Question

我是Python的新手，这是我的问题。我有一套词：

entities = ['blab', 'r1', 'zss']

我想检测它们，我想在没有

的情况下封装它们

For Instance：

this r1 is about zsse - ＆gt; this [r1] is about [zsse]

如果已经封装，我将不会更改任何内容，例如，[ blablab r1 blabala ]仍然是相同的。

我尝试了一些东西，但它不起作用：

for s in sentences:
    for e in entities:
        if re.search(r"\[\[%s\]\]" % e, s):
            pass
        else:
            s=s.replace(e,'[['+e+']]')

        New_sentences.append(s)

Answer 1

我就是这样做的。请注意，我使用两种不同的正则表达式：

(\[.*?])确定括号内的区域
'({})'.format('|'.join(entities))匹配非括号内区域内的任何实体。

import re

brackets = re.compile(r'(\[.*?])')
def rewrite(sentence, entities):
    sentence = brackets.split(sentence)
    entities = re.compile('({})'.format('|'.join(entities)))
    for i, phrase in enumerate(sentence):
        if not phrase.startswith('['):
            sentence[i] = entities.sub(r'[\1]', phrase)
    sentence = ''.join(sentence)
    return sentence

print rewrite('this r1 is about zsse', ['blab', 'r1', 'zss'])
print rewrite('[ blablab r1 blabala ]', ['blab', 'r1', 'zss'])

结果：

$ python x.py 
this [r1] is about [zss]e
[ blablab r1 blabala ]

Answer 2

如果您只关心entities是句子中单词的子串，那么您就不必使用正则表达式。

sentences = ['this r1 is about zsse', '[blablab r1 blabala]']
entities = ['blab', 'r1', 'zss']
new_sentences = []

for sentence in sentences:
    if sentence.startswith('[') and sentence.endswith(']'):
        new_sentences.append(sentence)
        continue

    sentence = sentence.split(' ')

    for index, word in enumerate(sentence):
        for entity in entities:
            if entity in word:
                sentence[index] = '[{w}]'.format(w=word)

    new_sentences.append(' '.join(sentence))

print new_sentences
>>> ['this [r1] is about [zsse]', '[blablab r1 blabala]']

Answer 3

试试这段代码：

import re

entities = ['blab', 'r1', 'zss']
sentences = ['this r1 is about zsse', 'this [r1] is about [zss]e']
new_sentences = []

for s in sentences:
    for e in entities:
        if re.search(r"\[%s\]" % e, s):
            pass
        else:
            s=s.replace(e,'['+e+']')
    new_sentences.append(s)

print(new_sentences)
# >>> ['this [r1] is about [zss]e', 'this [r1] is about [zss]e']

Demo on Ideone

您的代码唯一的问题是new_sentences.append(s)缩进太多了。每次new_sentences循环播放时，.append都被entities版。因此，对于每个sentence，有3 new_sentences。

您似乎还拥有[[]]或\[\[\]\]而非[]或\[\]

的所有内容

Answer 4

你可以试试这个：

In [48]: eentities = ['blab', 'r1', 'zss']

In [49]: s = 'this r1 is about zsse'

In [50]: import re

In [52]: rs = re.compile('|'.join(eentities))

In [60]: sl = list()

for se in s.split():
    if(rs.match(se)):
        sl.append('[{0}]'.format(se))
    else:
        sl.append(se)

In [62]: sl
Out[62]: ['this', '[r1]', 'is', 'about', '[zsse]']

In [63]: ' '.join(sl)
Out[63]: 'this [r1] is about [zsse]'

如何将特定单词封装在括号中？

4 个答案: