Regular expression parsed with grepl replacement

问题

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

Consider this example:

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

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"

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"

which is NOT the expected result:

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

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

回答1:

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" 

or

> 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" 

回答2:

You could also do

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

Or

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

回答3:

Try this:

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

来源：https://stackoverflow.com/questions/31625706/regular-expression-parsed-with-grepl-replacement