嵌套';如果';Python中的函数
作为Python的新手,我正在做的一件事是角色生成器。从我编写的代码中可以看出,我正在尝试进行种族选择嵌套';如果';Python中的函数,python,function,nested,Python,Function,Nested,作为Python的新手,我正在做的一件事是角色生成器。从我编写的代码中可以看出,我正在尝试进行种族选择 ########Race######## #.racechoose (label) hero_race = input("What is your hero's race? (Human / Elf / Dwarf / Orc) Don't forget to capitalize! ") if hero_race == 'Human': print ("Humans are wel
########Race########
#.racechoose (label)
hero_race = input("What is your hero's race? (Human / Elf / Dwarf / Orc) Don't forget to capitalize! ")
if hero_race == 'Human':
print ("Humans are well-rounded, average characters. They have a bonus of + 1 to Speed and + 1 to Int.")
yn = input ("Do you want to be a Human (Y/N)? ")
if yn == 'y' or yn == 'Y':
profile['Race'] = "Human"
print ("Your hero", profile['Name'], "is a human.")
else:
#goto racechoose
elif hero_race == 'Elf':
print("Elves are very fast, and they have a bonus of + 2 to Speed.")
yn = input("Do you want to be an Elf? (y/n) ")
if yn == 'y' or yn == 'Y':
profile['Race'] = "Elf"
print("Your hero ", profile['Name'], "is an Elf.")
else:
#goto racechoose
elif hero_race == 'Dwarf':
print("Dwarves are small, but strong. Dwarves get a bonus of + 2 Muscle.")
yn = input("Do you want to be a Dwarf? (Y/N) ")
if yn == 'y' or yn =='Y':
profile['Race'] = 'Dwarf'
print("Your hero ", profile['Name'], "is a Dwarf.")
else:
#goto racechoose
else: #orc
print("Orcs are brute muscle. Orcs get a bonus of + 3 to Muscle, but - 1 to Int.")
yn = input("Do you want to be an Orc? (Y/N) ")
if yn == 'y' or yn == 'Y':
profile['Race'] = 'Orc'
print("Your hero ", profile['Name'], "is an Orc.")
else:
#goto racechoose
请忽略goto和label注释-我最近刚刚停止使用blitzbasic,现在正在尝试查找python的label和goto命令
无论如何,我在elif hero race Elf行中得到了“预期缩进块”,我想知道如何正确缩进这段代码。谢谢 如果需要评论以外的内容,请使用pass
最好使用字典,而不是嵌套的if语句,即,如果您有一个配置文件键、文本输出、描述的字典,并使用默认值为“受害者”的get,则代码会更干净。如果您需要注释以外的其他内容,请在必要时使用pass
最好使用字典,而不是嵌套的if语句,即如果您有一个配置文件键、文本输出、描述的字典,并使用默认值为“受害者”的get,那么代码会更干净。出于某种原因,如果您离开一个块(需要语句的块)空,然后使用
pass
作为占位符。使用注释是行不通的
else:
pass
发件人:
pass
是一个空操作-当它被执行时,不会发生任何事情。它是
在语法上需要语句时用作占位符,
但不需要执行任何代码,例如:
示例:
def f(arg): pass # a function that does nothing (yet)
class C: pass # a class with no methods (yet)
而
如果yn=='y'或yn=='y':
可以减少为如果yn.lower()=='y':
出于某种原因,如果要将块(需要语句的块)保留为空,则使用pass
作为占位符。使用注释是行不通的
else:
pass
发件人:
pass
是一个空操作-当它被执行时,不会发生任何事情。它是
在语法上需要语句时用作占位符,
但不需要执行任何代码,例如:
示例:
def f(arg): pass # a function that does nothing (yet)
class C: pass # a class with no methods (yet)
如果yn=='y'或yn=='y':可以简化为
如果yn.lower()=='y':
为什么不在这样的函数中提取公共内容:
def get_race_description(race):
return {
'human': "Humans are well-rounded ...",
'elf': "Elves are very fast ...",
# all the other race descriptions
}.get(race.lower(), "WTF? I don't know that race")
def confirm_race_selection(race):
print (get_race-description(race))
yn = input ("Do you want to be {0} (Y/N)? ".format(race))
return (yn.lower() == 'y')
while True:
hero_race = input("What is your hero's race?")
if (confirm_race_selection(hero_race)):
profile['Race'] = hero_race
print ("Your hero {0} is {2}".format(profile['Name'], hero_race))
break
这段代码完全不需要使用
break
就可以重写,但我已经做了很多重构,所以现在就看你了为什么你不在这样的函数中提取常见的东西:
def get_race_description(race):
return {
'human': "Humans are well-rounded ...",
'elf': "Elves are very fast ...",
# all the other race descriptions
}.get(race.lower(), "WTF? I don't know that race")
def confirm_race_selection(race):
print (get_race-description(race))
yn = input ("Do you want to be {0} (Y/N)? ".format(race))
return (yn.lower() == 'y')
while True:
hero_race = input("What is your hero's race?")
if (confirm_race_selection(hero_race)):
profile['Race'] = hero_race
print ("Your hero {0} is {2}".format(profile['Name'], hero_race))
break
这段代码完全可以不用
break
重写,但是我已经做了很多重构,所以现在就看你了Python中没有label
和goto
。任何自尊的高级编程语言都不应该有这样的东西。这正是OP说忽略它们的原因@Matthias。他说“[…]我现在正试图为python找到label和goto命令”。python中没有label
和goto
。任何自尊的高级编程语言都不应该有这样的东西。这正是OP说忽略它们的原因@Matthias.And(s)他说“[…]我现在正试图为python找到label和goto命令”。