Add empty rows for gaps between subscriptions - r

Question

I have been struggling with this for a while now and I haven't been able to find a comparable question asked anywhere, hence my first question on here!
I'm fairly new to R so please excuse any obvious errors I have made.
I have a dataset which has a row for each subscription that a user has or has had. Some users have multiple rows, while some others only have one. Only active or previously active subscriptions are present.
I have two variables which state when the subscription has started and when it ended called, Begindate and Enddate respectively. I already have relationlength variables created which state the amount of days between these two variables for each type of subscription. This means that the relationlength variables only give the amount of days for when a subscription was active.
What I would like to do is create empty rows in between the different subscription rows for the time periods in which no subscription was active, starting from the earliest Begindate known for the specific user and ending on a given date where all subscriptions end (20-04-2022).
I have tried to compare the date difference from the first begindate known for a user and the final date and subtracting the relation length known for the other subscription types. However, I could not make this work.
An example of what the df currently looks like:
(rl standing for relationlength)
ID Begindate Enddate Subscrtype active rl_fixed rl_promotional Productgroup
1 2019-08-26 2022-04-20 fixed 1 968 0 1
1 2018-08-24 2019-08-23 fixed 0 364 0 1
1 2015-08-24 2016-08-23 promo 0 0 364 2
2 2019-08-26 2019-09-12 fixed 0 17 0 1
2 2018-08-24 2019-08-23 fixed 0 364 0 1
What I would like it to look like:
ID Begindate Enddate Subscrtype active rl_fixed rl_promo rl_none Productgroup
1 2019-08-26 2022-04-20 fixed 1 968 0 0 1
1 2019-08-24 2019-08-25 none 0 0 0 2 NA
1 2018-08-24 2019-08-23 fixed 0 364 0 0 1
1 2016-08-24 2018-08-23 none 0 0 0 729 NA
1 2015-08-24 2016-08-23 promo 0 0 364 0 2
2 2019-09-13 2022-04-20 none 0 0 0 950 NA
2 2019-08-26 2019-09-12 fixed 0 17 0 0 1
2 2019-08-24 2019-08-25 none 0 0 0 2 NA
2 2018-08-24 2019-08-23 fixed 0 364 0 0 1
The end goal is to aggregate and have a clear overview of the specific relation lengths for the different types of relations possible for a user.
Thank you in advance!
dput for one specific user in the real df:
structure(list(ï..CRM.relatienummer = structure(c(1L, 1L, 1L,
1L, 1L, 1L), .Label = "1", class = "factor"), Begindatum = c("2019-08-26",
"2018-08-24", "2017-08-24", "2016-08-24", "2015-08-20", "2016-06-01"
), Einddatum = c("2022-04-20", "2019-08-23", "2018-08-23", "2017-08-23",
"2016-05-31", "2016-08-19"), Type.abonnement = structure(c(1L,
1L, 1L, 1L, 1L, 1L), .Label = "Actie", class = "factor"), Status_dummy = c(1,
0, 0, 0, 0, 0), relationlength_fixed = c(0, 0, 0, 0, 0, 0), relationlength_promo = c(968,
364, 364, 364, 285, 79), relationlength_trial = c(0, 0, 0, 0,
0, 0), fixed_dummy = c(0, 0, 0, 0, 0, 0), trial_dummy = c(0,
0, 0, 0, 0, 0), promotional_dummy = c(1, 1, 1, 1, 1, 1)), row.names = c("1:20610",
"2:38646", "2:39231", "2:39232", "2:39248", "2:39837"), class = "data.frame")
Edit:
I have tried to run this code:
dfs <- split(testdata,testdata$ï..CRM.relatienummer)
r <- lapply(seq(length(dfs)), function(k){
v <- dfs[[k]]
vt <- data.frame(unique(v$ï..CRM.relatienummer),
as.character((as.Date(v$Einddatum)+1)[-1]),
as.character((as.Date(v$Begindatum)-1)[-nrow(v)]),
0,
0,
0,
0,
(as.Date(v$Begindatum)-1)[-nrow(v)] - (as.Date(v$Einddatum)+1)[-1],
NA,
0,
0,
0,
0,
0)
colnames(vt) <- c(colnames(v)[-ncol(v)],"rl_none",colnames(v)[ncol(v)])
(testdata <- rbind(data.frame(v[-ncol(v)],rl_none = 0,v[ncol(v)]),vt))[order(as.Date(testdata$Begindatum),decreasing = T),]
})
res <- data.frame(Reduce(rbind,r),row.names = NULL)
On this dataframe, with no luck unfortunately:
structure(list(ï..CRM.relatienummer = structure(c("d45248b8974dc4f8ff948779e0fd07e20f304e929ada4e14c0420aebed81e9b5",
"2ab04e80b3e64601147df977d6054c04ffa80014b3691b25dd1cc8ef85cea06a",
"2ab04e80b3e64601147df977d6054c04ffa80014b3691b25dd1cc8ef85cea06a",
"bcf2c99e6dc974380f967204b9623dce2c8a3fad694dc0b4430fcbf77f8f39f3",
"bcf2c99e6dc974380f967204b9623dce2c8a3fad694dc0b4430fcbf77f8f39f3",
"f8610cd0237858ac9384d6ba209759ae306860ffabb3f8e6c3d6fc68dbaddc51",
"e5b8b3f46165e48aec8bbe65ed1cb29d18a0492fbcac44803372f672348459db",
"c737815b2365b01a8a85c380364a0f721685a131de98cd7790b4d40bb8c4e05b",
"b9c0272caa8d5d3497d28cce3bda5d3d17c22f18c5f65c5e82c572b410a8ea71",
"b9c0272caa8d5d3497d28cce3bda5d3d17c22f18c5f65c5e82c572b410a8ea71",
"539c6c3e604245008daefbe500ff29357bee91f82a7896126bd0f69848524cb7",
"d361338bed51cb9c8aa73fd8914cbf392f4e05e7b073f637f7b150cf02b89c8c",
"505d3df3f1298e07aa96073490b72acd2391da06ad4cfbd5a9fbde3a3de79684",
"826443481cbb5b4e061040d443a0ce8d94322615d8ffae1e68b2ff7d896afcf7",
"2b59a1ec028c261c0f22cd6a49220dc7cec9a9fb0fabe2296b4ba77a60cfdaae"
), class = c("hash", "sha256")), Begindatum = c("2019-06-14",
"2019-03-01", "2019-09-02", "2019-03-03", "2019-04-01", "2019-09-21",
"2019-02-02", "2019-06-11", "2019-02-05", "2019-02-09", "2019-07-24",
"2019-05-08", "2019-09-27", "2019-08-03", "2019-04-03"), Einddatum = c("2022-04-20",
"2019-09-01", "2022-04-20", "2019-03-31", "2022-04-20", "2022-04-20",
"2019-02-14", "2019-07-08", "2019-02-11", "2020-02-08", "2019-09-03",
"2019-06-18", "2019-11-07", "2019-08-16", "2022-04-20"), Status_dummy = c(1,
0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1), relationlength_fixed = c(0,
184, 961, 28, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0), relationlength_promo = c(1041,
0, 0, 0, 1115, 942, 12, 0, 0, 364, 0, 0, 0, 0, 1113), relationlength_trial = c(0,
0, 0, 0, 0, 0, 0, 27, 0, 0, 41, 41, 41, 13, 0), rl_none = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), fixed_dummy = c(0,
1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0), trial_dummy = c(0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0), promotional_dummy = c(1,
0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1), active_subscr_dummy = c(3,
0, 5, 0, 3, 3, 0, 0, 0, 3, 0, 0, 1, 0, 3), hashedEmail = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c("1:1",
"1:2", "1:3", "1:4", "1:5", "1:6", "1:7", "1:8", "1:9", "1:10",
"1:11", "1:12", "1:13", "1:14", "1:15"), class = "data.frame")

Answer 1

Hopefully this is what you are expecting
dfs <- split(df,df$ID)
r <- lapply(seq(length(dfs)), function(k){
v <- dfs[[k]]
vt <- data.frame(unique(v$ID),
as.character((as.Date(v$Enddate)+1)[-1]),
as.character((as.Date(v$Begindate)-1)[-nrow(v)]),
"none",
0,
0,
0,
(as.Date(v$Begindate)-1)[-nrow(v)] - (as.Date(v$Enddate)+1)[-1],
NA)
colnames(vt) <- c(colnames(v)[-ncol(v)],"rl_none",colnames(v)[ncol(v)])
(df <- rbind(data.frame(v[-ncol(v)],rl_none = 0,v[ncol(v)]),vt))[order(as.Date(df$Begindate),decreasing = T),]
})
res <- data.frame(Reduce(rbind,r),row.names = NULL)
which gives
> res
ID Begindate Enddate Subscrtype active rl_fixed rl_promo rl_none Productgroup
1 1 2019-08-26 2022-04-20 fixed 1 968 0 0 1
2 1 2019-08-24 2019-08-25 none 0 0 0 1 NA
3 1 2018-08-24 2019-08-23 fixed 0 364 0 0 1
4 1 2016-08-24 2018-08-23 none 0 0 0 729 NA
5 1 2015-08-24 2016-08-23 promo 0 0 364 0 2
6 2 2019-08-26 2019-09-12 fixed 0 17 0 0 1
7 2 2019-08-24 2019-08-25 none 0 0 0 1 NA
8 2 2018-08-24 2019-08-23 fixed 0 364 0 0 1
DATA
structure(list(ID = c(1L, 1L, 1L, 2L, 2L), Begindate = structure(c(3L,
2L, 1L, 3L, 2L), .Label = c("2015-08-24", "2018-08-24", "2019-08-26"
), class = "factor"), Enddate = structure(c(4L, 2L, 1L, 3L, 2L
), .Label = c("2016-08-23", "2019-08-23", "2019-09-12", "2022-04-20"
), class = "factor"), Subscrtype = structure(c(1L, 1L, 2L, 1L,
1L), .Label = c("fixed", "promo"), class = "factor"), active = c(1L,
0L, 0L, 0L, 0L), rl_fixed = c(968L, 364L, 0L, 17L, 364L), rl_promo = c(0L,
0L, 364L, 0L, 0L), Productgroup = c(1L, 1L, 2L, 1L, 1L)), class = "data.frame", row.names = c(NA,
-5L))