Question

我有以下txt文件，其中包含每个单词的POS（Part of Speech）标记。

无需/ jj / /说/ vb，/，I / ppss / / bedz furious / jj at / in this / dt unparalleled / jj intrusion / nn on / in free / jj enterprise / nn ./。如何/ wrb dared / vbn他们/ ppss

有没有办法在没有POS标签的情况下读取文件，结果将是：

毋庸置疑，我对这种对自由企业的无与伦比的侵扰感到愤怒。多么大胆

所以，基本上我想删除/之后的任何字符。

words = re.findall('\w+',open(input_file).read())

上面的代码会删除/但是缩写如jj，ppss仍会出现。那么，如何删除/跟随任何角色。

Answer 1

这还不错吗？

>>> import re
>>> s = 'Needless/jj to/to say/vb ,/, I/ppss was/bedz furious/jj at/in this/dt unparalleled/jj intrusion/nn upon/in free/jj enterprise/nn ./.'
>>> re.sub(r'/[^\s]+','',s)
'Needless to say , I was furious at this unparalleled intrusion upon free enterprise .'

这只会删除以/开头的任何文本，直到找到空格。

Answer 2

正如Wooble建议的那样，你可以通过嵌套在列表理解中的两个分割来做到这一点：

s = 'Needless/jj to/to say/vb ,/, I/ppss was/bedz furious/jj at/in this/dt unparalleled/jj intrusion/nn upon/in free/jj enterprise/nn ./.'
print " ".join(word.split("/")[0] for word in s.split())

输出：

Needless to say , I was furious at this unparalleled intrusion upon free enterprise .

s.split()将句子分成单独的单词。 word.split("/")将英语单词（或puncutation mark）与其词性分开。 word.split("/")[0]仅选择英文单词并丢弃POS。 " ".join()将生成的英语单词列表合并为一个字符串。

Answer 3

此代码考虑了Wooble的评论以及您需要处理字符串列表，afaiu：

li = [ ('//Needless/jj to/to say/vb ,/, '
        'I/ppss was/bedz fur/ious/jj at/in this/dt '
        'unparalleled/jj intrusion/nn upon/in '
        'free/jj enterprise/nn ./. '
        'How/wrb dared/vbn they/ppss'),
       '/Before/jj to/to say/vb ,/, /I/ppss am/bedz h/a/p/p/y/jj']

import re

def clean(s,r=re.compile('(?<![\s/])/[^\s/]+(?![\S/])')):
    return r.sub('',s)

x = map(clean, li)

print '\n\n'.join(x)

结果

//Needless to say , I was fur/ious at this unparalleled intrusion upon free enterprise . How dared they

/Before to say , /I am h/a/p/p/y

PYTHON：从txt文件中删除POS标签

3 个答案: