Question

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

TypeError：Sub：'NoneType'和'int'
不支持的操作数类型

我理解这个错误。我的代码生成的player_number是None类型，这就是我无法减去它的原因。

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

你有这个分配

player_number = name_to_number(player_choice)

但if/elif/else中的name_to_number个案例都没有使用return关键字。要return来自此函数的值，您可以执行

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

number_to_name功能

也是如此

Answer 2

未经测试的修复。 'return'声明丢失了。对字符串和int使用相同的变量名称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：减法时没有类型错误

2 个答案: