试试这个：

str <- c("I have zero a accordance")

stopwords = c("a", "able", "about", "above", "abst", "accordance", "yourself",
"yourselves", "you've", "z", "zero")

x <- unlist(strsplit(str, " "))

x <- x[!x %in% stopwords]

paste(x, collapse = " ")

# [1] "I have"

• • 添加：**编写"removeWords"函数很简单，因此不需要为此加载外部包：

removeWords <- function(str, stopwords) {
  x <- unlist(strsplit(str, " "))
  paste(x[!x %in% stopwords], collapse = " ")
}

removeWords(str, stopwords)
# [1] "I have"

您可以使用tm库来执行此操作：

require("tm")
removeWords(str,stopwords)
#[1] "I have   "

如果你想让代码向量化很多句子，而不仅仅是一个句子，这里有另一个函数选项，它借用了Mikko最初答案的内容。

remove_words <- function(str, words) {
      
  purrr::map_chr(
    str, 
    function(sentence) {
      sentence_split <- unlist(strsplit(sentence, " "))
      paste(sentence_split[!sentence_split %in% words], collapse = " ")
    }
  )
      
}
    
remove_words(c('Hello world', 'This is another sentence', 'Test sentence 3'), c('world', 'sentence'))