从 R 中的字符串中删除指定的模式 [英] Remove specified pattern from string in R

问题描述

我有一个像下面这样的字符串

I have a string like the following

s <- "abc a%bc 1.2% 234 1.2 (1.4%)) %3ed"

我想删除所有带有 %.所以结果是

I would like to remove all "words" which have the %. So the result would be

"abc 234 1.2"

推荐答案

可以使用

> gsub("^\\s+|\\s+$", "", (gsub("\\s+", " " ,gsub("\\s+\\S*%\\S*(?=\\s+|$)", " ",input, perl=TRUE))))
#[1] "abc 234 1.2"

代码分解

gsub("^\\s+|\\s+$", "", (gsub("\\s+", " " ,gsub("\\s+\\S*%\\S*(?=\\s+|$)", " ",input, perl=TRUE))))
                                           <--------------------------------------------------->
                                                     Remove strings with %
                        <------------------------------------------------------------------------>
                        Substitute extra spaces with single space from resultant string obtained from above
<-------------------------------------------------------------------------------------------------->
      Trim initial and final whitespaces from the string obtained from above

正则表达式分解

\\s+ #Match whitespaces
\\S* #Match all non whitespace character before % if its there
% #Match % literally
\\S* #Match all non whitespace character after % if its there
(?=\\s+|$) #Lookahead to check whether there is a space or end of string after matching word with %

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