用grepl替换解析正则表达式 [英] Regular expression parsed with grepl replacement

问题描述

目标是解析正则表达式并替换匹配的模式.

The objective is to parse a regular expression and replace the matched pattern.

请考虑以下示例:

data <- c("cat 6kg","cat g250", "cat dog","cat 10 kg") 

我必须找到所有出现的cat和一个数字[0-9].为此:

I have to locate all occurrences of cat and a number [0-9]. To do this:

found <- data[grepl("(^cat.[a-z][0-9])|(^cat.[0-9])",data)]
found
[1] "cat 6kg"   "cat g250"  "cat 10 kg"

下一步是将found的每个元素替换为字符串cat.我已尝试根据堆栈问题20219311从软件包(gsubfn)中尝试 gsub ， sub 和 gsubfn():

The next step is to replace each element of found with string cat. I have attempted gsub, sub, and gsubfn() from package (gsubfn) according to Stack question 20219311:

gsubfn("((^cat.[a-z][0-9])|(^cat.[0-9]))", "cat",data)
[1] "catkg"   "cat50"   "cat dog" "cat0 kg"

这不是预期的结果:

[#] "cat" "cat" "cat dog" "cat"

我想我遗漏了一点.我将不胜感激.谢谢.

I think I'm missing a point. I would appreciate any help I could get. Thanks.

推荐答案

简单，只需将字符串cat分配给match元素.这将用cat

Simple,,,, Just assign the string cat to the match elements. This will replace all the chars present in the element with cat

> data <- c("cat 6kg","cat g250", "cat dog","cat 10 kg") 
> data[grepl("(^cat.[a-z][0-9])|(^cat.[0-9])",data)] <- "cat"
> data
[1] "cat"     "cat"     "cat dog" "cat" 

或

> data <- c("cat 6kg","cat g250", "cat dog","cat 10 kg") 
> data[grepl("^cat.[a-z]?[0-9]",data)] <- "cat"
> data
[1] "cat"     "cat"     "cat dog" "cat" 

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