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
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。