我们可以对"Employment_Status"应用rleid，将相同的相邻元素分组为一个组，获取"Person'、" grp "的distinct元素，然后按paste进行分组

library(dplyr)
library(data.table)
df1 %>%
   mutate(grp = rleid(Employment_Status)) %>%
   distinct(Person, grp, .keep_all = TRUE) %>%
   group_by(Person) %>%
   summarise(Job_Sequence = str_c(Employment_Status,
     collapse = '-'), .groups = 'drop')

• 输出

# A tibble: 3 × 2
  Person Job_Sequence                                      
  <chr>  <chr>                                             
1 Bob    High School Teacher-Freelancer-High School Teacher
2 James  Actor-Producer-Investor                           
3 Peter  Singer

或者使用base R

aggregate(cbind(Job_Sequence = Employment_Status) ~ Person, 
  subset(df1, !duplicated(with(rle(Employment_Status), 
   rep(seq_along(values), lengths)))), FUN = paste, collapse = '-')

• 输出

Person                                       Job_Sequence
1    Bob High School Teacher-Freelancer-High School Teacher
2  James                            Actor-Producer-Investor
3  Peter                                             Singer

数据

df1 <- structure(list(Year = c(1990L, 1991L, 1992L, 1993L, 1990L, 1991L, 
1990L, 1991L, 1992L, 1993L, 1994L), Person = c("Bob", "Bob", 
"Bob", "Bob", "Peter", "Peter", "James", "James", "James", "James", 
"James"), Employment_Status = c("High School Teacher", "High School Teacher", 
"Freelancer", "High School Teacher", "Singer", "Singer", "Actor", 
"Actor", "Producer", "Producer", "Investor")), 
class = "data.frame", row.names = c(NA, 
-11L))