我一直在挣扎于json中的嵌套结构,如何转换为正确的形式
{
"id": "0c576f35-d704-4fa8-8cbb-311c6be36358",
"employee_id": null,
"creator_id": "16ca2db9-206c-4e18-891d-a00a5252dbd3",
"closed_by_id": null,
"request_number": 23,
"priority": "2",
"form_id": "urlaub-weitere-abwesenheiten",
"status": "opened",
"name": "Urlaub & weitere Abwesenheiten",
"read_by_employee": false,
"custom_status": {
"id": 15793,
"name": "In Bearbeitung HR"
},
"due_date": null,
"created_at": "2021-03-29T15:18:37.572040+02:00",
"updated_at": "2021-03-29T15:22:15.590156+02:00",
"closed_at": null,
"archived_at": null,
"attachment_count": 1,
"category": {
"id": "payroll-time-management",
"name": "Payroll, Time & Attendance"
},
"public_comment_count": 0,
"form_data": [
{
"field_id": "subcategory",
"values": [
"Time & Attendance - Manage monthly/year-end consolidation and report"
]
},
{
"field_id": "separator-2",
"values": [
null
]
},
{
"field_id": "art-der-massnahme",
"values": [
"Fortbildung"
]
},
{
"field_id": "bezeichnung-der-schulung-kurses",
"values": [
"dfgzhujiko"
]
},
{
"field_id": "startdatum",
"values": [
"2021-03-26"
]
},
{
"field_id": "enddatum",
"values": [
"2021-03-27"
]
},
{
"field_id": "freistellung",
"values": [
"nein"
]
},
{
"field_id": "mit-bildungsurlaub",
"values": [
""
]
},
{
"field_id": "kommentarfeld_fortbildung",
"values": [
""
]
},
{
"field_id": "separator",
"values": [
null
]
},
{
"field_id": "instructions",
"values": [
null
]
},
{
"field_id": "entscheidung-hr-bp",
"values": [
"Zustimmen"
]
},
{
"field_id": "kommentarfeld-hr-bp",
"values": [
"wsdfghjkmhnbgvfcdxsybvnm,"
]
},
{
"field_id": "individuelle-abstimmung",
"values": [
""
]
}
],
"form_files": [
{
"id": 30129,
"filename": "empty_background.png",
"field_id": "anhang"
}
],
"visible_by_employee": false,
"organization_ids": [],
"need_edit_by_employee": false,
"attachments": []}
对熊猫使用一个简单的解决方案,
Request = pd.DataFrame.from_dict(pd.json_normalize(data), orient='columns')它几乎以正确的形式显示:

如何将字典从列form_data i form_files中分离出来,我已经做了很多研究,但是我仍然有很多困难来解决这个问题,如何分割列的form_data,没有元到ID的行
发布于 2021-03-30 16:47:28
你可以做这样的事。
将dataframe和列作为参数传递给函数
def explode_node(child_df, column_value):
child_df = child_df.dropna(subset=[column_value])
if isinstance(child_df[str(column_value)].iloc[0], str):
child_df[column_value] = child_df[str(column_value)].apply(ast.literal_eval)
expanded_child_df = (pd.concat({i: json_normalize(x) for i, x in child_df.pop(str(column_value)).items()}).reset_index(level=1,drop=True).join(child_df, how='right', lsuffix='_left', rsuffix='_right').reset_index(drop=True))
expanded_child_df.columns = map(str.lower, expanded_child_df.columns)
return expanded_child_dfhttps://stackoverflow.com/questions/66874360
复制相似问题