R: How do you apply grep() in lapply()

Question

I would like to apply grep() in R, but I am not really good in lapply(). I understand that lapply is able to take a list, apply function to each members and output a list. For instance, let x be a list consists of 2 members.

> x<-strsplit(docs$Text," ")
> 
> x
[[1]]
 [1] "I"         "lovehttp"  "my"        "mum."      "I"         "love"     
 [7] "my"        "dad."      "I"         "love"      "my"        "brothers."

[[2]]
 [1] "I"         "live"      "in"        "Eastcoast" "now."      "Job.I"    
 [7] "used"      "to"        "live"      "in"        "WestCoast."

I would like to apply grep() function to remove words consisting of http. So, I would apply:

> lapply(x,grep(pattern="http",invert=TRUE, value=TRUE))

But it does not work and it says

Error in grep(pattern = "http", invert = TRUE, value = TRUE) : 
argument "x" is missing, with no default

So, I tried

> lapply(x,grep(pattern="http",invert=TRUE, value=TRUE,x))

But it says

Error in match.fun(FUN) : 
'grep(pattern = "http", invert = TRUE, value = TRUE, x)' is not a 
function, character or symbol

A help please, and thanks!

@TimBiegeleisen Initially, I wanted to remove the entire word consisting of http. So since "lovehttp" consists of "http", it would be removed. In the case where I want to remove just "http" and retain "love", is it possible? — HNSKD, Mar 14 '16 at 05:51
You need to figure out what answer you want. You are changing the requirements now. — Tim Biegeleisen, Mar 14 '16 at 05:53

RHertel · Accepted Answer · 2016-03-14T07:35:19.927

This can be done in one line:

lst <- lapply(lst, grep, pattern="http", value=TRUE, invert=TRUE)

#lst
#[[1]]
# [1] "I"         "my"        "mum."      "I"         "love"      "my"        "dad."      "I"         "love"      "my"        "brothers."
#
#[[2]]
# [1] "I"          "live"       "in"         "Eastcoast"  "now."       "Job.I"      "used"       "to"         "live"       "in"         "WestCoast."

If you don't want to remove the entire word that contains the pattern and remove only the pattern itself while retaining the rest of the word (as discussed in the comments), you can use gsub instead of grep:

lapply(lst, gsub, pattern="http", replacement="")
#[[1]]
# [1] "I"         "love"      "my"        "mum."      "I"         "love"      "my"        "dad."      "I"         "love"      "my"        "brothers."
#
#[[2]]
# [1] "I"          "live"       "in"         "Eastcoast"  "now."       "Job.I"      "used"       "to"         "live"       "in"         "WestCoast."

Tim Biegeleisen · Answer 2 · 2016-03-14T05:51:37.847

The following line of code will remove all entries from vectors in your list which contain the substring http:

repx <- function(x) {
    y <- grep("http", x)
    vec <- rep(TRUE, length(x))
    vec[y] <- FALSE
    x <- x[vec]
    return(x)
}

lapply(lst, function(x) { repx(x) })

Data:

x1 <- c("I", "lovehttp", "my", "mum.", "I", "love", "my", "dad.", "I", "love", "my", "brothers.")
x2 <- c("I", "live", "in", "Eastcoast", "now.", "Job.I", "used", "to", "live", "in", "WestCoast.")
lst <- list(x1, x2)

R: How do you apply grep() in lapply()

2 Answers2