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
Python>=3.10
哇,Python 3.10+现在有一个match/case语法,类似于switch/case等等!
PEP 634——结构模式匹配
匹配/案例的选定特征
1-匹配值:
匹配值类似于另一种语言中的简单开关/大小写:
match something:
case 1 | 2 | 3:
# Match 1-3.
case _:
# Anything else.
#
# Match will throw an error if this is omitted
# and it doesn't match any of the other patterns.
2-匹配结构模式:
match something:
case str() | bytes():
# Match a string like object.
case [str(), int()]:
# Match a `str` and an `int` sequence
# (`list` or a `tuple` but not a `set` or an iterator).
case [_, _]:
# Match a sequence of 2 variables.
# To prevent a common mistake, sequence patterns don’t match strings.
case {"bandwidth": 100, "latency": 300}:
# Match this dict. Extra keys are ignored.
3-捕获变量
解析对象;将其保存为变量:
match something:
case [name, count]
# Match a sequence of any two objects and parse them into the two variables.
case [x, y, *rest]:
# Match a sequence of two or more objects,
# binding object #3 and on into the rest variable.
case bytes() | str() as text:
# Match any string like object and save it to the text variable.
捕获变量在解析数据(如JSON或HTML)时非常有用,这些数据可能以多种不同的模式之一出现。
捕获变量是一项功能。但这也意味着您只需要使用虚线常量(例如:COLOR.RED)。否则,常量将被视为捕获变量并被覆盖。
更多示例用法:
match something:
case 0 | 1 | 2:
# Matches 0, 1 or 2 (value).
print("Small number")
case [] | [_]:
# Matches an empty or single value sequence (structure).
# Matches lists and tuples but not sets.
print("A short sequence")
case str() | bytes():
# Something of `str` or `bytes` type (data type).
print("Something string-like")
case _:
# Anything not matched by the above.
print("Something else")
Python<=3.9
我最喜欢的switch/case Python配方是:
choices = {'a': 1, 'b': 2}
result = choices.get(key, 'default')
简单场景的简短和简单。
比较11行以上的C代码:
// C Language version of a simple 'switch/case'.
switch( key )
{
case 'a' :
result = 1;
break;
case 'b' :
result = 2;
break;
default :
result = -1;
}
您甚至可以通过使用元组来分配多个变量:
choices = {'a': (1, 2, 3), 'b': (4, 5, 6)}
(result1, result2, result3) = choices.get(key, ('default1', 'default2', 'default3'))
我一直喜欢这样做
result = {
'a': lambda x: x * 5,
'b': lambda x: x + 7,
'c': lambda x: x - 2
}[value](x)
从这里开始
class switch(object):
value = None
def __new__(class_, value):
class_.value = value
return True
def case(*args):
return any((arg == switch.value for arg in args))
用法:
while switch(n):
if case(0):
print "You typed zero."
break
if case(1, 4, 9):
print "n is a perfect square."
break
if case(2):
print "n is an even number."
if case(2, 3, 5, 7):
print "n is a prime number."
break
if case(6, 8):
print "n is an even number."
break
print "Only single-digit numbers are allowed."
break
测验:
n = 2
#Result:
#n is an even number.
#n is a prime number.
n = 11
#Result:
#Only single-digit numbers are allowed.
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。
虽然已经有了足够的答案,但我想指出一个更简单、更强大的解决方案:
class Switch:
def __init__(self, switches):
self.switches = switches
self.between = len(switches[0]) == 3
def __call__(self, x):
for line in self.switches:
if self.between:
if line[0] <= x < line[1]:
return line[2]
else:
if line[0] == x:
return line[1]
return None
if __name__ == '__main__':
between_table = [
(1, 4, 'between 1 and 4'),
(4, 8, 'between 4 and 8')
]
switch_between = Switch(between_table)
print('Switch Between:')
for i in range(0, 10):
if switch_between(i):
print('{} is {}'.format(i, switch_between(i)))
else:
print('No match for {}'.format(i))
equals_table = [
(1, 'One'),
(2, 'Two'),
(4, 'Four'),
(5, 'Five'),
(7, 'Seven'),
(8, 'Eight')
]
print('Switch Equals:')
switch_equals = Switch(equals_table)
for i in range(0, 10):
if switch_equals(i):
print('{} is {}'.format(i, switch_equals(i)))
else:
print('No match for {}'.format(i))
输出:
Switch Between:
No match for 0
1 is between 1 and 4
2 is between 1 and 4
3 is between 1 and 4
4 is between 4 and 8
5 is between 4 and 8
6 is between 4 and 8
7 is between 4 and 8
No match for 8
No match for 9
Switch Equals:
No match for 0
1 is One
2 is Two
No match for 3
4 is Four
5 is Five
No match for 6
7 is Seven
8 is Eight
No match for 9