Question

下面的代码片段可以正常工作，并且可以产生我想要的东西，但是我需要一些关于如何使Pythonic更加线条化的指针

if avail[employee, day, "Morning"].varValue==0 and    
    avail[employee, day, "Mid"].varValue==0 and 
    avail[employee, day, "Night"].varValue==0:

完整代码

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

    for employee in Employees:
        for day in Days:
            if avail[employee, day, "Morning"].varValue==0 and    
                avail[employee, day, "Mid"].varValue==0 and 
                avail[employee, day, "Night"].varValue==0:
                    print (f"{employee} on {day} is off.")
            else:
                for shift in Shift_pattern_Master:
                    if avail[employee, day, shift].varValue==1:
                        print (f"{employee} on {day} works in {shift}.")

所以我尝试使if avail[employee, day, shift].varValue==0 for shift in Shift_pattern_Master:成为通用条件，并且一直说for是无效语法。

我想我缺少了一些东西，但是我不知道是什么。感谢您的任何提前帮助。

Answer 1

怎么样：

if all(avail[employee, day, time].varValue==0 for time in ["Morning", "Mid", "Night"]):

另一种选择是重新包装条件：

if (
    avail[employee, day, "Morning"].varValue==0
    and avail[employee, day, "Mid"].varValue==0
    and avail[employee, day, "Night"].varValue==0
):

使用2d嵌套字典中的if / else语句进行列表理解循环

1 个答案: