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?
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
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
其他回答
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)
您可以使用分派的dict:
#!/usr/bin/env python
def case1():
print("This is case 1")
def case2():
print("This is case 2")
def case3():
print("This is case 3")
token_dict = {
"case1" : case1,
"case2" : case2,
"case3" : case3,
}
def main():
cases = ("case1", "case3", "case2", "case1")
for case in cases:
token_dict[case]()
if __name__ == '__main__':
main()
输出:
This is case 1
This is case 3
This is case 2
This is case 1
我要把我的两分钱放在这里。Python中没有case/switch语句的原因是因为Python遵循“只有一种正确的方法”的原则。很明显,您可以想出各种方法来重新创建switch/case功能,但实现这一点的Python方法是if/elf构造。即。,
if something:
return "first thing"
elif somethingelse:
return "second thing"
elif yetanotherthing:
return "third thing"
else:
return "default thing"
我只是觉得PEP 8应该在这里获得认可。Python的一个优点是它的简单和优雅。这在很大程度上源于PEP8中提出的原则,包括“只有一种正确的方法可以做某事。”
我发现一种常见的交换机结构:
switch ...parameter...
case p1: v1; break;
case p2: v2; break;
default: v3;
可以用Python表示如下:
(lambda x: v1 if p1(x) else v2 if p2(x) else v3)
或以更清晰的方式格式化:
(lambda x:
v1 if p1(x) else
v2 if p2(x) else
v3)
Python版本不是一个语句,而是一个表达式,其计算结果为一个值。
Python 3.10(2021)引入了match-case语句,该语句提供了Pythons“switch”的一流实现。例如:
def f(x):
match x:
case 'a':
return 1
case 'b':
return 2
case _:
return 0 # 0 is the default case if x is not found
match-case语句比这个简单的示例强大得多。
以下原始答案写于2008年,当时还未提供匹配案例:
你可以用字典:
def f(x):
return {
'a': 1,
'b': 2,
}[x]