[贝叶斯公式]Python案例实现

发布于:2024-05-19 ⋅ 阅读:(157) ⋅ 点赞:(0)

步骤分析:

1.  首先需要一个包含正确单词的语料库,用于计算单词的概率。
2.  当用户输入一个单词时,首先检查该单词是否在语料库中,如果在则认为是正确的,不需要纠错。
3.  如果用户输入的单词不在语料库中,需要计算输入单词与语料库中每个单词的编辑距离,找出编辑距离最小的单词作为纠错的候选。
4.  对候选单词进行概率计算,计算每个候选单词在语料库中出现的概率,选择概率最大的作为最终的纠正单词。

下面是一个简单的demo代码:

# 定义单词列表
words_list = [
    "apple",
    "banana",
    "orange",
    "grape",
    "watermelon",
    "kiwi",
    "pineapple",
    "strawberry"
]

# 将单词列表写入到"big.txt"文件中
with open('big.txt', 'w') as file:
    for word in words_list:
        file.write(word + '\n')

print("成功创建并写入单词到big.txt文件!")
import re
from collections import Counter


def words(text):
    return re.findall(r'\w+', text.lower())


WORDS = Counter(words(open('big.txt').read()))


def P(word, N=sum(WORDS.values())):
    return WORDS[word] / N


def candidates(word):
    return (known([word]) or known(edits1(word)) or known(edits2(word)) or [word])


def known(words):
    return set(w for w in words if w in WORDS)


def edits1(word):
    letters = 'abcdefghijklmnopqrstuvwxyz'
    splits = [(word[:i], word[i:]) for i in range(len(word) + 1)]
    deletes = [L + R[1:] for L, R in splits if R]
    transposes = [L + R[1] + R[0] + R[2:] for L, R in splits if len(R) > 1]
    replaces = [L + c + R[1:] for L, R in splits if R for c in letters]
    inserts = [L + c + R for L, R in splits for c in letters]
    return set(deletes + transposes + replaces + inserts)


def edits2(word):
    return (e2 for e1 in edits1(word) for e2 in edits1(e1))


def correction(word):
    return max(candidates(word), key=P)


def spell_check(input_word):
    if input_word in WORDS:
        return input_word + " is correct!"
    else:
        return "Did you mean: " + correction(input_word)


# 测试
input_word = input("请输入一个单词:")
print(spell_check(input_word))


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