day5_1 拼写检查器的实现

代码思想很好理解

• 首先读取并处理文件文件（全部小写），并将其转化成字典的格式（NWORDS）。单词为键 出现的次数为值
• 通过input输入需要判断的单词，并调用correct
• 在crrect中，判断四种情：
       1.（known）是否是NWORDS中的单词，如不满足到2
       2.（known(edits1(word))）经过补充，缺失，转换 后的单词是否属于NWORDS，如不满足到3
       3.（known_edits2(word）再重复2的操作，如果还不满足那么到4
       4.（[word]）将这个单词返回
• 将返回的set对象中的值从字典中取出，比较出现次数的大小。然后返回次数最多的单词

数据集下载地址

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import re, collections


def words(text):
    return re.findall('[a-z]+', text.lower())


def train(features):
    model = collections.defaultdict(lambda: 1)
    for f in features:
        model[f] += 1
    return model


def edits1(word):
    n = len(word)
    return set([word[0:i]+word[i+1:] for i in range(n)] +
    [word[0:i] + word[i+1] + word[i] + word[i+2:] for i in range(n-1)] +
    [word[0:i]+c + word[i + 1:] for i in range(n) for c in alphabet] +
    [word[0:i]+c + word[i:] for i in range(n + 1) for c in alphabet])


def known_edits2(word):
    return set(e2 for e1 in edits1(word) for e2 in edits1(e1) if e2 in NWORDS)


def known(words):
    x = set(w for w in words if w in NWORDS)
    for w in words:
        print(w)
    return x

def correct(word):
    candidates = known([word]) or known(edits1(word)) or known_edits2(word) or [word]
    # print(known([word]),'1', known(edits1(word)), known_edits2(word))
    return max(candidates, key=lambda w: NWORDS[w])


if __name__ == "__main__":
    NWORDS = train(words(open('big.txt').read()))
    alphabet = 'abcdefghijklmnopqrstuvwxy'
    while True:
        str1 = input('请输入要书写的单词')
        print(correct(str1))