获取下一个枚举常量/属性

Question

假设我有一个枚举器，是否可能获取其后面的属性？如果我有 today=Days.Sunday ，那么我能否做类似于 tomorrow=today.next() 这样的操作吗？

例如：

class Days(Enum):
     Sunday = 'S'
     Monday = 'M'
     ...
     Saturday = 'Sa'

我知道我可以使用元组（例如下面的代码）来执行类似于tomorrow=today[1]的操作，但我希望有一些内置的或更加优雅的方法。

class Days(Enum):
     Sunday = ('S','Monday')
     Monday = ('M','Tuesday')
     ...
     Saturday = ('Sa','Sunday')

- Parker

---

创建某种类型的迭代器？ - Padraic Cunningham

谢谢@PadraicCunningham，这比我现在的元组解决方案更有效吗？我喜欢你的方法，只是好奇它是否可以节省内存或运行时间 :) - Parker

Answer 1

当然可以。只需将所需的功能添加到您的 Days 类中：

class Days(Enum):

    Sunday = 'S'
    Monday = 'M'
    Tuesday = 'T'
    Wednesday = 'W'
    Thursday = 'Th'
    Friday = 'F'
    Saturday = 'Sa'

    def next(self):
        cls = self.__class__
        members = list(cls)
        index = members.index(self) + 1
        if index >= len(members):
            index = 0
        return members[index]

并且在使用中：

today = Days.Wednesday
print(today.next())
# Days.Thursday

---

虽然上面的方法可能更容易理解，但是也可以在__init__中通过给每个成员添加next属性（同时添加previous）来完成这项工作。

class Days(Enum):
    #
    Sunday = 'S'
    Monday = 'M'
    Tuesday = 'T'
    Wednesday = 'W'
    Thursday = 'Th'
    Friday = 'F'
    Saturday = 'Sa'
    #
    def __init__(self, value):
        if len(self.__class__):
            # make links
            all = list(self.__class__)
            first, previous = all[0], all[-1]
            previous.next = self
            self.previous = previous
            self.next = first

并且正在使用：

>>> Days.Tuesday.next
<Days.Wednesday: 'W'>

>>> Days.Tuesday.previous
<Days.Monday: 'M'>

>>> Days.Saturday.next
<Days.Sunday: 'S'>

>>> Days.Saturday.previous
<Days.Friday: 'F'>

NB 使用这种属性的方法意味着我们不再需要在next/previous后面加上()。

- Ethan Furman

Answer 2

您可以创建一个字典来查找下一天，如下所示：

您可以创建一个字典来查找下一天，如下所示：

In [10]: class Days(Enum):
    Sun = 'Su'
    Mon = 'M'
    Tue = 'Tu'
    Wed = 'W'
    Thu = 'Th'
    Fri = 'F'
    Sat = 'Sa'

In [11]: days = list(Days)

In [12]: nxt = dict((day, days[(i+1) % len(days)]) for i, day in enumerate(days))

快速测试：

In [13]: nxt[Days.Tue]
Out[13]: <Days.Wed: 'W'>

In [14]: nxt[Days.Sat]
Out[14]: <Days.Sun: 'Su'>

- Bas Swinckels

Answer 3

对我来说，这似乎是最优雅的解决方案，而无需额外的函数

day = Days.Sunday

day = Days((day.value + 1) % len(Days) + 1) # next day cycled

- vozman

Answer 4

#ENUM CLASS

#colors
import enum

class Color(enum.Enum):
    turquoise = 1
    indigo = 2
    magenta = 3
    cyan = 4
    teal = 5
    azure = 6
    rose = 7
    amber = 8
    vermillon = 9
    plum = 10
    russet = 11
    slate = 12

    def __iter__(self):
        self.idx_current = self.value
        return self

    def __next__(self):
        if (self.idx_current > 12):
            return None

        self.idx_current = self.idx_current + 1
        return Color (self.idx_current - 1)

#CLIENT CODE
    #iterate colors starting from cyan
    it = iter (Color.cyan)
    while True:
        #print (v.get_id())
        c = next (it)
        if c is None:
            break
        print(c)

#OUTPUT
Color.cyan
Color.teal
Color.azure
Color.rose
Color.amber
Color.vermillon
Color.plum
Color.russet
Color.slate

- jdhall

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更合理的做法是在__next__()中引发StopIteration，而不是返回None。 - ApproachingDarknessFish