假设我有这样的文字,
text<-c("[McCain]: We need tax policies that respect the wage earners and job creators. [Obama]: It's harder to save. It's harder to retire. [McCain]: The biggest problem with American healthcare system is that it costs too much. [Obama]: We will have a healthcare system, not a disease-care system. We have the chance to solve problems that we've been talking about... [Text on screen]: Senators McCain and Obama are talking about your healthcare and financial security. We need more than talk. [Obama]: ...year after year after year after year. [Announcer]: Call and make sure their talk turns into real solutions. AARP is responsible for the content of this advertising.")
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我想删除(编辑:删除)[和](和括号本身)之间的所有文本.最好的方法是什么?这是我使用正则表达式和stingr包的微弱尝试:
str_extract(text, "\\[[a-z]*\\]")
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谢谢你的帮助!
zx8*_*x81 21
有了这个:
gsub("\\[[^\\]]*\\]", "", subject, perl=TRUE);
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正则表达式意味着什么:
\[ # '['
[^\]]* # any character except: '\]' (0 or more
# times (matching the most amount possible))
\] # ']'
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以下应该做的伎俩.该?部队懒惰匹配,它匹配尽可能少的.前后面越好].
gsub('\\[.*?\\]', '', text)
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