如果您想要默认值，可以使用dictionary get（key[，default]）函数：

def f(x):
    return {
        'a': 1,
        'b': 2
    }.get(x, 9)    # 9 will be returned default if x is not found

class switch(object):
    value = None
    def __new__(class_, value):
        class_.value = value
        return True

def case(*args):
    return any((arg == switch.value for arg in args))

用法：

while switch(n):
    if case(0):
        print "You typed zero."
        break
    if case(1, 4, 9):
        print "n is a perfect square."
        break
    if case(2):
        print "n is an even number."
    if case(2, 3, 5, 7):
        print "n is a prime number."
        break
    if case(6, 8):
        print "n is an even number."
        break
    print "Only single-digit numbers are allowed."
    break

测验：

n = 2
#Result:
#n is an even number.
#n is a prime number.
n = 11
#Result:
#Only single-digit numbers are allowed.

在阅读了公认的答案后，我感到非常困惑，但这一切都清楚了：

def numbers_to_strings(argument):
    switcher = {
        0: "zero",
        1: "one",
        2: "two",
    }
    return switcher.get(argument, "nothing")

该代码类似于：

function(argument){
    switch(argument) {
        case 0:
            return "zero";
        case 1:
            return "one";
        case 2:
            return "two";
        default:
            return "nothing";
    }
}

有关字典映射到函数的详细信息，请查看源代码。

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

简单，未经测试；每个条件都是独立计算的：没有贯穿，但所有情况都会计算（尽管要打开的表达式只计算一次），除非有break语句。例如

for case in [expression]:
    if case == 1:
        print(end='Was 1. ')

    if case == 2:
        print(end='Was 2. ')
        break

    if case in (1, 2):
        print(end='Was 1 or 2. ')

    print(end='Was something. ')

指纹是1。是1或2。是什么。（该死！为什么在内联代码块中不能有尾随空格？）若表达式的计算结果为1，则为2。如果表达式的计算结果为2或Was某物。if表达式的计算结果为其他值。

如果您真的只是返回一个预定的固定值，那么可以创建一个字典，其中包含所有可能的输入索引作为键，以及它们的对应值。此外，您可能真的不希望函数执行此操作，除非您以某种方式计算返回值。

哦，如果你想做一些类似开关的事情，请看这里。