如何在R中选择重复字符串中最长的ngram？

我有一个类似于以下数据集的数据集（只是有更多行）：

正如您所看到的，有一个清晰的模式：在更改基础并再次重新启动进程之前，一次向上一个ngram添加一个单词。让我以第一个“块”为例：

 [1] "abov level"                                          
 [2] "abov level consist"                                  
 [3] "abov level consist price"                            
 [4] "abov level consist price stabil"                     
 [5] "abov level consist price stabil protract"            
 [6] "abov level consist price stabil protract period"     
 [7] "abov level consist price stabil protract period time"

对于上面的每个“块”，我只保留最长的句子/ngram。在上述情况下，我只保留第七行。对每个街区进行此操作，我会得到：

    
 [1] "abov level consist price stabil protract period time"           
 [2] "abov level consist price stabil sinc last autumn"    
 [3] "abov level consist price stabil some time"                                              
 [4] "abov over come month"                                      
 [5] "abov precis level depend futur energi price develop"

有人能帮我吗

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谢谢

您可以计算每个字符串中的字符数，并选择字符数小于上一个字符串的值

inds <- c(which(diff(nchar(x)) < 0), length(x))
x[inds]

#[1] "abov level consist price stabil protract period time"
#[2] "abov level consist price stabil sinc last autumn"    
#[3] "abov level consist price stabil some time"           
#[4] "abov over come month"                                
#[5] "abov precis level depend futur energi price develop" 

---

inds我们可以在
dplyr
中使用
filter
和
lead

library(dplyr)
tibble(x) %>%
     filter((nchar(lead(x, default = last(x))) - nchar(x)) <= 0)

库（dplyr）
tibble（x）%>%
过滤器（（nchar（lead（x，default=last（x）））-nchar（x））如果这个主题在这里被否决了，那么它可以说是代码高尔夫的主题！
library(dplyr)
tibble(x) %>%
     filter((nchar(lead(x, default = last(x))) - nchar(x)) <= 0)