Fill Dates based on Consecutive occurrences - r

Question

ID Date
1 1-1-2016
1 2-1-2016
1 3-1-2016
2 5-1-2016
3 6-1-2016
3 11-1-2016
3 12-1-2016
4 7-1-2016
5 9-1-2016
5 19-1-2016
5 20-1-2016
6 11-04-2016
6 12-04-2016
6 16-04-2016
6 04-08-2016
6 05-08-2016
6 06-08-2016
Expected Data Frame is based on consecutive dates pairwise
1st_Date is when he visited for first time
2nd_Date is the date after which he visited for 2 consecutive days
3rd_Date is the date after which he visited for 3 consecutive days
For e.g :
For ID = 1 , He visited first time on 1-1-2016 and his 2 consecutive visits also began on the 1-1-2016 as well as his 3rd one .
Similarly For ID = 2 , He only visited 1 time so rest will remain blank
For ID = 3 , he visited 1st Time on 6-1-2016 but visited for 2 consecutive days starting on 11-1-2016.
NOTE : This has to be done till earliest 3rd Date only
Expected Output
ID 1st_Date 2nd_Date 3rd_Date
1 1-1-2016 1-1-2016 1-1-2016
2 5-1-2016 NA NA
3 6-1-2016 11-1-2016 NA
4 7-1-2016 NA NA
5 9-1-2016 19-1-2016 NA
6 11-04-2016 11-04-2016 04-08-2016

Answer 1

Here is an attempt using dplyr and tidyr. The first thing to do is to convert your Date to as.Date and group_by the IDs. We next create a few new variables. The first one, new, checks to see which dates are consecutive. Date is then updated to give NA for those consecutive dates. However, If not all the dates are consecutive, then we filter out the ones that were converted to NA. We then fill (replace NA with latest non-na date for each ID), remove unwanted columns and spread.
library(dplyr)
library(tidyr)
df %>%
mutate(Date = as.Date(Date, format = '%d-%m-%Y')) %>%
group_by(ID) %>%
mutate(new = cumsum(c(1, diff.difftime(Date, units = 'days'))),
Date = replace(Date, c(0, diff(new)) == 1, NA),
new1 = sum(is.na(Date)),
new2 = seq(n())) %>%
filter(!is.na(Date)|new1 != 1) %>%
fill(Date) %>%
select(-c(new, new1)) %>%
spread(new2, Date) %>%
select(ID:`3`)
# ID `1` `2` `3`
#* <int> <date> <date> <date>
#1 1 2016-01-01 2016-01-01 2016-01-01
#2 2 2016-01-05 <NA> <NA>
#3 3 2016-01-06 2016-01-11 <NA>
#4 4 2016-01-07 <NA> <NA>
#5 5 2016-01-09 2016-01-09 2016-01-09
With your Updated Data set, It gives
# ID `1` `2` `3`
#* <int> <date> <date> <date>
#1 1 2016-01-01 2016-01-01 2016-01-01
#2 2 2016-01-05 <NA> <NA>
#3 3 2016-01-06 2016-01-11 <NA>
#4 4 2016-01-07 <NA> <NA>
#5 5 2016-01-09 2016-01-19 <NA>
DATA USED
dput(df)
structure(list(ID = c(1L, 1L, 1L, 2L, 3L, 3L, 3L, 4L, 5L, 5L,
5L), Date = structure(c(1L, 5L, 7L, 8L, 9L, 2L, 3L, 10L, 11L,
4L, 6L), .Label = c("1-1-2016", "11-1-2016", "12-1-2016", "19-1-2016",
"2-1-2016", "20-1-2016", "3-1-2016", "5-1-2016", "6-1-2016",
"7-1-2016", "9-1-2016"), class = "factor")), .Names = c("ID",
"Date"), class = "data.frame", row.names = c(NA, -11L))

Answer 2

Use reshape. Code below assumes z is your data frame where date is a numeric date/time variable, ordered increasingly.
# a "set" variable represents a set of consecutive dates
z$set <- unsplit(tapply(z$date, z$ID, function(x) cumsum(diff(c(x[1], x)) > 1)), z$ID)
# "first.date" represents the first date in the set (of consecutive dates)
z$first.date <- unsplit(lapply(split(z$date, z[, c("ID", "set")]), min), z[, c("ID", "set")])
# "occurence" is a consecutive occurence #
z$occurrence <- unsplit(lapply(split(seq(nrow(z)), z$ID), seq_along), z$ID)
reshape(z[, c("ID", "first.date", "occurrence")], direction = "wide",
idvar = "ID", v.names = "first.date", timevar = "occurrence")
The result:
ID first.date.1 first.date.2 first.date.3
1 1 2016-01-01 2016-01-01 2016-01-01
4 2 2016-01-05 <NA> <NA>
5 3 2016-01-06 2016-01-11 2016-01-11
8 4 2016-01-07 <NA> <NA>
9 5 2016-01-09 2016-01-09 2016-01-09