Question

假设我有以下data.frame：

df <- data.frame(string=c("word1 word2 word3 word4", "word1 word2", "word1"), stringsAsFactors = FALSE)

我想在列表中（或每行）派生前n个/后n个单词（从1到单词数的n个）的串联。预期结果：

list(
string1=c('left1'="word1", 'left2'= "word1 word2", 'left3'="word1 word2 word3",
          'left4'="word1 word2 word3 word4",
        'right1'="word4", 'right2'="word3 word4", 'right3'="word2 word3 word4"),
string2= c('left1'="word1", 'left2'="word1 word2", 'right1'="word2"),
string3="word1")

（根本不需要元素名称，但有助于理解）。

不需要：粘贴诸如“ word2 word3”之类的中间元素。

我目前使用strsplit(df$string)准备所需列表的第一步，然后可以通过双循环实现我想要的功能，但这远没有效率。

在基本R / data.table中首选方法，但使用tidyverse高效的解决方案就可以了。

Answer 1

基本R版本：

我们可以编写一个函数，将每次添加每个单词的值增量粘贴。

paste_words <- function(x) {
   sapply(seq_along(x), function(y) paste0(x[1:y], collapse = " "))
}

lapply(strsplit(df$string, " "), function(x) c(paste_words(x), paste_words(rev(x))))


#[[1]]
#[1] "word1"    "word1 word2"    "word1 word2 word3"   "word1 word2 word3 word4"
#[5] "word4"    "word4 word3"     "word4 word3 word2"  "word4 word3 word2 word1"

#[[2]]
#[1] "word1"    "word1 word2" "word2"    "word2 word1"

#[[3]]
#[1] "word1" "word1"

您可能希望包装unique以避免重复类似最后一个元素中的单词。

Answer 2

一个dplyr，tidyr和purrr选项可以是：

df %>%
 rowid_to_column() %>%
 separate_rows(string, sep = " ") %>%
 group_by(rowid) %>%
 transmute(concatenated = accumulate(string, ~ paste(.x, .y)),
           concatenated_rev = accumulate(rev(string), ~ paste(.x, .y)))

  rowid concatenated            concatenated_rev       
  <int> <chr>                   <chr>                  
1     1 word1                   word4                  
2     1 word1 word2             word4 word3            
3     1 word1 word2 word3       word4 word3 word2      
4     1 word1 word2 word3 word4 word4 word3 word2 word1
5     2 word1                   word2                  
6     2 word1 word2             word2 word1            
7     3 word1                   word1

或具有更多左/右信息：

df %>%
 rowid_to_column() %>%
 separate_rows(string, sep = " ") %>%
 group_by(rowid) %>%
 transmute(left = paste0("left", 1:n()),
           concatenated = accumulate(string, ~ paste(.x, .y)),
           right = paste0("right", 1:n()),
           concatenated_rev = accumulate(rev(string), ~ paste(.x, .y)))

  rowid left  concatenated            right  concatenated_rev       
  <int> <chr> <chr>                   <chr>  <chr>                  
1     1 left1 word1                   right1 word4                  
2     1 left2 word1 word2             right2 word4 word3            
3     1 left3 word1 word2 word3       right3 word4 word3 word2      
4     1 left4 word1 word2 word3 word4 right4 word4 word3 word2 word1
5     2 left1 word1                   right1 word2                  
6     2 left2 word1 word2             right2 word2 word1            
7     3 left1 word1                   right1 word1

Answer 3

感谢Ronak的方法（谢谢），我得到了以下代码。比我的循环更加优雅和高效。

paste_words_left <- function(x) {
 sapply(seq_along(x), function(y) paste0(x[1:y], collapse = " "))
}

paste_words_right <- function(x) {
 sapply(seq_along(x)[-1], function(y) paste0(x[y:length(x)], collapse = " "))
}

## lapply(strsplit(df$string, " "), function(x) c(paste_words_left(x), paste_words_right(x)))

lapply(strsplit(df$string, " "), function(x){
  if (length(x)==1) x else  c(paste_words_left(x), paste_words_right(x))})

串联字符串中的前n个/后n个单词

3 个答案: