根据R中不同id的前一个函数对下一行应用函数

Question

我想为不同的df-id应用一个函数，如果下一行不同于前一行，则返回1，否则返回0。

我完全不知道我该如何实现这一目标，无论如何都可以完成它吗？

数据结构如下:

df <- structure(list(
      id = c(2,2,3,3,4,4,5,5),
      job_position = c("Analyst", "Supervisor",
                       "HRBP", "HRBP", "Economist",
                       "Financial Planner", "Reporter",
                       "Reporter")),
      class = "data.frame",
      row.names = c(NA, -8L))
))

#  id    job_position
#   2       Analyst
#   2      Supervisor
#   3        HRBP
#   3        HRBP
#   4      Economist
#   4      Financial Planner
#   5       Reporter
#   5       Reporter

预期输出应如下所示：

#  id    job_position         changing_job_position
#   2       Analyst                   0
#   2      Supervisor                 1
#   3        HRBP                     0
#   3        HRBP                     0
#   4      Economist                  0
#   4      Financial Planner          1 
#   5       Reporter                  0
#   5       Reporter                  0

Answer 1

对于每个id，您可以检查当前值是否与前一个值不同。

library(dplyr)

df %>%
  group_by(id) %>%
  mutate(changing_job_position = +(job_position != lag(job_position, 
                                   default = first(job_position))))

#    id job_position      changing_job_position
#  <dbl> <chr>                             <int>
#1     2 Analyst                               0
#2     2 Supervisor                            1
#3     3 HRBP                                  0
#4     3 HRBP                                  0
#5     4 Economist                             0
#6     4 Financial Planner                     1
#7     5 Reporter                              0
#8     5 Reporter                              0

使用＜code＞数据。table可以执行以下操作：

library(data.table)
setDT(df)[, changing_job_position := +(job_position != shift(job_position,
                                       fill = first(job_position))), id]

Answer 2

我们可以按“id”分组，根据任何< code >重复或< code>n_distinct(job_position)创建“changing_job_position”

library(dplyr)
df %>%
    group_by(id) %>% 
    mutate(changing_job_position = + (any(duplicated(job_position)) &
          row_number() == 1)) %>% 
    ungroup %>% 
    mutate(changing_job_position = lead(changing_job_position, default = 0))

-输出

# A tibble: 8 x 3
#     id job_position      changing_job_position
#  <dbl> <chr>                             <dbl>
#1     2 Analyst                               0
#2     2 Supervisor                            1
#3     3 HRBP                                  0
#4     3 HRBP                                  0
#5     4 Economist                             0
#6     4 Financial Planner                     1
#7     5 Reporter                              0
#8     5 Reporter                              0