Question

在分析推文的过程中，我会在“单词”中运行，这些单词具有\或/（在一个“单词”中可能有多个外观）。我想完全删除这些词，但不能完全指出这个

这就是我的尝试：

sen = 'this is \re\store and b\\fre'
sen1 = 'this i\s /re/store and b//fre/'

slash_back =  r'(?:[\w_]+\\[\w_]+)'
slash_fwd = r'(?:[\w_]+/+[\w_]+)'
slash_all = r'(?<!\S)[a-z-]+(?=[,.!?:;]?(?!\S))'

strt = re.sub(slash_back,"",sen)
strt1 = re.sub(slash_fwd,"",sen1)
strt2 = re.sub(slash_all,"",sen1)
print strt
print strt1
print strt2

我想得到：

this is and
this i\s and
this and

但是，我收到了：

and 
this i\s / and /
i\s /re/store  b//fre/

要添加：在这种情况下，“单词”是由空格或标点符号（如常规文本）分隔的字符串

Answer 1

这是怎么回事？我添加了一些标点符号示例：

import re

sen = r'this is \re\store and b\\fre'
sen1 = r'this i\s /re/store and b//fre/'
sen2 = r'this is \re\store, and b\\fre!'
sen3 = r'this i\s /re/store, and b//fre/!'

slash_back =  r'\s*(?:[\w_]*\\(?:[\w_]*\\)*[\w_]*)'
slash_fwd = r'\s*(?:[\w_]*/(?:[\w_]*/)*[\w_]*)'
slash_all = r'\s*(?:[\w_]*[/\\](?:[\w_]*[/\\])*[\w_]*)'

strt = re.sub(slash_back,"",sen)
strt1 = re.sub(slash_fwd,"",sen1)
strt2 = re.sub(slash_all,"",sen1)
strt3 = re.sub(slash_back,"",sen2)
strt4 = re.sub(slash_fwd,"",sen3)
strt5 = re.sub(slash_all,"",sen3)
print(strt)
print(strt1)
print(strt2)
print(strt3)
print(strt4)
print(strt5)

输出：

this is and
this i\s and
this and
this is, and!
this i\s, and!
this, and!

Answer 2

没有re的一种方法就是使用join和理解。

sen = 'this is \re\store and b\\fre'
sen1 = 'this i\s /re/store and b//fre/'

remove_back = lambda s: ' '.join(i for i in s.split() if '\\' not in i)
remove_forward = lambda s: ' '.join(i for i in s.split() if '/' not in i)

>>> print(remove_back(sen))
this is and
>>> print(remove_forward(sen1))
this i\s and
>>> print(remove_back(remove_forward(sen1)))
this and

删除带有特殊字符“\”和“/”的单词

2 个答案: