我有以下json。
[
{
"TX_ID": "C592096166066790609734",
"FEE_1": 0.25,
"FEE_7": 0.95,
"USAGE": "Yzua12"
},
{
"TX_ID": "C636460166108115381222",
"FEE_1": 0.35,
"FEE_3": 1.39,
"USAGE": "Yzua13"
}
]
使用python，我需要下面的輸出值“Fee_*”和dict型別，但欄位“TX_ID”和“USAGE”除外
預期結果：
[
{
"TX_ID": "C592096166066790609734",
"FEE_1": {"set":0.25},
"FEE_7": {"set":0.95},
"USAGE": "Yzua12"
},
{
"TX_ID": "C636460166108115381222",
"FEE_1": {"set":0.35},
"FEE_3": {"set":1.39},
"USAGE": "Yzua13"
}
]
uj5u.com熱心網友回復：
bt=[
{
"TX_ID": "C592096166066790609734",
"FEE_1": 0.25,
"FEE_7": 0.95,
"USAGE": "Yzua12"
},
{
"TX_ID": "C636460166108115381222",
"FEE_1": 0.35,
"FEE_3": 1.39,
"USAGE": "Yzua13"
}
]
for item in bt:
# now song is a dictionary
for attribute, value in item.items():
if 'FEE' in attribute :
item[attribute] ={"set":value}
print(bt)
uj5u.com熱心網友回復：
從 JSON 中決議出原始字典后，您可以使用簡單的嵌套串列和字典理解來包裝所有這些FEE_值：
>>> [{k: {"set": v} if k.startswith("FEE_") else v for k, v in d.items()} for d in original]
[{'TX_ID': 'C592096166066790609734',
'FEE_1': {'set': 0.25},
'FEE_7': {'set': 0.95},
'USAGE': 'Yzua12'},
{'TX_ID': 'C636460166108115381222',
'FEE_1': {'set': 0.35},
'FEE_3': {'set': 1.39},
'USAGE': 'Yzua13'}]
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標籤：Python python-3.x