很好)：

    data = {
    "fact": {
        "UP": [{
            "SCODE": "CNB",
            "SNAME": "Kanpur Central"
        }, {
            "SCODE": "JHS",
            "SNAME": "Jhansi Junction"
        }],
        "MP": [{
            "SCODE": "BPL",
            "SNAME": "Bhopal Junction"
        }, {
            "SCODE": "JBP",
            "SNAME": "Jabalpur Junction"
        }]
    }
}

keys = data['fact']

   (pd.concat([jn(data['fact'][key]) for key in keys], 
              keys = keys)
      .droplevel(-1)
      .rename_axis(index='fact')
      .reset_index()
     )



fact SCODE              SNAME
0   UP   CNB     Kanpur Central
1   UP   JHS    Jhansi Junction
2   MP   BPL    Bhopal Junction
3   MP   JBP  Jabalpur Junction