來自字典的Pandas DataFrame，帶有嵌套的字典列表

Question

my_dict = { 'company_a': [],
            'company_b': [ {'gender': 'Male',
                            'investor': True,
                            'name': 'xyz',
                            'title': 'Board Member'} ],
            'company_c': [],
            'company_m': [ {'gender': 'Male',
                            'investor': None,
                            'name': 'abc',
                            'title': 'Advisor'}, 
                            {'gender': 'Male',
                            'investor': None,
                            'name': 'opq',
                            'title': 'Advisor'} ],
            'company_x': [],
            'company_y': [] }

如何將上述Python字典轉換為帶有以下列的Pandas數據框： company, gender, investor, name, title

列company將由my_dict的頂級鍵my_dict 。 其他列將填充數組中字典中的值。

我已經嘗試過pd.DataFrame.from_dict(my_dict, orient='index') ，但它沒有給我我想要的東西。

Answer 1

此版本使用None填充所有缺失值：

data = {'company': [], 'gender': [], 'investor': [], 'name': [], 'title': []}
for k, v in my_dict.items():
    for entry in v:
        data['company'].append(k)
    if not v:
        data['company'].append(k)
    for name in ['gender', 'investor', 'name', 'title']:
        has_entry = False
        for entry in v:
            has_entry = True
            data[name].append(entry.get(name))
        if not has_entry:
            data[name].append(None)
df = pd.DataFrame(data)
print(df)

輸出：

     company gender investor  name         title
0  company_a   None     None  None          None
1  company_y   None     None  None          None
2  company_b   Male     True   xyz  Board Member
3  company_c   None     None  None          None
4  company_x   None     None  None          None
5  company_m   Male     None   abc       Advisor
6  company_m   Male     None   opq       Advisor

您也可以用NaN替換所有None ：

print(df.fillna(np.nan))

輸出：

     company gender investor name         title
0  company_a    NaN      NaN  NaN           NaN
1  company_y    NaN      NaN  NaN           NaN
2  company_b   Male     True  xyz  Board Member
3  company_c    NaN      NaN  NaN           NaN
4  company_x    NaN      NaN  NaN           NaN
5  company_m   Male      NaN  abc       Advisor
6  company_m   Male      NaN  opq       Advisor

Answer 2

有點混亂，但這很靈活，取決於嵌套字典中的屬性，並將公司放在他們自己的列中。

df = pd.DataFrame(columns = ['company'])
i = 0

for company in my_dict:
    for nested_dict in my_dict[company]:
        df.loc[i,'company'] = company
        for attribute in nested_dict.keys():
            df.loc[i, attribute] = nested_dict[attribute]
        i += 1

輸出：

Out[46]:
    company     name  gender  title         investor
0   company_m   abc   Male    Advisor       NaN
1   company_m   opq   Male    Advisor       NaN
2   company_b   xyz   Male    Board Member  True