d %>%
  group_by(id) %>%
  mutate(Other_count = cumsum(rater == "Other"),
         rater = if_else(rater == "Other" & Other_count > 1,
                         "Other2",  # EDIT
                         rater)) %>%
  ungroup() %>%
  select(-Other_count)

结果如下。根据您的描述，我怀疑您最后两行所需的输出中有错别字。第10行似乎是ID 2的第三个其他，第11行是ID 3的第一个其他。

# A tibble: 11 × 3
      id rater  raterscore
   <int> <chr>       <dbl>
 1     1 rater2        3  
 2     1 rater1        2  
 3     1 Other         3  
 4     2 rater1        1.5
 5     2 rater2        3.2
 6     2 Other         1.1
 7     2 Other2        2  
 8     3 rater1        2.5
 9     2 rater2        2.7
10     2 Other2        2.1
11     3 Other         2

我怀疑排印错误是在id列第9行和第10行应该是3：
我们可以这样做：

library(dplyr)

df %>% 
  group_by(id) %>% 
  mutate( x = cumsum(rater == "Other"),
          x = ifelse(x == 2, 2, "")) %>% 
  mutate(rater = paste0(rater, x), .keep="unused")

使用dplyr >= 1.1.0更优雅

df %>% 
  mutate( x = cumsum(rater == "Other"),
          x = ifelse(x == 2, 2, ""), .by=id) %>% 
  mutate(rater = paste0(rater, x), .keep="unused")

id rater  raterscore
   <int> <chr>       <dbl>
 1     1 rater2        3  
 2     1 rater1        2  
 3     1 Other         3  
 4     2 rater1        1.5
 5     2 rater2        3.2
 6     2 Other         1.1
 7     2 Other2        2  
 8     3 rater1        2.5
 9     3 rater2        2.7
10     3 Other         2.1
11     3 Other2        2