Question

我在R中有一个停用词的字符向量：

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

假设我有字符串：

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

如何从str删除我定义的停用词？

我认为gsub或其他grep工具可能是一个很好的选择，尽管其他建议值得欢迎。

Answer 1

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，因此无需为此目的加载外部包：

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

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

Answer 2

您可以使用tm库：

date_select

Answer 3

如果stopwords长，则removeWords() solution应该比任何基于正则表达式的解决方案都快得多。

为完整起见，如果str是字符串的向量，则可以这样写：

library("magrittr")
library("stringr")
library("purrr")

remove_words <- function(x, .stopwords) {
  x %>%
    stringr::str_split(" ") %>%
    purrr::flatten_chr() %>%
    setdiff(.stopwords) %>%
    stringr::str_c(collapse = " ")
}
purrr::map_chr(str, remove_words, .stopwords = stopwords)

从字符串中删除字符向量中的单词

3 个答案: