在上面的代码中加入一个程序“每猜一次,根据提供的字符串(例如a _ _ le)与字典中所有可能的单词进行匹配,将这些可能的单词中出现的字母频率制成表格,然后猜测出现频率最高的尚未被猜测的字母。如果没有剩余的单词匹配,那么它将默认返回到整个字典的字符频率分布。“
时间: 2024-02-23 19:03:09 浏览: 80
好的,我理解了你的要求。下面是你想要的功能的代码实现:
```python
import random
from collections import Counter
def guess_word(word_list):
guessed_letters = set()
word = random.choice(word_list)
guessed_word = ['_'] * len(word)
while True:
print(' '.join(guessed_word))
letter_freq = Counter(''.join(word_list) + ''.join(guessed_letters))
sorted_freq = sorted(letter_freq.items(), key=lambda x: x[1], reverse=True)
for letter, freq in sorted_freq:
if letter not in guessed_letters:
guessed_letters.add(letter)
break
else:
print('No more letters to guess. The word is:', word)
break
guess = input('Guess a letter: ').lower()
if guess in guessed_letters:
print('You already guessed that letter. Try again.')
elif guess in word:
print('Good job! The letter is in the word.')
for i in range(len(word)):
if word[i] == guess:
guessed_word[i] = guess
if '_' not in guessed_word:
print('Congratulations! You guessed the word:', word)
break
else:
print('Oops! The letter is not in the word.')
new_word_list = []
for w in word_list:
if all(w[i] == guessed_word[i] or guessed_word[i]=='_' for i in range(len(w))) and guess not in w:
new_word_list.append(w)
if not new_word_list:
new_word_list = word_list
word_list = new_word_list
word = random.choice(word_list)
if len(word_list) == 1:
print('The word is:', word_list[0])
break
word_list = ['apple', 'banana', 'cherry', 'orange', 'pear', 'peach', 'pineapple', 'strawberry']
guess_word(word_list)
```
这个程序会在每次猜测后,根据提供的字符串与剩余的单词进行匹配,将这些可能的单词中出现的字母频率制成表格,然后猜测出现频率最高的尚未被猜测的字母。如果没有剩余的单词匹配,那么它将默认返回到整个字典的字符频率分布。
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