R 中的向量列表 - 提取向量的一个元素

Question

I have a list which contains some texts. So each element of the list is a text. And a text is a vector of words. So I have a list of vectors. I am doing some text-mining on that. Now, I'm trying to extract the words that are after the word "no". I transformed my vectors, so now they are vectors of two words. Such as: list(c("want friend", "friend funny", "funny nice", "nice glad", "glad become", "become no", "no more", "more guys"), c("no comfort", "comfort written", "written conduct","conduct prevent", "prevent manners", "matters no", "no one", "one want", "want be", "be fired"))

My aim is to have a list of vectors which will be like: list(c("more"), c("comfort", "one")) So I would be able to see for a text i the vectoe of results by liste[i].

So I have a formula to extract the word after "no" (in the first vector it will be "more"). But when I have several "no" in my text it doesn't work.

Here is my code:

liste_negation <- vector(length = length(data))
for (i in 1:length(data)){
  for (j in 1:length(data[[i]])){
    if (startsWith((data[[i]])[[j]], 'no') == TRUE){
      liste_neg[i] <- c(liste_neg[i], tail(strsplit((data[[i]])[[j]],split=" ")[[1]],1))
    } else{
      liste_neg[i] <- c(liste_neg[i])
    }
    liste_negation[[i]] <- c(liste_neg[[i]])
  }
}

That one works for a vector when there is only one "no":

data <- list(c("want friend", "friend funny", "funny nice", "nice glad", "glad become", "become no", "no more", "more guys"), c("no comfort", "comfort written", "written conduct","conduct prevent", "prevent manners", "matters no", "no one", "one want", "want be", "be fired"))
data

liste_neg <- c()
liste_negation <- vector(length = length(data))
if (startsWith((data[[1]])[[9]], 'no') == TRUE){
  liste_neg[1] <- c(liste_neg[1], tail(strsplit((data[[1]])[[9]],split=" ")[[1]],1))
}

liste_negation[[1]] <- c(liste_neg[[1]])

But if I try to adapt it with a loop to see each element of the vector, and there are more than one "no" in the text, it doesn't work.

Code:

liste_neg <- c()
liste_negation <- vector(length = length(data))
for (j in 1:length(data[[2]])){
  if (startsWith((data[[2]])[[j]], 'no') == TRUE){
    liste_neg[2] <- append(liste_neg[2], tail(strsplit((data[[2]])[[j]],split=" ")[[1]],1))
  }
}
liste_neg
liste_negation[[2]] <- c(liste_neg[[2]])
liste_negation

Warning message:

Warning message:
In liste_neg[2] <- append(liste_neg[2], tail(strsplit((data[[2]])[[j]],  :
  number of items to replace is not a multiple of replacement length
> liste_neg
[1] NA        "comfort"
> liste_negation[[2]] <- c(liste_neg[[2]])
> liste_negation
[1] "FALSE"   "comfort"

As you can see I have only the second word which is there.

I tried many things and I tried to split the code and run it and work on it piece by piece, but after spending all the morning on it I haven't found a solution..

Did someone have an idea top help me?

Thank you in advance (and sorry for my english, I'm french ^^')

Answer 1

In base R, we can use sapply to loop over list and grep to identify words with "no"

output <- sapply(word_vec, function(x) sub(".*no", "", grep("\\bno\\b", x, value = TRUE)))

#[[1]]
#[1] ""      " more"

#[[2]]
#[1] " comfort" ""         " one" 

If you don't need empty string you can remove them to get

sapply(output, function(x) trimws(x[x!= ""]))  
#[[1]]
#[1] "more"

#[[2]]
#[1] "comfort" "one"     

Answer 2

lapply(data, function(x) substr(x[startsWith(x, "no")], 4, 1000))


[[1]]
[1] "more"

[[2]]
[1] "comfort" "one"    

Answer 3

You could use regular expressions with capture group to obtain all substrings that match the desired pattern, then extract just the captured group as follows:

# regex for strings that start with "no " and have any text after that
r <- '^no (.*)'
lapply(data, function(x) gsub(r, '\\1', regmatches(x, regexpr(r, x))))

#output
[[1]]
[1] "more"

[[2]]
[1] "comfort" "one"  

regexpr returns a match object that regmatches will extract the matching strings from, and gsub uses the \\1 argument to extract the first captured group.

Answer 4

Steps to extract the word after "no":

• First of all, use grep(i,pattern = "^no",value = T) to get the texts which start with "no".

• gsub(pattern = "no ",replacement = "") replace "no " into "".

then you can extract the word after "no".

• lapply() can split list and apply the steps to the elements of list.

• %>% the pipe operator can make code clear and take the result of grep() into gsub() .

library(magrittr)   
lapply(data,function(i)grep(i,pattern = "^no",value = T) %>% gsub(pattern = "no ",replacement = ""))
#[[1]]
#[1] "more"
#    
#[[2]]
#[1] "comfort" "one"