使用diff创建"单位差异"，然后使用%%

library(dplyr)
df1 %>% 
   mutate(unit_diff = c(0, diff(i)),
    together = unit_diff %% 3 == 0 & unit_diff != 0)

-输出

# A tibble: 4 × 4
      i m     unit_diff together
  <dbl> <chr>     <dbl> <lgl>   
1    12 ac1           0 FALSE   
2    17 ac1           5 FALSE   
3    23 ac1           6 TRUE    
4    26 me0           3 TRUE    

如果我们想判断n个值附近的真值，请使用base R中的rleid或rle

library(数据.table)
n <- 2
df1 %>% 
   mutate(unit_diff = c(0, diff(i)),
    together = unit_diff %% 3 == 0 & unit_diff != 0) %>%
   group_by(grp = rleid(together)) %>%
   mutate(together = all(together) &  n() == n) %>%
   ungroup %>%
   select(-grp)

对于第二个更新的问题，我们可以使用正则表达式插入一些分隔符——也就是说，最初，我们捕获一个或多个非小写字母(([^a-z]+))的字符，并替换为捕获组的反向引用，后跟:(\\1:)，然后，我们在小写字母、数字和大写字母之间插入|，取下trimws结尾处的护套: ，最后将:替换为,，字母之间的一个或多个小写字母后跟一个或多个数字

gsub("([a-z]+\\d+):", "\\1,",
  trimws(gsub("(?<=[a-z][0-9])(?=[A-Z])", " | ", 
 gsub("([^a-z]+)", "\\1: ", str_), perl = TRUE), whitespace = ":\\s+"))
[1] "H3K9: me0 | S10: ph1 | K14: ac1, me0"

数据

df1 <- structure(list(i = c(12, 17, 23, 26), m = c("ac1", "ac1", "ac1", 
"me0")), class = c("tbl_df", "tbl", "数据.frame"), 
row.names = c(NA, 
-4L))