如何在2d嵌套的python中对纸浆进行循环理解

Question

我无法克服程序中的这个难题。 我想将此重复代码简化为更简单的代码。 简而言之，这些是纸浆的限制。

我有2种转换模式：“ Shift_Pattern_1”和“ Shift_Pattern_Master”

员工是一个列表，里面有名字。

 Days:["Monday", "Tuesday", "Wednesday", "Thursday", "Friday",    
 "Saturday", "Sunday"]

Shift_pattern_Master = ["Morning", "Mid", "Night"]
Shift_pattern_1 = ["Morning", "Night"]

Week1={"Monday":2, "Tuesday":2, "Wednesday":3, "Thursday":2, "Friday":2,    
"Saturday":3, "Sunday":2} # number a people needed a to day work.

for day in Days[0:2]: # Monday and Tuesday only
    for employee in Employees:
        prob += pulp.lpSum(avail[employee, day, shift] for shift in      
Shift_pattern_1)==requests[employee][day]

for day in Days[2:3]: #wednesday
    for employee in Employees:
        prob += pulp.lpSum(avail[employee, day, shift] for shift in     
 Shift_pattern_Master)==requests[employee][day]

 ....more code to finish the week.........

当我从上面完成整个代码时，我得到了35个约束。

我的尝试是使用if和else缩短代码，我只得到30个约束。 我知道问题是“如果Week1 [day] == 2”，因为缺少一些约束。

1. 我不知道将if语句放在哪里，或者

2. 有没有更好的方法来使用更多pythonic？

   对于天中的天：如果Week1 [day] == 2：对于员工中的员工：prob + = pulp.lpSum（可用[员工，天，班次]
   Shift_pattern_1）==请求[员工] [天]否则：
   概率+ = pulp.lpSum（在Shift_pattern_Master中可用[班次[员工，天，班]] ==请求[员工] [天]

提前致谢。

Answer 1

如果唯一的一天是星期三，则要执行以下操作：

for day in Days: 
   if day=="Wednesday": 
      for employee in Employees: 
          prob += pulp.lpSum(avail[employee, day, shift] for shift in Shift_pattern_1)==requests[employee][day] 
   else:
      for employee in Employees: 
          prob += pulp.lpSum(avail[employee, day, shift] for shift in Shift_pattern_Master)==requests[employee][day]

但是，我认为您实际上需要上述条件，因此您只需要包括员工循环

for day in Days: 
   if Week1[day]==2: 
      for employee in Employees: 
          prob += pulp.lpSum(avail[employee, day, shift] for shift in Shift_pattern_1)==requests[employee][day] 
   else:
      for employee in Employees: 
          prob += pulp.lpSum(avail[employee, day, shift] for shift in Shift_pattern_Master)==requests[employee][day]