Python: None 减法时输入错误

Question

我需要计算 'player_number' 和 'comp_number' 之间的差异，但它说：

TypeError: unsupported operand type(s) for Sub: 'NoneType' and 'int'

我理解这个错误。我的代码生成 player_number 是 None 类型，这就是为什么我不能减去它。

我该如何处理这个问题？有什么想法吗？

这是我的代码：

def name_to_number(name):
   if name == "rock":
       name =  0
   elif name == "paper":
       name = 1
   elif name == "Spock":
       name = 2
elif name == "lizard":
       name = 3
elif name == "scissors":
       name = 4
else:
    print 'Name is not listed:',name

def number_to_name(number):
if number == 0:
    print "rock"
elif number == 1:
    print "Spock"
elif number == 2:
    print "paper"
elif number == 3:
    print "lizard"
elif number == 4:
    print "scissors"
else:
    print 'Your number is not valid:',number


def rpsls(player_choice):
if player_choice == "rock":

    print 'Player choses', player_choice
    player_number = name_to_number(player_choice)
elif player_choice == "Spock":
    print 'Player choses',player_choice
    player_number = name_to_number(player_choice)
elif player_choice == "paper":
    print 'Player choses',player_choice
    player_number = name_to_number(player_choice)
elif player_choice == "lizard":
    print 'Player choses',player_choice
    player_number = name_to_number(player_choice)
elif player_choice == "scissors":
    print 'Player choses',player_choice
    player_number = name_to_number(player_choice)
else:
    print "Name not in list",player_choice

import random
comp_number = random.randrange(0,4)
if comp_number == 0:
    print "Computer choses",number_to_name(0)
elif comp_number == 1:
    print "Computer choses",number_to_name(1)
elif comp_number == 2:
    print "Computer choses",number_to_name(2)
elif comp_number == 3:
    print "Computer choses",number_to_name(3)
elif comp_number == 4:
    print "Computer choses",number_to_name(4)

diffrence =  player_number - comp_number

if  diffrence % 5 == 1 or 2:
    print 'Player wins!'
elif (diffrence % 5) == 3 or 4:
    print 'Computer wins!'
else:
    print 'Game tie'




rpsls("rock")

Answer 1

你有这个作业

player_number = name_to_number(player_choice)

但是 name_to_number 中 if/elif/else 个案例中的 none 使用了 return 关键字。 return 这个函数的值，你会做

def name_to_number(name):
    if name == "rock":
        return 0
    elif name == "paper":
        return 1
    elif name == "Spock":
        return 2
    elif name == "lizard":
        return 3
    elif name == "scissors":
        return 4
    else:
        print 'Name is not listed:',name

你的 number_to_name 函数也是如此

Answer 2

未经测试的修复。 'return' 语句丢失。对字符串和整数使用相同的变量名不是很好的编程风格。

def name_to_number(name):
   number = -1
   if name == "rock":
       number =  0
   elif name == "paper":
       number = 1
   elif name == "Spock":
       number = 2
   elif name == "lizard":
       number = 3
   elif name == "scissors":
       number = 4
   else:
       print 'Name is not listed:',name

   return number    #This line was missing

Python: None 减法时输入错误

Python: None type error while subtraction

python

nonetype