Number of continuous weeks by group - r

Question

How do I find number of continuous weeks by group but counted from the max date in the dataset?
Say I have this dataframe:
id Week
1 A 2/06/2019
2 A 26/05/2019
3 A 19/05/2019
4 A 12/05/2019
5 A 5/05/2019
6 B 2/06/2019
7 B 26/05/2019
8 B 12/05/2019
9 B 5/05/2019
10 C 26/05/2019
11 C 19/05/2019
12 C 12/05/2019
13 D 2/06/2019
14 D 26/05/2019
15 D 19/05/2019
16 E 2/06/2019
17 E 19/05/2019
18 E 12/05/2019
19 E 5/05/2019
My desired output is:
id count
1: A 5
2: B 2
3: D 3
4: E 1
I am currently converting dates into factor to get ordered number and checking against the reference number created based on the number of rows in each group.
library(data.table)
df <- structure(list(id = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 5L),
.Label = c("A", "B", "C", "D", "E"), class = "factor"),
Week = structure(c(3L, 4L, 2L, 1L, 5L, 3L, 4L, 1L, 5L, 4L, 2L, 1L, 3L, 4L, 2L, 3L, 2L, 1L, 5L),
.Label = c("12/05/2019", "19/05/2019", "2/06/2019", "26/05/2019", "5/05/2019"), class = "factor")),
class = "data.frame", row.names = c(NA, -19L))
dt <- data.table(df)
dt[, Week_no := as.factor(as.Date(Week, format = "%d/%m/%Y"))]
dt[, Week_no := factor(Week_no)]
dt[, Week_no := as.numeric(Week_no)]
max_no <- max(dt$Week_no)
dt[, Week_ref := max_no:(max_no - .N + 1), by = "id"]
dt[, Week_diff := Week_no - Week_ref]
dt[Week_diff == 0, list(count = .N), by = "id"]

Answer 1

Here's one way to do this:
dt <- dt[, Week := as.Date(Week, format = "%d/%m/%Y")]
ids_having_max <- dt[.(max(Week)), id, on = "Week"]
dt <- dt[.(ids_having_max), on = "id"
][order(-Week), .(count = sum(rleid(c(-7L, diff(Week))) == 1)), by = "id"]
Breaking it into steps:
We leave Week as a date because it can already be compared,
and you can subtract dates to get time differences.
We then get all the ids that contain the maximum date in the whole table.
This is using secondary indices.
We use secondary indices again to filter out those ids that were not part of the previous result
(the dt[.(ids_having_max), on = "id" part).
The last frame is tricky.
We group by id and make sure that rows are ordered by Week in descending order.
Then the logic is as follows.
When you have contiguous weeks,
diff(Week) is always -7 with the chosen sorting.
Computing diff returns a shorter vector because the first result is computed by subtracting the first input element from the second,
so we prepend a -7 to make sure that it is the first element in the input to rleid.
With rleid we assign a 1 to the first -7 and keep the 1 until we see something different from -7.
Something different means weeks stopped being contiguous.
The sum(rleid(c(-7L, diff(Week))) == 1) will simply return how many rows had a rleid equal to 1.
Example of the last part for B:
Differences: -7, -14, -7
After prepending -7: -7, -7, -14, -7
After rleid: 1, 1, 2, 3
From the previous, two had a rleid == 1

Answer 2

Apologies for dplyr solution, but I presume a similar approach can be achieved more concisely with data.table.
library(dplyr)
df$Week = lubridate::dmy(df$Week)
df %>%
group_by(id) %>%
arrange(id, Week) %>%
# Assign group to each new streak
mutate(new_streak = cumsum(Week != lag(Week, default = 0) + 7)) %>%
add_count(id, new_streak) %>%
slice(n()) # Only keep last week

Answer 3

So I would suggest converting the format of the data column to show week number "%W" as follows
dt[, Week_no := format(as.Date(Week, format = "%d/%m/%Y"),"%W")]
Then find the amount of unique week numbers for each id value
dt[,(length(unique(Week_no))),by="id"]
FULL DISCLOSURE
I realise that when I run this I get a different table than you present, as R counts the week by the week # for the given year.
If this doesnt answer your question just let me know and I can try to update