2 回答

TA貢獻1876條經驗 獲得超7個贊
作為@rpanai 解決方案的替代方案,我將處理轉移到 vanilla python 中:
將數據幀轉換為 dict :
M = df.to_dict("records")
為項目創建字典
items = [
{key: value
for key, value in entry.items()
if key not in ("user_id", "order_num")}
for entry in M
]
item_details = [{str(num + 1): entry}
for num, entry
in enumerate(items)]
print(item_details)
[{'1': {'item_id': 1, 'item_desc': 'red'}},
{'2': {'item_id': 2, 'item_desc': 'blue'}},
{'3': {'item_id': 3, 'item_desc': 'green'}}]
初始化dict并添加剩余數據
d = dict()
d['New-Data'] = item_details
d['order_number'] = M[0]['order_num']
d['user_id'] = M[0]['user_id']
wrapper = [d]
print(wrapper)
[{'New-Data': [{'1': {'item_id': 1, 'item_desc': 'red'}},
{'2': {'item_id': 2, 'item_desc': 'blue'}},
{'3': {'item_id': 3, 'item_desc': 'green'}}],
'order_number': 1,
'user_id': 1}]

TA貢獻1993條經驗 獲得超6個贊
您是否考慮過使用自定義功能
import pandas as pd
df = pd.DataFrame({'user_id': {0: 1, 1: 1, 2: 1},
'order_num': {0: 1, 1: 1, 2: 1},
'item_id': {0: 1, 1: 2, 2: 3},
'item_desc': {0: 'red', 1: 'blue', 2: 'green'}})
out = df.groupby(['user_id', 'order_num'])[["item_id", "item_desc"]]\
.apply(lambda x: x.to_dict("records"))\
.apply(lambda x: [{str(l["item_id"]):l for l in x}])\
.reset_index(name="New-Data")\
.to_dict("records")
在哪里out返回
[{'user_id': 1,
'order_num': 1,
'New-Data': [{'1': {'item_id': 1, 'item_desc': 'red'},
'2': {'item_id': 2, 'item_desc': 'blue'},
'3': {'item_id': 3, 'item_desc': 'green'}}]}]
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