Question

任何有关修复下面复杂arg的指针＆＃34;小时＆＃34;，我正在

SyntaxError：语法无效

我想要的只是减少尽可能多的代码。我希望它在最后给出一个对应于i值的值。

我的代码在

失败了

 (('($200 Hr) estimated at' if i ==1 )   or ('($130 Hr) estimated at' if i == 2 )   or ('($120 Hr) estimated at' if i == 3 )  or ('($125 Hr) estimated at' if i == 4)  or ('($90 Hr) estimated at'  if i == 5))

代码：

brs_items_list.append({                     
'desc' : kwargs["brs_d_{}".format(i)],
'hours' : kwargs["brs_hrs_{}".format(i)]+(' hour' if float(kwargs["brs_hrs_{}".format(i)])<2 else ' hours')+ (('($200 Hr) estimated at' if i ==1 )   or ('($130 Hr) estimated at' if i == 2 )   or ('($120 Hr) estimated at' if i == 3 )  or ('($125 Hr) estimated at' if i == 4)  or ('($90 Hr) estimated at'  if i == 5))   ,    
'cost' :  "${:,}".format(round(float(kwargs["brs_c_{}".format(i)]), 2))
})

Answer 1

为什么不使用字典？

{1:'($200 Hr) estimated at', 2:'($130 Hr) estimated at', 3:'($120 Hr) estimated at', 4:'($125 Hr) estimated at', 5:'($90 Hr) estimated at'}[i]

更好的是，你有很多重复的文字，所以你可以这样做：

'($%d Hr) estimated at' % {1:200, 2:130, 3:120, 4:125, 5:90}[i]

Answer 2

我认为我需要

brs_items_list.append({

'desc' : kwargs["brs_d_{}".format(i)],

'hours' : kwargs["brs_hrs_{}".format(i)]+(' hour' if float(kwargs["brs_hrs_{}".format(i)])<2 else ' hours')+ (('($200 Hr) estimated at' if int(i) ==1  else '')   or ('($130 Hr) estimated at' if int(i) == 2   else '')   or ('($120 Hr) estimated at' if int(i)== 3   else '' )  or ('($125 Hr) estimated at' if int(i) == 4  else '')  or ('($90 Hr) estimated at'  if int(i) == 5  else ''))   ,    

'cost' :  "${:,}".format(round(float(kwargs["brs_c_{}".format(i)]), 2))

})

简化if语句条件：＆＃39; result1＆＃39;如果a == b其他＆＃39;结果2＆＃39;要么

2 个答案: