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
我最喜欢的是一个非常好的食谱。这是我见过的最接近实际switchcase语句的语句,尤其是在特性中。
class switch(object):
def __init__(self, value):
self.value = value
self.fall = False
def __iter__(self):
"""Return the match method once, then stop"""
yield self.match
raise StopIteration
def match(self, *args):
"""Indicate whether or not to enter a case suite"""
if self.fall or not args:
return True
elif self.value in args: # changed for v1.5, see below
self.fall = True
return True
else:
return False
下面是一个示例:
# The following example is pretty much the exact use-case of a dictionary,
# but is included for its simplicity. Note that you can include statements
# in each suite.
v = 'ten'
for case in switch(v):
if case('one'):
print 1
break
if case('two'):
print 2
break
if case('ten'):
print 10
break
if case('eleven'):
print 11
break
if case(): # default, could also just omit condition or 'if True'
print "something else!"
# No need to break here, it'll stop anyway
# break is used here to look as much like the real thing as possible, but
# elif is generally just as good and more concise.
# Empty suites are considered syntax errors, so intentional fall-throughs
# should contain 'pass'
c = 'z'
for case in switch(c):
if case('a'): pass # only necessary if the rest of the suite is empty
if case('b'): pass
# ...
if case('y'): pass
if case('z'):
print "c is lowercase!"
break
if case('A'): pass
# ...
if case('Z'):
print "c is uppercase!"
break
if case(): # default
print "I dunno what c was!"
# As suggested by Pierre Quentel, you can even expand upon the
# functionality of the classic 'case' statement by matching multiple
# cases in a single shot. This greatly benefits operations such as the
# uppercase/lowercase example above:
import string
c = 'A'
for case in switch(c):
if case(*string.lowercase): # note the * for unpacking as arguments
print "c is lowercase!"
break
if case(*string.uppercase):
print "c is uppercase!"
break
if case('!', '?', '.'): # normal argument passing style also applies
print "c is a sentence terminator!"
break
if case(): # default
print "I dunno what c was!"
一些评论指出,使用foo作为case而不是foo中的case的上下文管理器解决方案可能更干净,对于大型switch语句,线性而不是二次行为可能是一个不错的选择。使用for循环的答案的一部分价值是有中断和突破的能力,如果我们愿意稍微使用我们选择的关键字,我们也可以在上下文管理器中实现这一点:
class Switch:
def __init__(self, value):
self.value = value
self._entered = False
self._broken = False
self._prev = None
def __enter__(self):
return self
def __exit__(self, type, value, traceback):
return False # Allows a traceback to occur
def __call__(self, *values):
if self._broken:
return False
if not self._entered:
if values and self.value not in values:
return False
self._entered, self._prev = True, values
return True
if self._prev is None:
self._prev = values
return True
if self._prev != values:
self._broken = True
return False
if self._prev == values:
self._prev = None
return False
@property
def default(self):
return self()
下面是一个示例:
# Prints 'bar' then 'baz'.
with Switch(2) as case:
while case(0):
print('foo')
while case(1, 2, 3):
print('bar')
while case(4, 5):
print('baz')
break
while case.default:
print('default')
break
其他回答
到目前为止,已经有很多回答说,“我们在Python中没有开关,这样做吧”。然而,我想指出,switch语句本身是一个容易被滥用的构造,在大多数情况下可以而且应该避免,因为它们促进了惰性编程。案例说明:
def ToUpper(lcChar):
if (lcChar == 'a' or lcChar == 'A'):
return 'A'
elif (lcChar == 'b' or lcChar == 'B'):
return 'B'
...
elif (lcChar == 'z' or lcChar == 'Z'):
return 'Z'
else:
return None # or something
现在,您可以使用switch语句(如果Python提供了switch语句)来执行此操作,但这会浪费您的时间,因为有些方法可以很好地执行此操作。或者,你有一些不太明显的东西:
def ConvertToReason(code):
if (code == 200):
return 'Okay'
elif (code == 400):
return 'Bad Request'
elif (code == 404):
return 'Not Found'
else:
return None
然而,这种操作可以而且应该用字典来处理,因为它会更快、更复杂、更不容易出错和更紧凑。
switch语句的绝大多数“用例”将属于这两种情况之一;如果你彻底考虑了你的问题,就没有什么理由使用它。
因此,与其问“我如何在Python中切换?”,或许我们应该问“我为什么要在Python中进行切换?”因为这往往是一个更有趣的问题,而且往往会暴露出您正在构建的任何设计中的缺陷。
现在,这并不是说也不应该使用开关。状态机、词法分析器、解析器和自动机都在某种程度上使用它们,一般来说,当你从对称输入开始到非对称输出时,它们会很有用;您只需要确保不要将开关用作锤子,因为您在代码中看到了一堆钉子。
我使用的解决方案:
这里发布的两个解决方案的组合,相对容易阅读,并支持默认值。
result = {
'a': lambda x: x * 5,
'b': lambda x: x + 7,
'c': lambda x: x - 2
}.get(whatToUse, lambda x: x - 22)(value)
哪里
.get('c', lambda x: x - 22)(23)
在dict中查找“lambda x:x-2”,并在x=23时使用它
.get('xxx', lambda x: x - 22)(44)
在dict中找不到它,使用默认的“lambda x:x-22”,x=44。
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')
""")
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.
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)