如何在python中检查名称是否在名称列表中？

Question

我正在写俄勒冈州小径游戏，这是我的代码，这是造成问题，我不知道为什么会有问题。我想做的是，如果他们输入一个包含列表中某个单词的名称，则会将变量easter_mode设置为1，如果他们不输入，则会将easter_mode设置为0。需要在列表中列出的单词是：（Sturtz，Sturtz，Nate，Nate） 多谢各位

#asking name
player_name = input('What is your name:')
while len(player_name) >= 0:
  if len(player_name) > 1:
    print("Weclome" + str(player_name))
    print('Which mode do you want to play?')
    mode_choice = input('(easy) More modes comming soon:')
    break
  if len(player_name) == 1:
    player_name_choice = input(str(player_name)+"? Are you kidding me? Only one letter? You might regreat it (Y/N):")
    if player_name_choice == "y" or player_name_choice == "Y":
      print("Ok Your Choice!!...")
      mode_choice = 'easter'
      break
    if player_name_choice == "n" or player_name_choice == "N":
      player_name = input('What is your name:')
  else:
    print("You do not type anything, try again")
    player_name = input('What is your name:')

#Check Easter Egg Names
easter_names = ["nate sturtz", "Nate Sturtz", "Nate", "nate", "Sturtz", "sturtz"]
if player_name in easter_names:
    easter_mode = 1
else:
    easter_mode = 0
#easter eggs for name

if easter_mode == 1:
  year_set = 2005
  mode_choice = 'easter'
else:
  year_set = input('Enter a year whatever you like:')
  if year_set.isdigit():
    return_num = 0
  else:
    return_num = 1
  while return_num == 1:
    print('Error,please try again!')
    year_set = input('Enter a year whatever you like:')
    if year_set.isdigit():
      return_num = 0
    else:
      return_num = 1
  year_set = int(year_set)

当我运行完整文件时，我得到

Traceback (most recent call last):
  File "Oregon.py", line 64, in <module>
    player_name = input('What is your name:')
  File "<string>", line 1, in <module>
NameError: name 'nate' is not defined

您可以在Github上查看完整代码
https://raw.githubusercontent.com/nsturtz/Oregon-Trail/master/Oregon.py

Answer 1

在Python2中会出现此错误。在Python2中，input()在输入值时使用精确的值

在您的示例中，您键入的是nate，而不是'nate'。前一个值是一个变量名（在代码中没有定义，因此是NameError），而后一个值是一个字符串

在Python3中，input()的行为与您假设的一样，并将字符串传递给您的代码

如果确定要使用Python 2，可以用raw_input()替换input()，它将把输入解释为字符串而不是变量名

Answer 2

在Python2下，可以使用raw_input而不是input来防止Python将用户输入解释为Python代码

但是，由于Python2已被弃用，我强烈建议不要使用它。改用Python 3，其中input按预期工作

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¹当然除了维护传统产品。但这在这里似乎并不重要