在这里,我试图将一个表与具有数百万个记录的2个表进行比较,所以我想分别比较dem并在主表“ tabl”名称状态中创建一个单独的列,以便我可以自己更新
步骤:在主表中的电子邮件名称“ status”旁边创建一个新列
步骤2:必须在与tab1,tab2表进行比较时更新该列
list_of_tables=['tab1','tab2']
for tab in list_of_tables:
cursor.execute("select main.*,if({}.email is not null ,'MATCH','NONMATCH') stataus from main left join {} on main.email={}.email".format(tab,tab,tab))
data_2 = cursor.fetchall()
print data_2
data3=list(data_2)
data_3=pd.DataFrame(data3)
upload(ftp,data_3,FILEPATH)
def upload(ftp,data_3,FILEPATH):
data_4=data_3.to_csv(Out_file,index=False,header=None)
main:
abc@gamil.com
xyz@email.com
ijk@gmail.com
ghi@gmail.com
pqr@gmail.com
yup@gmail.com
tab1:
ijk@gmail.com
yup@gmail.com
tab2:
xyz@email.com
pqr@gmail.com
要求的结果
email valid
abc@gamil.com non-match
xyz@email.com match
ijk@gmail.com match
ghi@gmail.com non-match
pqr@gmail.com match
yup@gmail.com match
but getting like dis:
abc@gamil.com non-match
xyz@email.com non-match
ijk@gmail.com match
ghi@gmail.com non-match
pqr@gmail.com non-match
yup@gmail.com match
abc@gamil.com nonmatch
xyz@email.com match
ijk@gmail.com nonmatch
ghi@gmail.com nonmatch
pqr@gmail.com nonmatch
yup@gmail.com nonmatch
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