Identify number of unique days of service use by group

One approach is to filter the data for each id, get and combine the date sequences for each row, then count the number of unique dates. Not sure what you meant about needing the output as a data frame with date variables, but I converted the result to a data frame, in the hope that it is close to what you are after. Note that in your data, row ten has a start date after the end date, so that needs fixing before the following will work. I assumed they were back-to-front.

DayTotals <- sapply(seq_along(unique(eg_data$id)), function(id_index) {
  Current_id <- unique(eg_data$id)[id_index]
  Current_id_data <- eg_data %>% filter(id == Current_id)
  Current_id_dates <- apply(Current_id_data,1,function(row) {
    seq.Date(from = as.Date(row['start_dt'],format="%m/%d/%Y"),
                      to=as.Date(row['end_dt'],format="%m/%d/%Y"),
             by="day")})
  Current_id_No_Of_Days <- Current_id_dates %>% unlist %>% unique %>% length
})

DayTotalsDF <- data.frame(id=unique(eg_data$id),
                          NoOfDays=DayTotals)

> DayTotalsDF
  id NoOfDays
1  1      402
2  2      333
3  3      298
4  4      232
5  5      268

Min and Max dates are calculated to ensure consistency of start <= end. Then mapply() used to generate a sequence of dates from seq.Date() function. These date sequences are combined into a vector using unlist() then duplicate dates removed using unique(). The length of the vector is then calculated to determine the total number of days with activity for each ID.

see: https://www.mycompiler.io/view/Jau8tbboisq

library(dplyr)

eg_data <- data.frame(
id = c(1,1,1,  2,2,  3,3,3,3,3,3,  4,4,  5,5,5,5),
start_dt = c("01/01/2016", "12/02/2016", "03/12/2017",  "02/01/2016", 
"03/04/2016",  "01/01/2016", "03/05/2016", "05/07/2016", "07/01/2016", 
"09/04/2016", "10/10/2016",  "01/01/2016", "05/28/2016",  "01/01/2016", 
"06/05/2016", "08/25/2016", "11/01/2016"),  
end_dt =   c("12/01/2016", "12/02/2016", "05/15/2017",  "05/15/2016", 
"12/29/2016",  "03/02/2016", "04/29/2016", "06/29/2016", "08/31/2016", 
"03/04/2016", "11/29/2016",  "05/31/2016", "08/19/2016",  "06/10/2016", 
"07/25/2016", "08/25/2016", "12/30/2016"))
eg_data$row_n <- 1:nrow(eg_data)

eg_data %>%
  mutate(
    start_dt = as.Date(start_dt, format = "%m/%d/%Y"),
    end_dt = as.Date(end_dt, format = "%m/%d/%Y"),
    min_date = pmin(start_dt, end_dt),
    max_date = pmax(start_dt, end_dt)
  ) %>%
  group_by(id) %>%
  summarize(
    total_days = length(unique(unlist(mapply(seq.Date, min_date, max_date, by = "day"))))
  )

the result from this is;

     id total_days
  <dbl>      <int>
1     1        402
2     2        333
3     3        298
4     4        232
5     5        268

If this isn't the wanted result please provide the wanted result and if possible explain how you arrived at that (from the sample data only)

I think this is actually a fantastic case for ivs, you just need to adjust your thinking a little bit to shift from closed intervals like [ ] to half-open ones like [ ). All you need to do is add 1 to your end dates, which "just works" in this case.

Using half-open intervals also ends up making the math work out nicely too.

(This requires dplyr 1.1.0 or higher)

library(dplyr, warn.conflicts = FALSE)
library(ivs)

df <- tibble(
  id = c(1,1,1,  2,2,  3,3,3,3,3,3,  4,4,  5,5,5,5),
  start_dt = c(
    "01/01/2016", "12/02/2016", "03/12/2017", "02/01/2016", 
    "03/04/2016", "01/01/2016", "03/05/2016", "05/07/2016", 
    "07/01/2016", "09/04/2016", "10/10/2016", "01/01/2016",
    "05/28/2016", "01/01/2016", "06/05/2016", "08/25/2016", 
    "11/01/2016"
  ),  
  end_dt = c(
    "12/01/2016", "12/02/2016", "05/15/2017", "05/15/2016", 
    "12/29/2016", "03/02/2016", "04/29/2016", "06/29/2016",
    "08/31/2016", "09/04/2016", "11/29/2016", "05/31/2016", 
    "08/19/2016", "06/10/2016", "07/25/2016", "08/25/2016", 
    "12/30/2016"
  )
)

df <- df %>%
  mutate(
    start_dt = as.Date(start_dt, format = "%m/%d/%Y"),
    end_dt = as.Date(end_dt, format = "%m/%d/%Y") + 1L
  ) %>%
  mutate(
    range = iv(start_dt, end_dt),
    .keep = "unused"
  )

df %>%
  reframe(range = iv_groups(range), .by = id) %>%
  mutate(days = as.integer(iv_end(range) - iv_start(range))) %>%
  summarise(count = sum(days), .by = id)
#> # A tibble: 5 × 2
#>      id count
#>   <dbl> <int>
#> 1     1   402
#> 2     2   333
#> 3     3   286
#> 4     4   232
#> 5     5   268

You'll notice that my answer for id 3 is different from other solutions. That is because I think you have a typo in your original dataset in row 10, where the end date is significantly before the start date:

df[10,]
#> # A tibble: 1 × 3
#>      id start_dt   end_dt    
#>   <dbl> <chr>      <chr>     
#> 1     3 09/04/2016 03/04/2016

ivs detected this automatically for me:

#> Error in `mutate()`:
#> ℹ In argument: `range = iv(start_dt, end_dt)`.
#> Caused by error in `iv()`:
#> ! `start` must be less than `end`.
#> ℹ `start` is not less than `end` at locations: `10`.