想要从列表中删除配对时遇到问题

Question

我有一段代码允许用户删除已经在线索列表中的配对。然而，当我尝试运行这段代码，一个错误出现，我就如何解决其删除配对是显示了这一点... 我的代码不能确定......

def delete_pairing(clues): 
    found = True 
    #USER INPUTS A LETTER AND SYMBOL 
    letter=input("What letter would you like to delete? ").upper 
    symbol=input("\nWhat symbol would you like to delete? ") 
    #THE LETTER AND SYMBOL THE USER INPUTS BECOMES ONE STRING 
    delClue = letter + symbol 
    #IF THE delClue exists in clues, it will delete the pairing 
    if delClue in clues: 
    #CODE FOR REMOVING THE CLUE 
     clues.remove(delClue) 
    # LETS THE USER KNOW WHAT CLUES HAS BEEN DELETED 
     print("\nClue ",(delClue)," has been deleted") 
     print("\nYour clues are now...") 
     print (clues) 
    #If delClue doesn't exist in clues, it will print an error message  
    else: 
     print("That clue does not exist ") 
    return clues 

成果应如果用户输入的字母和符号配对在线索列表中，那么它应该被删除。否则，万一发生说，这封信和符号配对的用户输入了错误信息，不线索的列表中存在....这我有

错误......

delClue = letter + symbol 
TypeError: unsupported operand type(s) for +: 'builtin_function_or_method' and 'str' 

Answer 1

由于the comment of Ashwini Chaudhary状态，你忘了()在该行

信=输入（“你想什么字母删除吗？”）结束.upper

因此letter不是您所期望的字符串类型，而是builtin_function_or_method（即upper() method of python's string type）。这就是你不能将它连接到另一个字符串的方式。