此代碼獲取數據并將數據制成一個循環并運行直到循環完成。所以我需要在每個過程完成后將數據附加到存儲數據的 df代碼 ：a = "SELECT id FROM USER WHERE time >'2018-03-01'"dataa = pd.read_sql_query(a, con=engine)print(dataa)for userid in dataa: x=f"SELECT idbody FROM col1 WHERE user_id='{userid}'" data = pd.read_sql_query(x,con = engine)所以這里的數據被處理并且每次產生的數據都不同需要將數據附加到存儲所有被處理數據的df
3 回答
慕無忌1623718
TA貢獻1744條經驗 獲得超4個贊
在循環中或通過列表理解將值附加到list并且僅使用一次concat：
a = "SELECT id FROM USER WHERE time >'2018-03-01'"
dataa = pd.read_sql_query(a, con=engine)
dfs = []
for userid in dataa:
x=f"SELECT idbody FROM col1 WHERE user_id='{userid}'"
data = pd.read_sql_query(x,con = engine)
dfs.append(data)
df = pd.concat(dfs, ignore_index=True)
dfs = [pd.read_sql_query(f"SELECT idbody FROM col1 WHERE user_id='{userid}'",con = engine)
for userid in dataa]
df = pd.concat(dfs, ignore_index=True)
慕村9548890
TA貢獻1884條經驗 獲得超4個贊
您還可以使用concat：
a = "SELECT id FROM USER WHERE time >'2018-03-01'"
dataa = pd.read_sql_query(a, con=engine)
print(dataa)
df = pd.DataFrame()
for userid in dataa:
x=f"SELECT idbody FROM col1 WHERE user_id='{userid}'"
data = pd.read_sql_query(x,con = engine)
df = pd.concat([df_all, data])
現在：
print(df)
將是所需的輸出。
慕神8447489
TA貢獻1780條經驗 獲得超1個贊
我假設您獲得相同數量的列，并且這些列具有相同的名稱。例如這是基本思想：
df = pd.DataFrame() # this will hold your all data
df1 = pd.DataFrame([(1, 2, 3)], columns=['a', 'b', 'c']) # 1st iteration data
df2 = pd.DataFrame([(11, 22, 33)], columns=['a', 'b', 'c']) # 2nd iteration data
df3 = pd.DataFrame([(111, 222, 333)], columns=['a', 'b', 'c']) # 3rd iteratin data etc.
for data in [df1, df2, df3]:
df = df.append(df1)
a b c
0 1 2 3
1 11 22 33
2 111 222 333
你需要做的是：
a = "SELECT id FROM USER WHERE time >'2018-03-01'"
dataa = pd.read_sql_query(a, con=engine)
print(dataa)
df_all = pd.DataFrame() # create an empty df to store all returns
for userid in dataa:
x=f"SELECT idbody FROM col1 WHERE user_id='{userid}'"
data = pd.read_sql_query(x,con = engine)
df_all = df_all.append(data) # update df with new dframes
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