CODEWITHSUNDEEP
rtime-seriesxtszoo
Eric W.
Eric W.

Reputation: 814

Rolling window over irregular time series

I have an irregular time series of events (posts) using xts, and I want to calculate the number of events that occur over a rolling weekly window (or biweekly, or 3 day, etc). The data looks like this:

                    postid
2010-08-04 22:28:07    867
2010-08-04 23:31:12    891
2010-08-04 23:58:05    901
2010-08-05 08:35:50    991
2010-08-05 13:28:02   1085
2010-08-05 14:14:47   1114
2010-08-05 14:21:46   1117
2010-08-05 15:46:24   1151
2010-08-05 16:25:29   1174
2010-08-05 23:19:29   1268
2010-08-06 12:15:42   1384
2010-08-06 15:22:06   1403
2010-08-07 10:25:49   1550
2010-08-07 18:58:16   1596
2010-08-07 21:15:44   1608

which should produce something like

                    nposts
2010-08-05 00:00:00     10
2010-08-06 00:00:00      9
2010-08-07 00:00:00      5

for a 2-day window. I have looked into rollapply, apply.rolling from PerformanceAnalytics, etc, and they all assume regular time series data. I tried changing all of the times to just the day the the post occurred and using something like ddply to group on each day, which gets me close. However, a user might not post every day, so the time series will still be irregular. I could fill in the gaps with 0s, but that might inflate my data a lot and it's already quite large.

What should I do?

Upvotes: 11

Views: 3360

Answers (3)

GoGonzo
GoGonzo

Reputation: 2877

With runner one can apply any R function on rolling windows. What OP requires is to calculate function (length) on rolling window only at specified time-points. Using runner user needs to specify at argument to indicate on what time-points output should be calculated. We can just pass vector of time-points to runner which we created on a side as a POSIXt sequence.
To make a runner time-dependent one has to specify idx by dates corresponding to x object. Length of the window can be set as k = "2 days"

at <- seq(as.POSIXct("2010-08-05 00:00:00"), 
          by = "1 days",
          length.out = 4)

# [1] "2010-08-05 CEST" "2010-08-06 CEST" "2010-08-07 CEST" "2010-08-08 CEST"


runner::runner(
  x = x$postid,
  k = "2 days",
  idx = x$datetime,
  at = at,
  f = length
)
# [1]  3 10  9  5

Upvotes: 0

Joshua Ulrich
Joshua Ulrich

Reputation: 176728

Here's a solution using xts:

x <- structure(c(867L, 891L, 901L, 991L, 1085L, 1114L, 1117L, 1151L, 
  1174L, 1268L, 1384L, 1403L, 1550L, 1596L, 1608L), .Dim = c(15L, 1L),
  index = structure(c(1280960887, 1280964672, 1280966285, 
  1280997350, 1281014882, 1281017687, 1281018106, 1281023184, 1281025529, 
  1281050369, 1281096942, 1281108126, 1281176749, 1281207496, 1281215744),
  tzone = "", tclass = c("POSIXct", "POSIXt")), class = c("xts", "zoo"),
  .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct", "POSIXt"),
  .indexTZ = "", tzone = "")
# first count the number of observations each day
xd <- apply.daily(x, length)
# now sum the counts over a 2-day rolling window
x2d <- rollapply(xd, 2, sum)
# align times at the end of the period (if you want)
y <- align.time(x2d, n=60*60*24)  # n is in seconds

Upvotes: 5

Dag Hjermann
Dag Hjermann

Reputation: 2210

This seems to work:

# n = number of days
n <- 30
# w = window width. In this example, w = 7 days
w <- 7

# I will simulate some data to illustrate the procedure
data <- rep(1:n, rpois(n, 2))

# Tabulate the number of occurences per day:
# (use factor() to be sure to have the days with zero observations included)
date.table <- table(factor(data, levels=1:n))  

mat <- diag(n)
for (i in 2:w){
  dim <- n+i-1
  mat <- mat + diag(dim)[-((n+1):dim),-(1:(i-1))]
  }

# And the answer is.... 
roll.mean.7days <- date.table %*% mat

Seems to be not too slow (although the mat matrix will get dimensions n*n). I tried to replace n=30 with n=3000 (which creates a matrix of 9 million elements = 72 MB) and it still was reasonable fast on my computer. For very big data sets, try on a subset first.... It will also be faster to use some of the functions in the Matrix package (bandSparse) to create the mat matrix.

Upvotes: 4

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