R 有没有办法将基于每个值中出现的两个关键字或短语的字符串向量重新编码为具有这两个值的新向量

从你的描述听起来，你很乐意go 掉第一个单词，除非有一个以上的单词.如果没有空格，我们可以假设只有一个单词.

remove_first_word  <- function(x) {
    ifelse(
        grepl("\\s", x),
        sub(".+\\s(*?)", "\\1", x),
        x
    )  |>
    # Make first letter upper case
    gsub("^([a-z])", "\\U\\1", x = _, perl = TRUE)
}

然后您可以根据需要在mutate()中使用它:

data  |>
    mutate(
        across(w:z, remove_first_word)
    )
# # A tibble: 6 × 4
#   w        x     y                z            
#   <chr>    <chr> <chr>            <chr>        
# 1 Disagree True  Not doing enough Comfortable  
# 2 Disagree True  Doing enough     Comfortable  
# 3 Disagree False Not doing enough Comfortable  
# 4 Agree    False Doing enough     Uncomfortable
# 5 Agree    True  Not doing enough Uncomfortable
# 6 Agree    False Doing enough     Comfortable  

tidyverse version

作为对您的 comments 的回应，以下是原始函数的stringr版本:

remove_first_word_tidy  <- function(x) {
    dplyr::if_else(
        stringr::str_detect(x, "\\s"),
        stringr::str_replace(x, "\\w+\\s", ""),
        x
    )  |>
    stringr::str_to_title()
}

您可以创建一个函数，该函数获取数据框和列的列表并应用此函数.如果要使用tidyverse，我们可以使用整齐的SELECT函数，并使用purrr::map()将其应用于所有需要的列，并生成向量列表:

make_dicho  <- function(dat, cols) {

    out  <- dat  |>
        select({{cols}})  |>
        purrr::map(remove_first_word_tidy)
    
    # Return vector if only one column supplied
    if(length(out)==1) return(out[[1]])
    # Otherwise return list of vectors
    out
}


make_dicho(data, w) 
# [1] "Disagree" "Disagree" "Disagree" "Agree"    "Agree"    "Agree"   

make_dicho(data, y:z)
# $y
# [1] "Not Doing Enough" "Doing Enough"     "Not Doing Enough" "Doing Enough"     "Not Doing Enough" "Doing Enough"    

# $z
# [1] "Comfortable"   "Comfortable"   "Comfortable"   "Uncomfortable" "Uncomfortable" "Comfortable"