用grepl替换解析的正则表达式

Question

目的是解析正则表达式并替换匹配的模式。

考虑这个例子:

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

我必须找到所有出现的cat和数字[0-9]。要做到这一点:

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。我已经尝试过gsub， sub和gsubfn()从包(gsubfn)根据堆栈问题20219311:

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

不是预期的结果:

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

我想我漏掉了一点。我将感激任何我能得到的帮助。谢谢。

Answer 1

简单，，，，只需将字符串cat分配给匹配元素。这将用cat

替换元素中存在的所有字符

> 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" 

Answer 2

你也可以做

 sub('\s*dog(*SKIP)(*F)|(?<=cat).*', '', data, perl=TRUE)
 #[1] "cat"     "cat"     "cat dog" "cat"    

或

 sub('(cat)\s*([0-9]|[a-z][0-9]).*$', '\1', data)
 #[1] "cat"     "cat"     "cat dog" "cat"    

Answer 3

试试这个:

gsub('(\w?[0-9].*)','',data)
#[1] "cat "    "cat "    "cat dog" "cat "