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

作为Mark Biek答案的一个小变化,对于像这样的不常见情况,用户有一堆函数调用要延迟,而参数要打包(而且不值得构建一堆不符合逻辑的函数),而不是这样:

d = {
    "a1": lambda: a(1),
    "a2": lambda: a(2),
    "b": lambda: b("foo"),
    "c": lambda: c(),
    "z": lambda: z("bar", 25),
    }
return d[string]()

…您可以这样做:

d = {
    "a1": (a, 1),
    "a2": (a, 2),
    "b": (b, "foo"),
    "c": (c,)
    "z": (z, "bar", 25),
    }
func, *args = d[string]
return func(*args)

这当然更短,但它是否更可读是一个悬而未决的问题…


我认为从lambda转换为partial可能更容易理解(虽然不是更简单):

d = {
    "a1": partial(a, 1),
    "a2": partial(a, 2),
    "b": partial(b, "foo"),
    "c": c,
    "z": partial(z, "bar", 25),
    }
return d[string]()

…它的优点是可以很好地处理关键字参数:

d = {
    "a1": partial(a, 1),
    "a2": partial(a, 2),
    "b": partial(b, "foo"),
    "c": c,
    "k": partial(k, key=int),
    "z": partial(z, "bar", 25),
    }
return d[string]()

其他回答

class Switch:
    def __init__(self, value):
        self.value = value

    def __enter__(self):
        return self

    def __exit__(self, type, value, traceback):
        return False # Allows a traceback to occur

    def __call__(self, *values):
        return self.value in values


from datetime import datetime

with Switch(datetime.today().weekday()) as case:
    if case(0):
        # Basic usage of switch
        print("I hate mondays so much.")
        # Note there is no break needed here
    elif case(1,2):
        # This switch also supports multiple conditions (in one line)
        print("When is the weekend going to be here?")
    elif case(3,4):
        print("The weekend is near.")
    else:
        # Default would occur here
        print("Let's go have fun!") # Didn't use case for example purposes

switch语句只是if/elif/else的语法糖。任何控制语句所做的都是基于某个条件(即决策路径)来授权作业。为了将其包装到模块中并能够基于其唯一id调用作业,可以使用继承和Python中的任何方法都是虚拟的这一事实来提供派生类特定的作业实现,作为特定的“case”处理程序:

#!/usr/bin/python

import sys

class Case(object):
    """
        Base class which specifies the interface for the "case" handler.
        The all required arbitrary arguments inside "execute" method will be
        provided through the derived class
        specific constructor

        @note in Python, all class methods are virtual
    """
    def __init__(self, id):
        self.id = id

    def pair(self):
        """
            Pairs the given id of the "case" with
            the instance on which "execute" will be called
        """
        return (self.id, self)

    def execute(self): # Base class virtual method that needs to be overridden
        pass

class Case1(Case):
    def __init__(self, id, msg):
        self.id = id
        self.msg = msg
    def execute(self): # Override the base class method
        print("<Case1> id={}, message: \"{}\"".format(str(self.id), self.msg))

class Case2(Case):
    def __init__(self, id, n):
        self.id = id
        self.n = n
    def execute(self): # Override the base class method
        print("<Case2> id={}, n={}.".format(str(self.id), str(self.n)))
        print("\n".join(map(str, range(self.n))))


class Switch(object):
    """
        The class which delegates the jobs
        based on the given job id
    """
    def __init__(self, cases):
        self.cases = cases # dictionary: time complexity for the access operation is 1
    def resolve(self, id):

        try:
            cases[id].execute()
        except KeyError as e:
            print("Given id: {} is wrong!".format(str(id)))



if __name__ == '__main__':

    # Cases
    cases=dict([Case1(0, "switch").pair(), Case2(1, 5).pair()])

    switch = Switch(cases)

    # id will be dynamically specified
    switch.resolve(0)
    switch.resolve(1)
    switch.resolve(2)

如果您不担心丢失案例套件中的语法高亮显示,可以执行以下操作:

exec {
    1: """
print ('one')
""", 
    2: """
print ('two')
""", 
    3: """
print ('three')
""",
}.get(value, """
print ('None')
""")

其中value是值。在C中,这将是:

switch (value) {
    case 1:
        printf("one");
        break;
    case 2:
        printf("two");
        break;
    case 3:
        printf("three");
        break;
    default:
        printf("None");
        break;
}

我们还可以创建一个helper函数来实现这一点:

def switch(value, cases, default):
    exec cases.get(value, default)

因此,我们可以将其用于一、二和三的示例:

switch(value, {
    1: """
print ('one')
    """, 
    2: """
print ('two')
    """, 
    3: """
print ('three')
    """,
}, """
print ('None')
""")

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。

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。