13  Manipulation of Longitudinal Data

flowchart LR
sf[Storage Formats]
sf1[Tree]
sf2[Short and Wide]
sf3[Tall and Thin]
sf3a[Uniform Number<br>of Rows,<br>With NAs]
sf3b[Variable Number<br>of Rows,<br>With Few NAs]
sf --> sf1 & sf2 & sf3
sf3 --> sf3a & sf3b
locf[Last<br>Observation<br>Carried<br>Forward]
crr[Carry<br>Forward<br>by Adding<br>Rows]
fi[Find First<br>Observation<br>Meeting Criteria]
cf[Using Functions<br>That Count<br>Criteria Met]
im[Inexact<br>Matching]
sf3a --> locf
sf3b --> crr & fi & cf & im

The data.table package provides exceptional capabilities for manipulating longitudinal data, especially when performing operations by subject. Before showing a variety of examples that typify common tasks, consider that there are many ways to store longitudinal data, including

13.1 Uniform Number of Rows

Consider first the case where most of the subjects have the same number of rows and NA is used as a placeholder with a certain measurement is not made on a given time. Though highly questionable statistically, last observation carried forward (LOCF) is sometimes used to fill in NAs so that simple analyses can be performed.

LOCF is a form of missing value imputation where imputed values are treated the same as real measurements, resulting in highly deflated estimates of standard errors and much higher than nominal \(\alpha\) in frequentist statistical testing.

data.table has an efficient built-in functions for LOCF (and for last observation carried backward and fill-in using a constant value): nafill and setnafill. Consider a longitudinal data table L with 5 observations per each of two subjects.

require(Hmisc)
require(data.table)
L <- data.table(id  = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
                day = c(1, 2, 3, 4, 5, 1, 2, 3, 4, 5),
                y1  = 1:10,
                y2  = c(NA, 1, NA, NA, 2, 1, 2, 3, 4, NA),
                key = .q(id, day))
# .q(id, day) is the Hmisc version of c('id', 'day')
L
    id day y1 y2
 1:  1   1  1 NA
 2:  1   2  2  1
 3:  1   3  3 NA
 4:  1   4  4 NA
 5:  1   5  5  2
 6:  2   1  6  1
 7:  2   2  7  2
 8:  2   3  8  3
 9:  2   4  9  4
10:  2   5 10 NA
setnafill(L, "locf", cols=.q(y1, y2))
L
    id day y1 y2
 1:  1   1  1 NA
 2:  1   2  2  1
 3:  1   3  3  1
 4:  1   4  4  1
 5:  1   5  5  2
 6:  2   1  6  1
 7:  2   2  7  2
 8:  2   3  8  3
 9:  2   4  9  4
10:  2   5 10  4

setnafill changed the data table in place. Note that y1 is unchanged since it contained no NAs.

13.2 Variable Number of Rows

Consider the somewhat more frequently occuring situation where there is one row per subject per time at which a measurement is made. Consider a different data table L, and create records to fill out the observations, carrying forward to 4 days the subject’s last observation on y if it was assessed earlier than day 4.

L <- data.table(id  = c(2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5),
                day = c(1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3),
                y    =  1 : 12, key='id')
w <- copy(L)   # fresh start with no propagation of changes back to L
# Only needed if will be using := to compute variables in-place and
# you don't want the new variables also added to L
# This is related to data.table doing things by reference instead of 
# making copies.  w <- L does not create new memory for w.
# Compute each day's within-subject record number and last record number
# Feed this directly into a data.table operation to save last records 
# when the last is on a day < 4
u <- w[, .q(seq, maxseq) := .(1 : .N, .N), by=id][seq == maxseq & day < 4,]
# Extra observations to fill out to day 4
u <- u[, .(day = (day + 1) : 4, y = y), by=id]
u
   id day  y
1:  2   4  3
2:  4   3  9
3:  4   4  9
4:  5   4 12
w <- rbind(L, u, fill=TRUE)
setkey(w, id, day)  # sort and index
w
    id day  y
 1:  2   1  1
 2:  2   2  2
 3:  2   3  3
 4:  2   4  3
 5:  3   1  4
 6:  3   2  5
 7:  3   3  6
 8:  3   4  7
 9:  4   1  8
10:  4   2  9
11:  4   3  9
12:  4   4  9
13:  5   1 10
14:  5   2 11
15:  5   3 12
16:  5   4 12

Find the first time at which y >= 3 and at which y >= 7.

day[y >= 3] is read as “the value of day when y >= 3”. It is a standard subscripting operation in R for two parallel vectors day and y. Taking the minimum value of day satisfying the condition gives us the first qualifying day.
L[, .(first3 = min(day[y >= 3]),
      first7 = min(day[y >= 7])), by=id]
   id first3 first7
1:  2      3    Inf
2:  3      1      4
3:  4      1      1
4:  5      1      1

Same but instead of resulting in an infinite value if no observations for a subject meet the condition, make the result NA.

mn <- function(x) if(length(x)) min(x) else as.double(NA)
# as.double needed because day is stored as double precision
# (type contents(L) to see this) and data.table requires
# consistent storage types
L[, .(first3 = mn(day[y >= 3]),
      first7 = mn(day[y >= 7])), by=id]
   id first3 first7
1:  2      3     NA
2:  3      1      4
3:  4      1      1
4:  5      1      1

Add a new variable z and compute the first day at which z is above 0.5 for two days in a row for the subject. Note that the logic below looks for consecutive days for which records exist for a subject. To also require the days to be one day apart add the clause day == shift(day) + 1 after shift(z) > 0.5.

set.seed(1)
w <- copy(L)
w[, z := round(runif(.N), 3)]
u <- copy(w)
u
    id day  y     z
 1:  2   1  1 0.266
 2:  2   2  2 0.372
 3:  2   3  3 0.573
 4:  3   1  4 0.908
 5:  3   2  5 0.202
 6:  3   3  6 0.898
 7:  3   4  7 0.945
 8:  4   1  8 0.661
 9:  4   2  9 0.629
10:  5   1 10 0.062
11:  5   2 11 0.206
12:  5   3 12 0.177
mn <- function(x)
  if(! length(x) || all(is.na(x))) as.double(NA) else min(x, na.rm=TRUE)
u[, consecutive := z > 0.5 & shift(z) > 0.5, by=id][, 
    firstday    := mn(day[consecutive]),     by=id]
u
    id day  y     z consecutive firstday
 1:  2   1  1 0.266       FALSE       NA
 2:  2   2  2 0.372       FALSE       NA
 3:  2   3  3 0.573       FALSE       NA
 4:  3   1  4 0.908          NA        4
 5:  3   2  5 0.202       FALSE        4
 6:  3   3  6 0.898       FALSE        4
 7:  3   4  7 0.945        TRUE        4
 8:  4   1  8 0.661          NA        2
 9:  4   2  9 0.629        TRUE        2
10:  5   1 10 0.062       FALSE       NA
11:  5   2 11 0.206       FALSE       NA
12:  5   3 12 0.177       FALSE       NA

In general, using methods that involve counters makes logic more clear, easier to incrementally debug, and easier to extend the condition to any number of consecutive times. Create a function that computes the number of consecutive TRUE values or ones such that whenever the sequence is interrupted by FALSE or 0 the counting starts over. As before we compute the first day at which two consecutive z values exceed 0.5.

nconsec is modified from code found here.
nconsec <- function(x) x * sequence(rle(x)$lengths)
# rle in base R: run length encoding
# Example:
x <- c(0,0,1,1,0,1,1,0,1,1,1,1)
nconsec(x)
 [1] 0 0 1 2 0 1 2 0 1 2 3 4
# To require that the time gap between measurements must be <= 2 time
# units use the following example
t <- c(1:9, 11, 14, 15)
rbind(t=t, x=x)
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
t    1    2    3    4    5    6    7    8    9    11    14    15
x    0    0    1    1    0    1    1    0    1     1     1     1
nconsec(x & t <= shift(t) + 2)
 [1] 0 0 1 2 0 1 2 0 1 2 0 1
u <- copy(w)
# nconsec(z > 0.5) = number of consecutive days (counting current
# day) for which the subject had z > 0.5
u[, firstday := mn(day[nconsec(z > 0.5) == 2]), by=id] 
#               |  |  |                    |
#         minimum  |  |                    |
#                day  |                    |
#             such that                    |
#  it's the 2nd consecutive day with z > 0.5
u
    id day  y     z firstday
 1:  2   1  1 0.266       NA
 2:  2   2  2 0.372       NA
 3:  2   3  3 0.573       NA
 4:  3   1  4 0.908        4
 5:  3   2  5 0.202        4
 6:  3   3  6 0.898        4
 7:  3   4  7 0.945        4
 8:  4   1  8 0.661        2
 9:  4   2  9 0.629        2
10:  5   1 10 0.062       NA
11:  5   2 11 0.206       NA
12:  5   3 12 0.177       NA

13.3 Overlap Joins and Non-equi Merges

The foverlaps function in data.table provides an amazingly fast way to do complex overlap joins. Our first example is modified from an example in the help file for foverlaps. An annotation column is added to describe what happened.

d1 <- data.table(w     =.q(a, a, b, b, b),
                 start = c( 5, 10, 1, 25, 50),
                 end   = c(11, 20, 4, 52, 60))
d2 <- data.table(w     =.q(a, a, b),
                 start = c(1, 15,  1),
                 end   = c(4, 18, 55),
                 name  = .q(dog, cat, giraffe),
                 key   = .q(w, start, end))
f <- foverlaps(d1, d2, type="any")
ann <- c('no a overlap with d1 5-11 & d2 interval',
         'a 10-20 overlaps with a 16-18',
         'b 1-4 overlaps with b 1-55',
         'b 25-62 overlaps with b 1-55',
         'b 50-60 overlaps with b 1-55')
f[, annotation := ann]
f
   w start end    name i.start i.end                              annotation
1: a    NA  NA    <NA>       5    11 no a overlap with d1 5-11 & d2 interval
2: a    15  18     cat      10    20           a 10-20 overlaps with a 16-18
3: b     1  55 giraffe       1     4              b 1-4 overlaps with b 1-55
4: b     1  55 giraffe      25    52            b 25-62 overlaps with b 1-55
5: b     1  55 giraffe      50    60            b 50-60 overlaps with b 1-55
# Don't include record for non-match
foverlaps(d1, d2, type='any', nomatch=NULL)
   w start end    name i.start i.end
1: a    15  18     cat      10    20
2: b     1  55 giraffe       1     4
3: b     1  55 giraffe      25    52
4: b     1  55 giraffe      50    60
# Require the d1 interval to be within the d2 interval
foverlaps(d1, d2, type="within")
   w start end    name i.start i.end
1: a    NA  NA    <NA>       5    11
2: a    NA  NA    <NA>      10    20
3: b     1  55 giraffe       1     4
4: b     1  55 giraffe      25    52
5: b    NA  NA    <NA>      50    60
# Require the intervals to have the same starting point
foverlaps(d1, d2, type="start")
   w start end    name i.start i.end
1: a    NA  NA    <NA>       5    11
2: a    NA  NA    <NA>      10    20
3: b     1  55 giraffe       1     4
4: b    NA  NA    <NA>      25    52
5: b    NA  NA    <NA>      50    60

Now consider an example where there is an “events” dataset e with 0 or more rows per subject containing start (s) and end (e) times and a measurement x representing a daily dose of something given to the subject from s to e. The base dataset b has one record per subject with times c and d. Compute the total dose of drug received between c and d for the subject. This is done by finding all records in e for the subject such that the interval [c,d] has any overlap with the interval [s,e]. For each match compute the number of days in the interval [s,e] that are also in [c,d]. This is given by min(e,d) + 1 - max(c,s). Multiply this duration by x to get the total dose given in [c,d]. For multiple records with intervals touching [c,d] add these products.

base   <- data.table(id    = .q(a,b,c), low=10, hi=20)
events <- data.table(id    = .q(a,b,b,b,k),
                     start = c( 8,  7, 12, 19, 99),
                     end   = c( 9,  8, 14, 88, 99),
                     dose  = c(13, 17, 19, 23, 29))
setkey(base,   id, low,   hi)
setkey(events, id, start, end)
w <- foverlaps(base, events,
               by.x = .q(id, low,   hi),
               by.y = .q(id, start, end ),
               type = 'any', mult='all', nomatch=NA)

w[, elapsed := pmin(end, hi) + 1 - pmax(start, low)]
w[, .(total.dose = sum(dose * elapsed, na.rm=TRUE)), by=id]
   id total.dose
1:  a          0
2:  b        103
3:  c          0

Similar things are can be done with non-equi merges. For those you can require exact subject matches but allow inexact matches on other variables. The following example is modified from here. A medication dataset holds the start and end dates for a patient being on a treatment, and a second dataset visit has one record per subject ID per doctor visit. For each visit look up the drug in effect if there was one.

medication <-
  data.table(ID         = c( 1, 1, 2, 3, 3),
             medication = .q(a, b, a, a, b),
             start      = as.Date(c("2003-03-25","2006-04-27","2008-12-05",
                                    "2004-01-03","2005-09-18")),
             stop       = as.Date(c("2006-04-02","2012-02-03","2011-05-03",
                                    "2005-06-30","2010-07-12")),
             key        = 'ID')
medication
   ID medication      start       stop
1:  1          a 2003-03-25 2006-04-02
2:  1          b 2006-04-27 2012-02-03
3:  2          a 2008-12-05 2011-05-03
4:  3          a 2004-01-03 2005-06-30
5:  3          b 2005-09-18 2010-07-12
set.seed(123)
visit <- data.table(
  ID   = rep(1:3, 4),
  date = sample(seq(as.Date('2003-01-01'), as.Date('2013-01-01'), 1), 12),
  sbp  = round(rnorm(12, 120, 15)),
  key  = c('ID', 'date'))
visit
    ID       date sbp
 1:  1 2004-06-09 113
 2:  1 2006-06-06 147
 3:  1 2008-01-16 126
 4:  1 2009-09-28 127
 5:  2 2003-07-14 138
 6:  2 2006-02-15 122
 7:  2 2006-06-21 127
 8:  2 2009-11-15 101
 9:  3 2005-11-03  91
10:  3 2009-02-04 110
11:  3 2011-03-05 125
12:  3 2012-03-24 112
# Variables named in inequalities need to have variables in
# medication listed first
m <- medication[visit, on = .(ID, start <= date, stop > date)]
m
    ID medication      start       stop sbp
 1:  1          a 2004-06-09 2004-06-09 113
 2:  1          b 2006-06-06 2006-06-06 147
 3:  1          b 2008-01-16 2008-01-16 126
 4:  1          b 2009-09-28 2009-09-28 127
 5:  2       <NA> 2003-07-14 2003-07-14 138
 6:  2       <NA> 2006-02-15 2006-02-15 122
 7:  2       <NA> 2006-06-21 2006-06-21 127
 8:  2          a 2009-11-15 2009-11-15 101
 9:  3          b 2005-11-03 2005-11-03  91
10:  3          b 2009-02-04 2009-02-04 110
11:  3       <NA> 2011-03-05 2011-03-05 125
12:  3       <NA> 2012-03-24 2012-03-24 112
# start and stop dates are replaced with actual date of visit
# drop one of them and rename the other
m[, stop := NULL]
setnames(m, 'start', 'date')
m
    ID medication       date sbp
 1:  1          a 2004-06-09 113
 2:  1          b 2006-06-06 147
 3:  1          b 2008-01-16 126
 4:  1          b 2009-09-28 127
 5:  2       <NA> 2003-07-14 138
 6:  2       <NA> 2006-02-15 122
 7:  2       <NA> 2006-06-21 127
 8:  2          a 2009-11-15 101
 9:  3          b 2005-11-03  91
10:  3          b 2009-02-04 110
11:  3       <NA> 2011-03-05 125
12:  3       <NA> 2012-03-24 112