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
运行函数的解决方案:
result = {
'case1': foo1,
'case2': foo2,
'case3': foo3,
}.get(option)(parameters_optional)
其中foo1()、foo2()和foo3()是函数
示例1(带参数):
option = number['type']
result = {
'number': value_of_int, # result = value_of_int(number['value'])
'text': value_of_text, # result = value_of_text(number['value'])
'binary': value_of_bin, # result = value_of_bin(number['value'])
}.get(option)(value['value'])
示例2(无参数):
option = number['type']
result = {
'number': func_for_number, # result = func_for_number()
'text': func_for_text, # result = func_for_text()
'binary': func_for_bin, # result = func_for_bin()
}.get(option)()
示例4(仅限值):
option = number['type']
result = {
'number': lambda: 10, # result = 10
'text': lambda: 'ten', # result = 'ten'
'binary': lambda: 0b101111, # result = 47
}.get(option)()
其他回答
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
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.
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]
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')
""")
# 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