blocks|key|975198|text|您现在遇到的主要问题是两重一，即replace()方法替换字符串中任何给定输入的所有实例，而不是第一个，第二个，您目前无法判断您已经发现了哪些字母。调用replace("_"，let)将总是替换“_”的每个实例，并且考虑到将其应用于仅由下划线组成的字符串，它将始终覆盖整个字符串。在每次使用猜测字母调用hangman()时，您似乎也在重新生成hidden_let，这意味着现在您的设计中最好的情况只是显示用户刚刚猜到的每一个字母，而另一些字母则是下划线。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975199|您想要做的是有两个值，correct_word和current_guess。correct_word将是玩家必须猜测的单词，而current_guess将是他们猜单词的进展，开始时只有与correct_word一样长的下划线。|975200|下面是一个简短的例子。我冒昧地删除了您的全局引用--全局引用通常是不允许的--并将行为封装在一个小类中。您可能希望用任意单词替换hangmanner.play_hangman()中的值。|975201|class+Hangmanner:
++++correct_word+=+''++++
++++current_guess+=+''

++++def+play_hangman(self,+word):
++++++++self.correct_word+=+word

++++++++self.current_guess+=+'_'+*+len(self.correct_word)

++++++++while+self.current_guess+!=+self.correct_word:
++++++++++++self.guess_letter(input("Please+guess+a+letter:+"))

++++def+guess_letter(self,+guessed_letter):
++++++++for+i+in+range(len(self.correct_word)):
++++++++++++if+self.correct_word[i]+==+guessed_letter:
++++++++++++++++if+i+==+0:
++++++++++++++++++++self.current_guess+=+self.correct_word[i]+%2B+self.current_guess[1:]
++++++++++++++++else:
++++++++++++++++++++self.current_guess+=+self.current_guess[:i]+%2B+self.correct_word[i]+%2B+self.current_guess[i+%2B+1:]
++++++++print(self.current_guess)

if+__name__+==+"__main__":
++++hangmanner+=+Hangmanner()
++++hangmanner.play_hangman("test")|code-block|syntax|javascript|975202|这使用python中的切片函数，可以使用括号和第一个语法来访问任意给定集合的任意范围。如果第一次或最后一次丢失，则切片将分别继续到集合的开始或结束。上面，current_guess1:从第二个索引返回current_guess到最后一个索引。current_guess:i从第一个索引到i之前的索引返回current_guess，因为最后是排它的结束界。|975203|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The main problem you're having right now is two-fold - one, that the replace() method replaces all instances of any given input within a string, not the first one, and two, that you don't currently have any way of telling which letters you've already uncovered. Calling replace("_", let) will always replace every single instance of "_", and given that you're applying that to a string that is only composed of underscores, it'll always overwrite the entire string. It seems like you're also regenerating hidden_let every time hangman() is called with a guess letter, meaning that best-case with your design now you're only going to ever show every letter the user just guessed and a bunch of underscores otherwise.

What you'd want to do is have two values, correct_word and current_guess. correct_word will be the word the player has to guess, and current_guess will be their progress in guessing the word, starting with a string of only underscores of the same length as correct_word.

Here's a short example. I've taken the liberty of removing your global references - globals are generally frowned upon - and encapsulated the behavior in a small class. You'd want to replace the value in hangmanner.play_hangman() with whatever your random word is.

<pre><code>class Hangmanner:
 correct_word = '' 
 current_guess = ''

 def play_hangman(self, word):
 self.correct_word = word

 self.current_guess = '_' * len(self.correct_word)

 while self.current_guess != self.correct_word:
 self.guess_letter(input("Please guess a letter: "))

 def guess_letter(self, guessed_letter):
 for i in range(len(self.correct_word)):
 if self.correct_word[i] == guessed_letter:
 if i == 0:
 self.current_guess = self.correct_word[i] + self.current_guess[1:]
 else:
 self.current_guess = self.current_guess[:i] + self.correct_word[i] + self.current_guess[i + 1:]
 print(self.current_guess)

if __name__ == "__main__":
 hangmanner = Hangmanner()
 hangmanner.play_hangman("test")
</code></pre>

This uses the slicing function in python, where you can use the brackets and the [first:last] syntax to access an arbitrary range of any given collection. If either first or last is missing, the slice continues to the beginning or end of the collection, respectively. Above, current_guess[1:] returns current_guess from the second index to the last. current_guess[:i] returns current_guess from the first index up to the index preceding i, given that last is the exclusive end bound.

blocks|key|1669538|text|hiddenLet.replace('_',let)用_所代表的任何内容替换_的所有出现。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1669539|newWordList+=+[x+if+x==let+else+'_'+for+x+in+randWord]+
newWord+=+''.join(newWordList)|code-block|syntax|javascript|1669540|entityMap^0|0|Q|R|1|12|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

<code>hiddenLet.replace('_',let)</code> replaces all occurrences of <code>_</code> with whatever <code>let</code> represents.

<pre><code>newWordList = [x if x==let else '_' for x in randWord] 
newWord = ''.join(newWordList)
</code></pre>

Here is the problematic piece of my function:

<pre><code>def hangman1(word):
global guessesMade
global guessesLeft
currentGuess = '_ ' * len(word)
let = print(input('Please guess a letter: '))
for i in range(len(word)):
 if word[i] == let:
 print('{} is contained in the word.'.format(let))
 if i == 0:
 currentGuess = word[0] + currentGuess[1:]
 else:
 currentGuess = currentGuess[:i] + word[i] + currentGuess[i + 1:]
print(currentGuess)
</code></pre>

The user enters a letter at the prompt and it checks if the letter is in the randomWord that was generated outside of the function from a list of words. I can get it to print the blanks correctly, but if the user enters a letter that is in the word it prints out a line of the correct letter instead of the blanks with the correct letter mixed in between. 

Any help is appreciated.

Implementing a hangman function in Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面是我的功能中有问题的部分：def hangman1(word):global guessesMadeglobal guessesLeftcurrentGuess = '_ ' * len(word)let = print(input('Please guess a letter: '))for i in range...

问用Python实现hangman函数
EN

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