我想用Python编写一个函数,根据输入索引的值返回不同的固定值。

在其他语言中,我会使用switch或case语句,但Python似乎没有switch语句。在此场景中,推荐的Python解决方案是什么?


当前回答

如果你有一个复杂的大小写块,你可以考虑使用函数字典查找表。。。

如果您以前没有这样做过,那么最好进入调试器并查看字典如何查找每个函数。

注意:不要在大小写/字典查找中使用“()”,否则将在创建字典/大小写块时调用每个函数。记住这一点,因为您只想使用哈希样式查找调用每个函数一次。

def first_case():
    print "first"

def second_case():
    print "second"

def third_case():
    print "third"

mycase = {
'first': first_case, #do not use ()
'second': second_case, #do not use ()
'third': third_case #do not use ()
}
myfunc = mycase['first']
myfunc()

其他回答

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

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。

def f(x):
     return 1 if x == 'a' else\
            2 if x in 'bcd' else\
            0 #default

简短易读,具有默认值,支持条件和返回值中的表达式。

然而,它的效率不如字典解决方案。例如,Python必须在返回默认值之前扫描所有条件。

我发现Python文档中的以下答案非常有用:

你可以通过一系列if…elif。。。埃利夫。。。其他的关于switch语句语法已经有了一些建议,但对于是否以及如何进行范围测试还没有达成共识。有关完整详细信息和当前状态,请参见PEP 275。

对于需要从大量可能性中进行选择的情况,可以创建一个字典,将大小写值映射到要调用的函数。例如:

def function_1(...):
    ...

functions = {'a': function_1,
             'b': function_2,
             'c': self.method_1, ...}

func = functions[value]
func()

对于在对象上调用方法,可以通过使用内置的getattr()来检索具有特定名称的方法来进一步简化:

def visit_a(self, ...):
    ...
...

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

建议为方法名称使用前缀,例如本例中的visit_。如果没有这样的前缀,如果值来自不受信任的源,攻击者将能够调用对象上的任何方法。

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.