How to calculate sum of subsequent rows in a dataframe changing the reference row under condition

Using purrr::accumulate to find the anchordate based on your logic (if more than 28 days, use that):

library(dplyr)
df %>%
  mutate(dateric = as.Date(dateric)) %>%
  arrange(ID, dateric, .by = ID) %>%
  mutate(
    anchordate = purrr::accumulate(dateric, 
                                    ~ if (.y - .x > 28) .y else .x),
    diff.dd = as.numeric(dateric - lag(anchordate, 
                                        default = first(dateric))),
    keep = ifelse(dateric == anchordate, 1, 0),
    IDn = rownumber(),
    .by = ID
  ) %>%  
  select(ID, dateric, IDn, diff.dd, keep)
#>   ID   dateric IDn diff.dd keep
#> 1  1 2018-03-09    1       0    1
#> 2  1 2018-03-21    2      12    0
#> 3  2 2003-11-30    1       0    1
#> 4  2 2003-12-16    2      16    0
#> 5  2 2004-02-05    3      67    1
#> 6  2 2004-02-10    4       5    0
#> 7  3 2006-02-23    1       0    1
#> 8  3 2006-02-24    2       1    0
#> 9  3 2014-09-25    3    3136    1

^{Created on 2026-02-03 with reprex v2.1.1}

library(dplyr)
df |> mutate(dateric = as.Date(dateric), 
              diff.dd = as.numeric(dateric - min(dateric)), 
              IDn = 1:length(dateric),
              keep = ifelse(duplicated((br = \(d, m = min(d), r = 1)
               if (!is.na(m))
                 br(d, d[s  28][1], r + s)
               else r)(dateric)), 0, 1), .by=ID)
 ID   dateric diff.dd IDn keep

#1  1 2018-03-09       0    1    1
#2  1 2018-03-21      12    2    0
#3  2 2003-11-30       0    1    1
#4  2 2003-12-16      16    2    0
#5  2 2004-02-05      67    3    1
#6  2 2004-02-10      72    4    0
#7  3 2006-02-23       0    1    1
#8  3 2006-02-24       1    2    0
#9  3 2014-09-25    3136    3    1

df |> 
   mutate(dateric = as.Date(dateric),
          diff.dd = as.numeric(dateric - min(dateric)),
          IDn = 1:length(dateric),
          keep = {r=c(); m=min(dateric); for(j in dateric){if(as.numeric(j)-as.numeric(m) > 28) m = j; r= c(r, m)}; ifelse(duplicated(r), 0, 1)}, .by = ID)
 ID   dateric diff.dd ID_n keep
#1  1 2018-03-09       0    1    1
#2  1 2018-03-21      12    2    0
#3  2 2003-11-30       0    1    1
#4  2 2003-12-16      16    2    0
#5  2 2004-02-05      67    3    1
#6  2 2004-02-10      72    4    0
#7  3 2006-02-23       0    1    1
#8  3 2006-02-24       1    2    0
#9  3 2014-09-25    3136    3    1