If 语句简单

Question

有没有办法使这个 ELIF 语句更简单，顺便说一句，这段代码不起作用。改变的变量是 APPOINTMENT

elif (rut == '80010900-0' and agental_launch != "" and 'TAL'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'IQQ'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'ANF'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'MJS'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'QTV'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'SVE'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'PMC'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'CHB'  in appointment):
        df.at[idx,'REBATE'] = round(int(monto_neto)*0.35)

Answer 1

使用 any 作为最后一个条件，成为单行：

elif rut == '80010900-0' and agental_launch != "" and any(x in appointment for x in ['TAL','IQQ','rest of strings to match...']):

Answer 2

你可以这样做：

elif rut == '80010900-0' and agental_launch != "" and any(elem in appointment for elem in ['TAL','IQQ','ANF','MJS','QTV','SVE','PMC','CHB']):
     df.at[idx,'REBATE'] = round(int(monto_neto)*0.35)

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