Python编写石头剪刀布游戏时报错generator与str无法相加如何解决

报错详情 #

运行代码时出现如下类型错误：

TypeError: unsupported operand type(s) for +: 'generator' and 'str'

该报错在加入随机逻辑后触发，完整报错堆栈如下：

Traceback (most recent call last)

<ipython-input-4-dea4c40cfdcd> in <module>()
 49   if key == 'q':
 50     break
---> 51   player_score, comp_score = gameplay(player_score, comp_score)
 52   print('Score is: YOU ', player_score, ' - ', comp_score, ' COMPUTER')
 53   print('')

 <ipython-input-4-dea4c40cfdcd> in gameplay(player_score, comp_score)
 12 
 13   comp_move = moves[randomize]
 ---> 14   battle = key + '-' + comp_move
 15 
 16   comp_print = (word for position, word in ['rock', 'paper', 'scissors'] if position == randomize - 1)

相关疑问 #

不清楚报错中的generator对象是什么，搜索结果显示是通过循环生成的序列，但代码中仅用到while循环，并未编写生成器相关逻辑，不确定是否是while循环导致的报错。另外代码逻辑仅需要取序列对应位置的单个字符，不需要使用整个字符序列，不理解报错触发原因，不确定自己对变量使用的认知是否有误。

相关代码 #

报错位置代码 #

from random import randint

player_score = 0
comp_score = 0
key = None

# Setting the rules
choices = {'r-s': 'rock breaks scissors', 'p-r': 'paper envelopes rock', 's-p': 'scissors cut paper'}
moves = {1: 'r', 2: 'p', 3: 's'}

def gameplay(player_score, comp_score):

  comp_move = moves[randomize]
  battle = key + '-' + comp_move

补充代码（随机数初始化逻辑） #

while True:
  randomize = randint(1, 3)
  print('<r>ock, <p>aper or <s>cissors? Press any key to continue; <q> for exit')
  key = check_input()
  if key == 'q':
    break
  player_score, comp_score = gameplay(player_score, comp_score)
  print('Score is: YOU ', player_score, ' - ', comp_score, ' COMPUTER')
  print('')
  key = None

可行解决方案 #

• 方案一：使用for循环配合数组索引简化实现

items = ['rock', 'paper', 'scissors']

for word in items:
  if items.index(word) == randomize - 1:
    print('Computer played: ', word)

• 方案二：使用enumerate函数优化原生成器表达式
  原错误生成器写法：

item = (word for position, word in ['rock', 'paper', 'scissors'] if position == randomize - 1)

优化后写法：

item = [word for position, word in enumerate(['rock', 'paper', 'scissors']) if position == randomize - 1]
print('Computer played: ', item[0])

问题根源 #

Python列表没有自带的显式索引匹配逻辑，需要自行实现对应功能，上述两种方案均可正常运行。

内容的提问来源于stack exchange，提问作者Marko Kolaksazov