Creating intervals from time series data - r

Question

I have a data frame of users and access times. Access times can be duplicated.
I am trying to create a list of users grouped and named by a given time interval, e.g. year.
timestamp user
1 2013-03-06 01:00:00 1
2 2014-07-06 21:00:00 1
3 2014-07-31 23:00:00 2
4 2014-08-09 17:00:00 2
5 2014-08-14 20:00:00 2
6 2014-08-14 22:00:00 3
7 2014-08-16 15:00:00 3
8 2014-08-19 02:00:00 1
9 2014-12-28 18:00:00 1
10 2015-01-17 17:00:00 1
11 2015-01-22 22:00:00 2
12 2015-01-22 22:00:00 3
13 2015-03-23 15:00:00 4
14 2015-04-05 18:00:00 1
15 2015-04-06 01:00:00 2
My code example already creates a list of users grouped by year.
My problem is that I need to modify the table in this approach, which becomes a problem with my tables of a million entries.
test <- structure(list(timestamp = c("2013-03-06 01:00:00", "2014-07-06 21:00:00",
"2014-07-31 23:00:00", "2014-08-09 17:00:00", "2014-08-14 20:00:00",
"2014-08-14 22:00:00", "2014-08-16 15:00:00", "2014-08-19 02:00:00",
"2014-12-28 18:00:00", "2015-01-17 17:00:00", "2015-01-22 22:00:00",
"2015-01-22 22:00:00", "2015-03-23 15:00:00", "2015-04-05 18:00:00",
"2015-04-06 01:00:00"), user = c(1L, 1L, 2L, 2L, 2L, 3L, 3L,
1L, 1L, 1L, 2L, 3L, 4L, 1L, 2L)), .Names = c("timestamp", "user"
), class = "data.frame", row.names = c(NA, -15L))
require(lubridate)
#Creating "POSIXct" object from string timestamp
timestamp <- lapply(test$timestamp,
function(x)parse_date_time(x, "y-m-d H:M:S"))
test$timestamp <- do.call(c,timestamp)
print(class(test$timestamp))
#Adding column for year
test <- cbind(test,sapply(timestamp, function(x)year(x)))
colnames(test)[3]<- "year"
#Creating list of year time intervals for users
intervals <- names(table(test$year))
users <- lapply(intervals, function(x)test[test$year %in% x,"user"])
names(users) <- intervals

Answer 1

without timestamps
treat the timestamp as a character. Only works if for every timestap, the first 4 digits represent the year.
library(dplyr)
test %>%
group_by( user, substr(timestamp,1,4 ) ) %>%
summarise( )
# user `substr(timestamp, 1, 4)`
# <int> <chr>
# 1 1 2013
# 2 1 2014
# 3 1 2015
# 4 2 2014
# 5 2 2015
# 6 3 2014
# 7 3 2015
# 8 4 2015
dplyr + lubridate
will extract the year from the timestamp
library( dplyr )
library( lubridate )
test %>%
mutate( timestamp = as.POSIXct( timestamp, format = "%Y-%m-%d %H:%M:%S" ) ) %>%
group_by( user, lubridate::year( timestamp ) ) %>%
summarise( )
# # Groups: user [?]
# user `year(timestamp)`
# <int> <dbl>
# 1 1 2013
# 2 1 2014
# 3 1 2015
# 4 2 2014
# 5 2 2015
# 6 3 2014
# 7 3 2015
# 8 4 2015
table
a frequency table is also quickly made
table( test$user, substr( test$timestamp, 1, 4 ) )
# 2013 2014 2015
# 1 1 3 2
# 2 0 3 2
# 3 0 2 1
# 4 0 0 1
there are any more alternatives... pick one
edit
if speed is an issue, ty data.table
dcast(
setDT( test )[, timestamp := as.POSIXct( timestamp, format = "%Y-%m-%d %H:%M:%S" )][, .N, by = list( user, data.table::year(timestamp) )],
user ~ data.table,
value.var = "N")
# user 2013 2014 2015
# 1: 1 1 3 2
# 2: 2 NA 3 2
# 3: 3 NA 2 1
# 4: 4 NA NA 1

Answer 2

Another option using the lightning fast data.table package:
library(data.table)
setDT(test) # make `test` a data.frame 'by reference' (no copy is made at all)
test[, j=.(users=list(unique(user))),
by=.(year=substr(test$timestamp,1,4))]
year users
1: 2013 1
2: 2014 1,2,3
3: 2015 1,2,3,4
Again assuming your test$timestamp column is a character vector - otherwise substitute lubridate::year() as needed.
Update:
Simple change to show grouping instead by month (just as it was mentioned in a comment):
test[, j=.(users=list(unique(user))),
by=.(ym=substr(test$timestamp,1,7))]
ym users
1: 2013-03 1
2: 2014-07 1,2
3: 2014-08 2,3,1
4: 2014-12 1
5: 2015-01 1,2,3
6: 2015-03 4
7: 2015-04 1,2
Or group by day, to help demonstrate how to subset with chaining:
test[, j=.(users=list(unique(user))),
by=.(ymd=substr(test$timestamp,1,11))][ymd>='2014-08-01' & ymd<= '2014-08-21']
ymd users
1: 2014-08-09 2
2: 2014-08-14 2,3
3: 2014-08-16 3
4: 2014-08-19 1
Note for filtering/subsetting, if you are only interested in a subset of dates for a "one off" calculation (and not otherwise saving the whole aggregated set to be stored for other purposes) it will likely be more efficient to do the subset in i of DT[i, j, by] for the "one off" calculation.

Answer 3

You could also use base (stats) function aggregate() as follows:
aggregate( x = test$user,
by = list(year=substr(test$timestamp,1,4)),
FUN = unique )
Result:
year x
1 2013 1
2 2014 1, 2, 3
3 2015 1, 2, 3, 4
Above working on assumption that your timestamp column is initially just a character vector exactly as included in your structured example data. In which case you may directly substr out the year with substr(test$timestamp,1,4) avoiding the need to first convert to dates.
However, if you have the timestamp column already as a date, simply substitute the lubridate::year() function you demonstrated in your attempted solution.