S <- df %>% 
  group_by(animal_id, glatos_array, station_no, station,  
           deploy_lat, deploy_long,
           season_year, time_bin, 
           season_length) %>% 
  summarise(S = n()) %>% 
  ungroup %>% 
  # minumum amount of dets needed to determine if fish was resident  
  # filter(!S < 10) %>% 
  complete(animal_id,
           nesting(
             glatos_array, station_no, station,  
             deploy_lat, deploy_long,
             season_year, time_bin, 
             season_length
           ), 
           fill = list(S = 0)
  ) 
arrange(animal_id, time_bin)

S

I found another way to do this, if you are interested in having `S` equals `0` for time periods where the fish were not heard at all and considered absent. You can add in `complete()` from `tidyr` and remove `filter(!S < 10)` when you create `S`. When you use `left join()` you'll have the variable `t` which will end up with many `NA` values which can be replaced to `0` using the following `ri_daily[is.na(ri_daily)] <- 0`. ``` S <- df %>% group_by(animal_id, glatos_array, station_no, station, deploy_lat, deploy_long, season_year, time_bin, season_length) %>% summarise(S = n()) %>% ungroup %>% # minumum amount of dets needed to determine if fish was resident # filter(!S < 10) %>% complete(animal_id, nesting( glatos_array, station_no, station, deploy_lat, deploy_long, season_year, time_bin, season_length ), fill = list(S = 0) ) arrange(animal_id, time_bin) S ```