我有一系列值IDs轉換為dfA輸出為的資料幀：
| 身份證 | |
|---|---|
| 0 | ID_3456789065 |
| 1 | ID_4546567657 |
| 2 | ID_1765768878 |
我正在嘗試轉換dfA['IDs']為輸出為的 URL 字串串列：
'https://api.names.io/v1/ids/ID_3456789065/IDAccounts'
'https://api.names.io/v1/ids/ID_4546567657/IDAccounts'
'https://api.names.io/v1/ids/ID_1765768878/IDAccounts'
如果我手動輸入一個 ID，這會得到我想要的：
acctID = 'ID_3456789065'
f'https://api.names.io/v1/ids/{acctID}/IDAccounts'
輸出：
'https://api.names.io/v1/ids/ID_3456789065/IDAccounts'
dfA當我嘗試使用不同版本的回圈時，我不斷得到錯誤的輸出或錯誤：
urlList=[]
for i in dfA:
acctID = dfA[i]
urlList = f'https://api.names.io/v1/ids/{acctID}/IDAccounts'
urlList.append(urlList)
輸出：
https://api.names.io/v1/ids/0 ID_3456789065
1 ID_4546567657
2 ID_1765768878
Name: data_bettor, dtype: object/IDAccounts
我也嘗試過 .concat 并且拋出TypeError: cannot concatenate object of type '<class 'str'>'; only Series and DataFrame objs are valid
請幫忙！
uj5u.com熱心網友回復：
您可以使用：
urlList = list('https://api.names.io/v1/ids/' df['IDs'] '/IDAccounts')
print(urlList)
# Output
['https://api.names.io/v1/ids/ID_3456789065/IDAccounts',
'https://api.names.io/v1/ids/ID_4546567657/IDAccounts',
'https://api.names.io/v1/ids/ID_1765768878/IDAccounts']
uj5u.com熱心網友回復：
您可以使用迭代“ID”列的串列推導：
out = [f'https://api.names.io/v1/ids/{acctID}/IDAccounts' for acctID in df['IDs']]
輸出：
['https://api.names.io/v1/ids/ID_3456789065/IDAccounts',
'https://api.names.io/v1/ids/ID_4546567657/IDAccounts',
'https://api.names.io/v1/ids/ID_1765768878/IDAccounts']
uj5u.com熱心網友回復：
您可以apply()在IDs列上使用并用于to_list從系列中獲取串列：
df["IDs"].apply(lambda acctID: f'https://api.names.io/v1/ids/{acctID}/IDAccounts').to_list()
這輸出：
['https://api.names.io/v1/ids/ID_3456789065/IDAccounts',
'https://api.names.io/v1/ids/ID_4546567657/IDAccounts',
'https://api.names.io/v1/ids/ID_1765768878/IDAccounts']
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