循环python时列出索引

Question

我有一个问题，以下代码。当我输入确切的秘密字字符时，下面的代码将与密码不匹配。

#initiate words for guessing secretWords =['cat','mouse','donkey','ant','lion']

#Generate a random word to be guessed generateWord = (random.choice(secretWords))

# User have attempts only to the random generate words LeftCount = 6

generateWord = ["_"] * len(secretWords) userInput="" LetterNumber=0 RightGuess =0 

while (LeftCount !=0):
     print ("Word Contains", len(generateWord),"letters")
     print ("You have", str(LeftCount),"attempts remaining")
     print ("\n")
     print(generateWord)
     print ("\n")
     userInput=input("Enter word: ")
     while(LetterNumber< len(generateWord)):
         if(generateWord[LetterNumber] == userInput):
             generateWord[LetterNumber]= userInput

             RightGuess +=1
         LetterNumber +=1
     LeftCount -=1
     LetterNumber=0

     if (RightGuess == len(generateWord)):
         print ("Congratulations")
         break

     if(LeftCount ==0):
         print ("Game over")

Answer 1

您正在重写所选单词。generateWord同时是密文，也是用户输入。这不行。其他问题我也要纠正：

import random

secretWords = ["cat", "mouse", "donkey", "ant", "lion"]

generatedWord = random.choice(secretWords)
leftCount = 6
userWord = ["_"] * len(generatedWord)
refusedLetters = ""

#returns all positions of character ch in the string s
def findOccurences(s, ch):
    return [i for i, letter in enumerate(s) if letter == ch]

print("Word contains", len(generatedWord), "letters")
while(leftCount > 0 and generatedWord != "".join(userWord)):
    print ("\n")
    print ("You have", str(leftCount), "attempts remaining")
    print ("Letters not present in your word:", "".join(sorted(refusedLetters)))
    print ("Your word so far: ","".join(userWord))
    print ("\n")

    #checks that the user enters something
    userInput = ""
    while not len(userInput):
        userInput=input("Enter letter or word: ")
    userInput = userInput.lower()

    #if len > 1, then the user has tried to guess the whole word:
    if len(userInput) > 1:
        if generatedWord == userInput:
            print("Congratulations")
            break
        else:
            print("Wrong word, sorry")
    #len == 1, thus the user asked for one letter
    else:
        #if letter isn't already found
        if not userInput in userWord:
            #for all occurences of the letter in the secret word
            occurences = findOccurences(generatedWord, userInput)
            for occurence in occurences:
                userWord[occurence] = userInput
            #if the letter was not found
            if not occurences:
                #if the letter has already been proposed
                if userInput in refusedLetters:
                    print("You have already tried this letter")
                    continue
                else:
                    refusedLetters += userInput
        else:
            print("You have already tried this letter")
            continue

    leftCount -= 1

#the else statement will only be entered if we did not exit the previous loop via a break.
#Thus, if the word has already been found, nothing happens here.
else:
    if generatedWord == "".join(userWord):
        print("Congratulations")
    else:
        print("Game over")

Answer 2

为什么要将generateWord中的一个字母与整个userInput进行比较？在

if(generateWord[LetterNumber] == userInput):

该行将“LetterNumber”索引处的字符与userInput进行比较，因此如果用户输入一个单词，它将永远不会返回true。在

如果你想计算用户猜测中正确字母的数量，你不应该将用户输入的每个字母与“generateWord”中对应的字母进行比较吗。在

^{pr2}$

还有一些一般性的观点，变量名不应该以大写字母开头，根据Python标准，它应该是“字母号码”。试着改进你的变量名，也许叫它“生成的单词”，而不是“生成单词”Generate_word“意味着它是一个函数，因为Generate是一个动词。在

if语句后面的行也将整个userInput重新分配到字母索引处的generateWord值中，为什么要这样做？在

最后，您需要在while循环的末尾或开始处生成一个新词，因为现在您只在开头生成一个词，然后它将在每次迭代中使用相同的词。在

尝试使用print打印出一些变量，这将有助于调试程序，因为它肯定没有达到预期效果。