python - 如何根据位置替换字符串中多个相同的字符？

Question

所以我正在为刽子手制作一个程序，到目前为止就是这样，

word = input("Enter a word: ").upper()
blanks = (len(word)*"_")
print(blanks)
chances = 5
while blanks.count("_") > 0:

    letter = input("Enter a letter: ").upper()

    if len(letter) > 1:
        print("Please enter one letter only")

    if letter in list(word):
        blanks = blanks[:word.index(letter)] + letter + blanks[word.index(letter)+1:]
        print(blanks)
        continue

    else:
        chances -= 1
        if chances > 0:
            print("That letter isn't in the word, you have", chances, "chance(s) left")

    if chances == 0:
        print("You lost. The word was", word.lower())
        exit()
print("You win! The word was", word.lower())

我只是在尝试代码，所以我会给出这个词来猜测自己，但是如果我曾经给出过一个重复字母的单词，比如“doggo”。

如果我把字母写成“o”，即使我再次猜到“o”，它也会给出“_ O _ _ _”而不是“_ O _ _ O”。

我可以解决这个问题吗？

Answer 1

这对我有用:

word = input("Enter a word: ").upper()
blanks = (len(word) * "_")
print(blanks)
chances = 5
while blanks.count("_") > 0:

    letter = input("Enter a letter: ").upper()

    if len(letter) > 1:
        print("Please enter one letter only")

    if letter in list(word):
        for count, character in enumerate(word):
            if letter == character:
                blanks = blanks[:count] + letter + blanks[count + 1:]
        print(blanks)
    else:
        chances -= 1
        if chances > 0:
            print("That letter isn't in the word, you have",
                  chances, "chance(s) left")

    if chances == 0:
        print("You lost. The word was", word.lower())
        exit()
print("You win! The word was", word.lower())

不同之处在于，我们不是进行一次替换，而是迭代该单词以查找所有巧合并对同一字母进行所有必要的替换。