虽然已经有了足够的答案，但我想指出一个更简单、更强大的解决方案：

class Switch:
    def __init__(self, switches):
        self.switches = switches
        self.between = len(switches[0]) == 3

    def __call__(self, x):
        for line in self.switches:
            if self.between:
                if line[0] <= x < line[1]:
                    return line[2]
            else:
                if line[0] == x:
                    return line[1]
        return None


if __name__ == '__main__':
    between_table = [
        (1, 4, 'between 1 and 4'),
        (4, 8, 'between 4 and 8')
    ]

    switch_between = Switch(between_table)

    print('Switch Between:')
    for i in range(0, 10):
        if switch_between(i):
            print('{} is {}'.format(i, switch_between(i)))
        else:
            print('No match for {}'.format(i))


    equals_table = [
        (1, 'One'),
        (2, 'Two'),
        (4, 'Four'),
        (5, 'Five'),
        (7, 'Seven'),
        (8, 'Eight')
    ]
    print('Switch Equals:')
    switch_equals = Switch(equals_table)
    for i in range(0, 10):
        if switch_equals(i):
            print('{} is {}'.format(i, switch_equals(i)))
        else:
            print('No match for {}'.format(i))

输出：

Switch Between:
No match for 0
1 is between 1 and 4
2 is between 1 and 4
3 is between 1 and 4
4 is between 4 and 8
5 is between 4 and 8
6 is between 4 and 8
7 is between 4 and 8
No match for 8
No match for 9

Switch Equals:
No match for 0
1 is One
2 is Two
No match for 3
4 is Four
5 is Five
No match for 6
7 is Seven
8 is Eight
No match for 9

我倾向于使用字典的解决方案是：

def decision_time( key, *args, **kwargs):
    def action1()
        """This function is a closure - and has access to all the arguments"""
        pass
    def action2()
        """This function is a closure - and has access to all the arguments"""
        pass
    def action3()
        """This function is a closure - and has access to all the arguments"""
        pass

   return {1:action1, 2:action2, 3:action3}.get(key,default)()

这样做的优点是它不需要每次都对函数求值，您只需确保外部函数获得内部函数所需的所有信息。

我使用的解决方案：

这里发布的两个解决方案的组合，相对容易阅读，并支持默认值。

result = {
  'a': lambda x: x * 5,
  'b': lambda x: x + 7,
  'c': lambda x: x - 2
}.get(whatToUse, lambda x: x - 22)(value)

哪里

.get('c', lambda x: x - 22)(23)

在dict中查找“lambda x:x-2”，并在x=23时使用它

.get('xxx', lambda x: x - 22)(44)

在dict中找不到它，使用默认的“lambda x:x-22”，x=44。

仅仅将一些键映射到一些代码并不是一个真正的问题，正如大多数人在使用字典时所展示的那样。真正的诀窍是尝试模仿整个直通和中断过程。我认为我从来没有写过一个案例陈述，其中我使用了“功能”。这里有一个直通车。

def case(list): reduce(lambda b, f: (b | f[0], {False:(lambda:None),True:f[1]}[b | f[0]]())[0], list, False)

case([
    (False, lambda:print(5)),
    (True, lambda:print(4))
])

我真的把它想象成一个单独的陈述。我希望你能原谅这种愚蠢的格式。

reduce(
    initializer=False,
    function=(lambda b, f:
        ( b | f[0]
        , { False: (lambda:None)
          , True : f[1]
          }[b | f[0]]()
        )[0]
    ),
    iterable=[
        (False, lambda:print(5)),
        (True, lambda:print(4))
    ]
)

我希望这是有效的Python代码。它应该能让你通过。当然，布尔检查可以是表达式，如果您希望它们被延迟求值，那么可以将它们全部封装在lambda中。在执行了列表中的一些项目之后，也不难让它被接受。只需创建元组（bool，bool，function），其中第二个bool指示是否突破或放弃。

虽然已经有了足够的答案，但我想指出一个更简单、更强大的解决方案：

class Switch:
    def __init__(self, switches):
        self.switches = switches
        self.between = len(switches[0]) == 3

    def __call__(self, x):
        for line in self.switches:
            if self.between:
                if line[0] <= x < line[1]:
                    return line[2]
            else:
                if line[0] == x:
                    return line[1]
        return None


if __name__ == '__main__':
    between_table = [
        (1, 4, 'between 1 and 4'),
        (4, 8, 'between 4 and 8')
    ]

    switch_between = Switch(between_table)

    print('Switch Between:')
    for i in range(0, 10):
        if switch_between(i):
            print('{} is {}'.format(i, switch_between(i)))
        else:
            print('No match for {}'.format(i))


    equals_table = [
        (1, 'One'),
        (2, 'Two'),
        (4, 'Four'),
        (5, 'Five'),
        (7, 'Seven'),
        (8, 'Eight')
    ]
    print('Switch Equals:')
    switch_equals = Switch(equals_table)
    for i in range(0, 10):
        if switch_equals(i):
            print('{} is {}'.format(i, switch_equals(i)))
        else:
            print('No match for {}'.format(i))

输出：

Switch Between:
No match for 0
1 is between 1 and 4
2 is between 1 and 4
3 is between 1 and 4
4 is between 4 and 8
5 is between 4 and 8
6 is between 4 and 8
7 is between 4 and 8
No match for 8
No match for 9

Switch Equals:
No match for 0
1 is One
2 is Two
No match for 3
4 is Four
5 is Five
No match for 6
7 is Seven
8 is Eight
No match for 9

扩展“dict as switch”思想。如果要为交换机使用默认值：

def f(x):
    try:
        return {
            'a': 1,
            'b': 2,
        }[x]
    except KeyError:
        return 'default'