Resolved: Create dummies based on a window of observations (in R or Stata)

structure(list(id = c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 
2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3), month = c(201701, 201702, 
201703, 201704, 201705, 201706, 201707, 201708, 201809, 201810, 
201811, 201812, 201901, 201902, 201903, 201705, 201706, 201707, 
201708, 201709, 201710, 201711, 201712, 201801, 201809), desired_output = c(0, 
0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 
0, 0, 0)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-25L))

* Example generated by -dataex-. To install: ssc install dataex
clear
input str2 A str6 B str14 C
"id" "month " "desired_output"
"1"  "201701" "0"             
"1"  "201702" "0"             
"1"  "201703" "0"             
"1"  "201704" "1"             
"1"  "201705" "1"             
"1"  "201706" "0"             
"1"  "201707" "0"             
"1"  "201708" "0"             
"2"  "201809" "0"             
"2"  "201810" "0"             
"2"  "201811" "0"             
"2"  "201812" "1"             
"2"  "201901" "0"             
"2"  "201902" "0"             
"2"  "201903" "0"             
"3"  "201705" "0"             
"3"  "201706" "0"             
"3"  "201707" "0"             
"3"  "201708" "1"             
"3"  "201709" "1"             
"3"  "201710" "1"             
"3"  "201711" "1"             
"3"  "201712" "0"             
"3"  "201801" "0"             
"3"  "201809" "0"             
end

library(dplyr) # >= 1.1.0, otherwise use "group_by"

rnge <- 1:3

df %>% 
  mutate(out = if_else(row_number() %in% 
                 c(!!rnge, n() - (!!rnge - 1)), 0, 1), .by = id) %>% 
  print(n = Inf)
# A tibble: 25 × 4
      id  month desired_output   out
   <dbl>  <dbl>          <dbl> <dbl>
 1     1 201701              0     0
 2     1 201702              0     0
 3     1 201703              0     0
 4     1 201704              1     1
 5     1 201705              1     1
 6     1 201706              0     0
 7     1 201707              0     0
 8     1 201708              0     0
 9     2 201809              0     0
10     2 201810              0     0
11     2 201811              0     0
12     2 201812              1     1
13     2 201901              0     0
14     2 201902              0     0
15     2 201903              0     0
16     3 201705              0     0
17     3 201706              0     0
18     3 201707              0     0
19     3 201708              1     1
20     3 201709              1     1
21     3 201710              1     1
22     3 201711              1     1
23     3 201712              0     0
24     3 201801              0     0
25     3 201809              0     0

rnge <- 1:3

df %>% 
  group_by(id) %>% 
  mutate(out = if_else(row_number() %in% c(!!rnge, n() - (!!rnge - 1)), 0, 1)) %>% 
  ungroup()