如何用用户输入替换字符串的元素答案

【问题标题】：How do I replace elements of a string with user input如何用用户输入替换字符串的元素
【发布时间】：2019-11-05 01:29:13
【问题描述】：

下面是一个刽子手游戏的代码。到目前为止，一切都运行良好，我遇到的唯一麻烦是弄清楚当用户正确猜测时如何将带下划线的空格转换为正确的字母。我没有为它编写代码，正在寻求帮助。它目前所做的只是返回空白下划线，即使你猜对了一个数字

import random

responses = {'title': 'Welcome to Hangman the Game!', 'rules':'Once you guess \
wrong 6 times you lose. Using the same letter twice does not count as a guess.', \
             'correct':'Well done, your guess is correct!', \
             'incorrect':'Sorry, your guess is incorrect...', \
             'win':'Well done, you win!', \
             'lose':'Out of guesses. You lose.'}

words = ['cat', 'dog', 'work', 'school', 'game', 'one', 'hangman', 'apple',
         'orange', 'list', 'words', 'bicycle', 'four', 'snowing', 'backpack',
         'computer', 'house', 'water', 'plant', 'hour']

game = random.choice(words)

print(game) # just in for ease of programming

guesses = 0

letter = []

length = len(game)

numletter = game.replace(game, '_ '*length) # display number of letters

while guesses < 6:
    print(numletter) # need to replace this with the code I will hopefully learn from this
    user = input('Guess a letter here: ')
    if user in game:
        if user not in letter:
            print(responses['correct'])

    if user not in game:
        if user not in letter:
            print(responses['incorrect'])
            guesses += 1


    if user not in letter:
        letter.append(user)

        print('You have guessed these letters', letter)

    else: 
        print('You have already guessed that letter, try again.')


    if user == game:
        print(responses['win'])
        break

else:
    print(responses['lose'])

这是我得到的一些输出

water
_ _ _ _ _ 
Guess a letter here: a
Well done, your guess is correct!
You have guessed these letters ['a']
_ _ _ _ _ 
Guess a letter here: w
Well done, your guess is correct!
You have guessed these letters ['a', 'w']
_ _ _ _ _ 
Guess a letter here: w
You have already guessed that letter, try again.

如您所见，空白保持不变]

【问题讨论】：

标签： python list random

【解决方案1】：

维护一个布尔值列表visibility，使得visibility[5] 为真当且仅当您希望字符串中的字符#5 可见时。

import io
class HiddenString:

    def __init__(self, word):
        self._word = list(str(word))
        self._visibility = [False]*len(self._word)

    def make_all_instances_visible(self, char):
        char = str(char)
        assert(len(char) == 1)
        for idx in range(len(self._word)):
            if self._word[idx] == char:
                self._visibility[idx] = True
        return

    def __str__(self):
        with io.StringIO() as string_stream:
            for idx in range(len(self._word)):
                char = "_"
                if self._visibility[idx]:
                    char = self._word[idx]
                print(char, end="", file=string_stream)
            stryng = string_stream.getvalue()
        return stryng

###################################################################

secret = HiddenString("secret")
print(secret)
secret.make_all_instances_visible("e")
print(secret)

下面是控制台输出：

______
_e__e_

【讨论】：

【解决方案2】：

一个有趣的字符串函数是.translate()。它需要一个字符序数值字典，并将它们转换为字典的值。因此，如果您将解决方案单词的所有序数放入翻译表中，将它们翻译为下划线，然后翻译该单词，您将得到所有内容的下划线：

>>> word = 'apple'
>>> xlat = {ord(c):'_' for c in word}
>>> xlat
{97: '_', 112: '_', 108: '_', 101: '_'}
>>> word.translate(xlat)
'_____'

现在，当您有猜测时，测试该序号键是否在翻译表中，如果是，则将其删除。当您再次翻译时，被删除的字母不会发生替换：

>>> guess = 'p'
>>> ord(guess) in xlat
True
>>> del xlat[ord(guess)]
>>> word.translate(xlat)
'_pp__'

当word.translate(xlat) == word 获胜时。

我将把它作为一个练习，将其合并到您的代码中。

【讨论】：