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我已将单词更改为仍然存在空格的框,但我不确定如何制作,以便当玩家猜出正确的字母时,框会变成字母。

继承人我是如何把它变成盒子的。

currentphrase = random.choice(WOFphrases)
hiddenphrase = re.sub('[A-Z]', u'\u2610',currentphrase)

我试过这样做,但即使是空格也被盒子替换了。

for letter in currentphrase:
    if letter in completedletters:
        print(letter,end='')
    else:
        print(u'\u2610',end='')
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1 回答 1

0

不要为此使用正则表达式。您的第二个解决方案应该可以工作,只需为空格添加一个条件

# Sample input
current_phrase = 'I am a phrase with letters and whitespaces'
completed_letters = ['a', 'e', 'w']  # make sure the letters here are already in lower case so it's easy to test later

# Solution
for letter in current_phrase:
    if letter.isspace() or letter.lower() in completed_letters:
        print(letter, end='')
    else:
        print(u'\u2610', end='')

输出:

☐ a☐ a ☐☐☐a☐e w☐☐☐ ☐e☐☐e☐☐ a☐☐ w☐☐☐e☐☐a☐e☐
于 2020-04-08T11:03:35.967 回答