如何从字符串中获取重复字符的索引？

Question

I'm working with a hangman like project whereas if the user inputs a letter and matches with the solution, it replaces a specific asterisk that corresponds to the position of the letter in the solution. I'm trying to do this by getting the index of the instance of that letter in the solution then replacing the the matching index in the asterisk.

The thing here is that I only get the first instance of a recurring character when I used var.index(character) whereas I also have to replace the other instance of that letter. Here's the code:

word = 'turtlet'
astk = '******'

for i in word:
    if i == t:
        astk[word.index('i')] = i

Here it just replaces the first instance of 't' every time. How can I possibly solve this?

Answer 1

index() gives you only the index of the first occurrence of the character (technically, substring) in a string. You should take advantage of using enumerate() . Also, instead of a string, your guess (hidden word) should be a list, since strings are immutable and do not support item assignment, which means you cannot reveal the character if the user's guess was correct. You can then join() it when you want to display it. Here is a very simplified version of the game so you can see it in action:

word = 'turtlet'
guess = ['*'] * len(word)

while '*' in guess:

    print(''.join(guess))

    char = input('Enter char: ')

    for i, x in enumerate(word):
        if x == char:
            guess[i] = char

print(''.join(guess))
print('Finished!')

Answer 2

Note the the find method of the string type has an optional parameter that tells where to start the search. So if you are sure that the string word has at least two t s, you can use

firstindex = word.find('t')
secondindex = word.find('t', firstindex + 1)

I'm sure you can see how to adapt that to other uses.

Answer 3

I believe there's a better way to do your specific task.

Simply keep the word (or phrase) itself and, when you need to display the masked phrase, calculate it at that point. The following snippet shows how you can do this:

>>> import re
>>> phrase = 'My hovercraft is full of eels'
>>> letters = ' mefl'
>>> display = re.sub("[^"+letters+"]", '*', phrase, flags=re.IGNORECASE)
>>> display
'M* ***e****f* ** f*ll *f eel*'

Note that letters should start with the characters you want displayed regardless (space in my case but may include punctuation as well). As each letter is guessed, add it to letters and recalculate/redisplay the masked phrase.

The regular expression substitution replaces all characters that are not in letters , with an asterisk.

Answer 4

for i in range(len(word)):
    if word[i] == "t":
        astk = astk[:i] + word[i] + astk[i + 1:]