如何在2d嵌套的python中對紙漿進行循環理解

Question

我無法克服程序中的這個難題。 我想將此重復代碼簡化為更簡單的代碼。 簡而言之，這些是紙漿的限制。

我有2種轉換模式：“ Shift_Pattern_1”和“ Shift_Pattern_Master”

員工是一個列表，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.........

當我從上面完成整個代碼時，我得到了35個約束。

我的嘗試是使用if和else縮短代碼，我只得到30個約束。 我知道問題是“如果Week1 [day] == 2”，因為缺少一些約束。

1. 我不知道將if語句放在哪里，或者

2. 有沒有更好的方法來使用更多pythonic？

   對於天中的天：如果Week1 [day] == 2：對於員工中的員工：prob + = pulp.lpSum（可用[員工，天，班次]
   Shift_pattern_1）==請求[員工] [天]否則：
   概率+ = pulp.lpSum（在Shift_pattern_Master中可用[班次[員工，天，班]] ==請求[員工] [天]

提前致謝。

Answer 1

如果唯一的一天是星期三，則要執行以下操作：

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，我認為您實際上需要上述條件，因此您只需要包括員工循環

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]