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。

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

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。

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。

# simple case alternative

some_value = 5.0

# this while loop block simulates a case block

# case
while True:

    # case 1
    if some_value > 5:
        print ('Greater than five')
        break

    # case 2
    if some_value == 5:
        print ('Equal to five')
        break

    # else case 3
    print ( 'Must be less than 5')
    break

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]()