使用更好的条件缩短Python中的长if-else检查

Shortening a long if-else check in Python using better conditions

我正在编写一个python程序，在这里我需要一个if-else案例来选择一个介于1和9之间的数字，每个数字都分配给一个类。关于如何缩短此代码有什么建议吗？

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import randint

variable1 = randint(1, 9)
if variable1 >= 9:
print ("Your class is Tiefling")
else:
if variable1 >= 8:
print ("Your class is Half-Orc")
else:
if variable1 >= 7:
print ("Your class is Half-Elf")
else:
if variable1 >= 6:
print ("Your class is Gnome")
else:
if variable1 >= 5:
print ("Your class is Dragonborn")
else:
if variable1 >= 4:
print ("Your class is Human")
else:
if variable1 >= 3:
print ("Your class is Halfling")
else:
if variable1 >= 2:
print ("Your class is Elf")
else:
if variable1 >= 1:
print ("Your class is Dwarf")

相关讨论

使用字典的不完整版本：

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val_msg = {3: 'Your class is Halfling',
2: 'Your class is Elf',
1: 'Your class is Dwarf'}

from random import randint

variable1 = randint(1, 3)
print(val_msg[variable1])

注意，randint生成整数，我将其用作字典的键。

如果您需要做更复杂的事情，您可以将函数放入字典并调用它们(当然，您也可以在这里使用基于list的解决方案)：

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def do_halfling_stuff():
print('Your class is Halfling')
def do_elf_stuff():
print('Your class is Elf')
def do_dwarf_stuff():
print('Your class is Dwarf')

val_func = {3: do_halfling_stuff,
2: do_elf_stuff,
1: do_dwarf_stuff}

variable1 = randint(1, 3)
func = val_func[variable1]
func()

希望这有帮助！

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import random
classList = ['Dwarf','Elf','Halfling','Human','Dragonborn','Gnome','Half-Elf','Half-Orc','Tiefling']
print 'Your class is ' + random.choice(classList)

如果您要多次使用此功能，我建议您使用以下功能：

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def race_definer():
races = {1: 'Tiefling', 2: 'Half-Orc', 3: 'Half-Elf', 4: 'Dragonborn',
5: 'Elf', 6: 'Gnome', 7: 'Human', 8: 'Halfling', 9: 'Elf', 0: 'Dwarf'}

print('Your race is {}.'.format(races[randint(0,9)]))

在需要的时候调用函数：

1	race_definer()

在使用函数之前，请不要忘记将randint导入程序：

1	from random import randint

我想你可以用：

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from random import randint
variable1 = randint(1, 9)
my_dict = {1:"Tiefling", 2:"Half-Orc", 3:"Half-Elf", 4:"Gnome", 5:"Dragonborn", 6:"Human", 7:"Halfling", 8:"Elf", 9:"Dwarf", }
base ="Your class is"
print("{}{}".format(base, my_dict[variable1]))