还可以使用列表存储案例，并通过select调用相应的函数-

cases = ['zero()', 'one()', 'two()', 'three()']

def zero():
  print "method for 0 called..."
def one():
  print "method for 1 called..."
def two():
  print "method for 2 called..."
def three():
  print "method for 3 called..."

i = int(raw_input("Enter choice between 0-3 "))

if(i<=len(cases)):
  exec(cases[i])
else:
  print "wrong choice"

也在螺丝台上进行了解释。

这里的大多数答案都很陈旧，尤其是那些被接受的答案，因此似乎值得更新。

首先，官方的Python常见问题解答涵盖了这一点，并为简单案例推荐elif链，为更大或更复杂的案例推荐dict。它还建议在某些情况下使用一组visit_方法（许多服务器框架使用的样式）：

def dispatch(self, value):
    method_name = 'visit_' + str(value)
    method = getattr(self, method_name)
    method()

FAQ还提到了PEP275，它是为了让官方一劳永逸地决定添加C风格切换语句而编写的。但PEP实际上被推迟到了Python 3，它只是作为一个单独的提案PEP3103被正式拒绝。答案当然是否定的，但如果你对原因或历史感兴趣的话，这两位政治公众人物可以获得更多信息。

有一件事多次出现（在PEP 275中可以看到，尽管它是作为实际推荐删除的），那就是如果你真的为处理4种情况而烦恼的是8行代码，而不是C或Bash中的6行代码，你总是可以这样写：

if x == 1: print('first')
elif x == 2: print('second')
elif x == 3: print('third')
else: print('did not place')

这并不是PEP 8所鼓励的，但它是可读的，并不是太单一。

自PEP 3103被拒绝以来的十多年里，C风格的案例陈述，甚至围棋中稍微更强大的版本，都被认为已经过时；每当有人提出python想法或-dev时，他们都会参考旧的决定。

然而，完全ML样式的模式匹配的想法每隔几年就会出现一次，特别是在Swift和Rust等语言采用它之后。问题是，如果没有代数数据类型，很难充分利用模式匹配。虽然圭多一直赞同这个想法，但没有人提出一个非常适合Python的方案。（你可以阅读我2014年的strawman作为一个例子。）这可能会随着3.7中的dataclass和一些零星的建议而改变，比如使用更强大的枚举来处理sum类型，或者使用不同类型的语句本地绑定的各种建议（如PEP3150，或者当前正在讨论的一组建议-ideas）。但到目前为止，它还没有。

偶尔也会有关于Perl 6样式匹配的建议，这基本上是从elif到regex到单分派类型切换的混合。

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

我在谷歌搜索上找不到简单的答案。但我还是想通了。这真的很简单。决定把它贴出来，也许可以防止别人的头上少刮几下。关键是简单的“in”和元组。下面是带有直通的switch语句行为，包括RANDOM直通。

l = ['Dog', 'Cat', 'Bird', 'Bigfoot',
     'Dragonfly', 'Snake', 'Bat', 'Loch Ness Monster']

for x in l:
    if x in ('Dog', 'Cat'):
        x += " has four legs"
    elif x in ('Bat', 'Bird', 'Dragonfly'):
        x += " has wings."
    elif x in ('Snake',):
        x += " has a forked tongue."
    else:
        x += " is a big mystery by default."
    print(x)

print()

for x in range(10):
    if x in (0, 1):
        x = "Values 0 and 1 caught here."
    elif x in (2,):
        x = "Value 2 caught here."
    elif x in (3, 7, 8):
        x = "Values 3, 7, 8 caught here."
    elif x in (4, 6):
        x = "Values 4 and 6 caught here"
    else:
        x = "Values 5 and 9 caught in default."
    print(x)

提供：

Dog has four legs
Cat has four legs
Bird has wings.
Bigfoot is a big mystery by default.
Dragonfly has wings.
Snake has a forked tongue.
Bat has wings.
Loch Ness Monster is a big mystery by default.

Values 0 and 1 caught here.
Values 0 and 1 caught here.
Value 2 caught here.
Values 3, 7, 8 caught here.
Values 4 and 6 caught here
Values 5 and 9 caught in default.
Values 4 and 6 caught here
Values 3, 7, 8 caught here.
Values 3, 7, 8 caught here.
Values 5 and 9 caught in default.

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。

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'