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
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。
其他回答
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
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]
定义:
def switch1(value, options):
if value in options:
options[value]()
允许您使用相当简单的语法,并将案例绑定到映射中:
def sample1(x):
local = 'betty'
switch1(x, {
'a': lambda: print("hello"),
'b': lambda: (
print("goodbye," + local),
print("!")),
})
我一直试图用一种能让我摆脱“lambda:”的方式重新定义开关,但我放弃了。调整定义:
def switch(value, *maps):
options = {}
for m in maps:
options.update(m)
if value in options:
options[value]()
elif None in options:
options[None]()
允许我将多个案例映射到同一代码,并提供默认选项:
def sample(x):
switch(x, {
_: lambda: print("other")
for _ in 'cdef'
}, {
'a': lambda: print("hello"),
'b': lambda: (
print("goodbye,"),
print("!")),
None: lambda: print("I dunno")
})
每个复制的案例都必须在自己的字典中;switch()在查找值之前合并字典。它仍然比我想象的更丑,但它的基本效率是对表达式使用散列查找,而不是循环所有键。
我喜欢Mark Bies的回答
由于x变量必须使用两次,我将lambda函数修改为无参数。
我必须运行结果[value](value)
In [2]: result = {
...: 'a': lambda x: 'A',
...: 'b': lambda x: 'B',
...: 'c': lambda x: 'C'
...: }
...: result['a']('a')
...:
Out[2]: 'A'
In [3]: result = {
...: 'a': lambda : 'A',
...: 'b': lambda : 'B',
...: 'c': lambda : 'C',
...: None: lambda : 'Nothing else matters'
...: }
...: result['a']()
...:
Out[3]: 'A'
编辑:我注意到我可以在字典中使用None类型。因此,这将模拟交换机;其他情况
还可以使用列表存储案例,并通过select调用相应的函数-
cases = ['zero()', 'one()', 'two()', 'three()']
def zero():
print "method for 0 called..."
def one():
print "method for 1 called..."
def two():
print "method for 2 called..."
def three():
print "method for 3 called..."
i = int(raw_input("Enter choice between 0-3 "))
if(i<=len(cases)):
exec(cases[i])
else:
print "wrong choice"
也在螺丝台上进行了解释。