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我有一个 Python 脚本，它接收“.html”文件，删除停用词并返回 python 字典中的所有其他词。但是如果同一个词出现在多个文件中，我希望它只返回一次。即包含不间断的单词，每个单词只有一次。 

def run():
filelist = os.listdir(path)
regex = re.compile(r'.*<div class="body">(.*?)</div>.*', re.DOTALL | re.IGNORECASE)
reg1 = re.compile(r'<\/?[ap][^>]*>', re.DOTALL | re.IGNORECASE)
quotereg = re.compile(r'&quot;', re.DOTALL | re.IGNORECASE)
puncreg = re.compile(r'[^\w]', re.DOTALL | re.IGNORECASE)
f = open(stopwordfile, 'r')
stopwords = f.read().lower().split()
filewords = {}

htmlfiles = []
for file in filelist:
    if file[-5:] == '.html':
        htmlfiles.append(file)
        totalfreq = {}


for file in htmlfiles:
    f = open(path + file, 'r')
    words = f.read().lower()
    words = regex.findall(words)[0]
    words = quotereg.sub(' ', words)
    words = reg1.sub(' ', words)
    words = puncreg.sub(' ', words)
    words = words.strip().split()

    for w in stopwords:
        while w in words:
            words.remove(w)


    freq = {}
    for w in words:
            words=words
    print words

if __name__ == '__main__':
run()
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1 回答 1

6

使用一套。只需将您找到的每个单词都添加到集合中；它忽略重复项。

假设您有一个迭代器，它返回文件中的每个单词（这是纯文本；HTML 会更复杂）：

def words(filename):
    with open(filename) as wordfile:
        for line in wordfile:
            for word in line.split():
                yield word

然后将它们放入 aset很简单：

wordlist = set(words("words.txt"))

如果您有多个文件，请这样做：

wordlist = set()
wordfiles = ["words1.txt", "words2.txt", "words3.txt"]

for wordfile in wordfiles:
    wordlist |= set(words(wordfile))

您还可以使用一组作为停用词。然后，您可以在事后简单地从单词列表中减去它们，这可能比在添加之前检查每个单词是否是停用词更快。

stopwords = set(["a", "an", "the"])
wordlist -= stopwords
于 2011-03-29T19:48:01.107 回答