13 Manipulation of Longitudinal Data
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
 as an R
list
hierarchical tree with one branch per subject (not considered here)  “short and wide” with one column per time point (not considered here because this setup requires much more metadata setup, more storage space, and more coding)
 “tall and thin” with one row per subject per time point observed (the primary format considered in this chapter; typically there are few
NA
s unless many response variables are collected at a single time and some of them areNA
)  “tall and thin” with one row per subject per time point potentially observed, with missing values (
NA
) for unobserved measurements (considered in the first example)
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 NA
s so that simple analyses can be performed.
data.table
has an efficient builtin functions for LOCF (and for last observation carried backward and fillin 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)
require(ggplot2)
< data.table(id = c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
L 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 NA
s.
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.
< data.table(id = c(2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5),
L day = c(1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3),
y = 1 : 12, key='id')
< copy(L)
w < w[, .q(seq, maxseq) := .(1 : .N, .N),
u =id][
by== maxseq & day < 4,]
seq < u[, .(day = (day + 1) : 4, y = y), by=id]
u
u< rbind(L, u, fill=TRUE)
w setkey(w, id, day)
w
 1

fresh start with no propagation of changes back to
L
. Only needed if will be using:=
to compute variables inplace and you don’t want the new variables also added toL
. This is related to data.table doing things by reference instead of making copies.w < L
does not create new memory forw
.  2

within one subject compute consecutive record numbers
1 : .N
and last record number.N
 3
 separately by subject
 4

feed this result directly into a
data.table
operation to save last records when the last record is on a day before day 4  5
 fill out to day 4 by adding extra observations, separately by subject
 6

vertically combine observations in
L
andu
, filling variables missing from one of the data tables with missing values (NA
)  7

sort by subject
id
andday
withinid
and set these variables as dataset keys
id day y
1: 2 4 3
2: 4 3 9
3: 4 4 9
4: 5 4 12
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.first3 = min(day[y >= 3]),
L[, .(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
.
< function(x) if(length(x)) min(x) else as.double(NA)
mn # as.double needed because day is stored as double precision
# (type contents(L) to see this) and data.table requires
# consistent storage types
first3 = mn(day[y >= 3]),
L[, .(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)
< copy(L)
w := round(runif(.N), 3)]
w[, z < copy(w)
u 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
< function(x)
mn if(! length(x)  all(is.na(x))) as.double(NA) else min(x, na.rm=TRUE)
:= z > 0.5 & shift(z) > 0.5, by=id][,
u[, consecutive := mn(day[consecutive]), by=id]
firstday 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.< function(x) x * sequence(rle(x)$lengths)
nconsec # rle in base R: run length encoding
# Example:
< c(0,0,1,1,0,1,1,0,1,1,1,1)
x nconsec(x)
# To require that the time gap between measurements must be <= 2 time
# units use the following example
< c(1:9, 11, 14, 15)
t rbind(t=t, x=x)
nconsec(x & t <= shift(t) + 2)
< copy(w)
u # 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[,:=
firstday mn(day
[nconsec(z > 0.05)
== 2]),
=id]
by u
 1

for all rows in data table
u
 2

add new column
firstday
 3

whose value is the minimum
day
(firstday
)  4
 such that
 5

the number of consecutive days (previous or current) for which
z > 0.05
 6
 equals 2
 7

run separately by subject
id
[1] 0 0 1 2 0 1 2 0 1 2 3 4
[,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
[1] 0 0 1 2 0 1 2 0 1 2 0 1
id day y z firstday
1: 2 1 1 0.266 2
2: 2 2 2 0.372 2
3: 2 3 3 0.573 2
4: 3 1 4 0.908 2
5: 3 2 5 0.202 2
6: 3 3 6 0.898 2
7: 3 4 7 0.945 2
8: 4 1 8 0.661 2
9: 4 2 9 0.629 2
10: 5 1 10 0.062 2
11: 5 2 11 0.206 2
12: 5 3 12 0.177 2
13.2.1 Summarizing Multiple Baseline Measurements
Suppose that subjects have varying numbers of records, with some having only two measurements, and that at least one measurement occurs before an intervention takes place at day zero. Our goal is to summarize the preintervention measurements, adding three new variables to the data and doing away with baseline records but instead carrying forward the three baseline summarizes merged with the postintervention responses. The three summary measures are
 the average response measurement over all measurements made between day 7 and day 0
 the average response over all measurements made earlier than day 7 if there are any (
NA
otherwise)  the slope over all measurements made on day 0 or earlier if there are at least two such measurements that are at least 2 days apart (
NA
otherwise)
Simulate data, starting with simulating a random number npre
of baseline measures from 15 and a possibly different random number npost
of followup measurements from 15. Sample days without replacement so that measurements within subject are always on different days. The dataset has a subjectspecific variable age
that is not used in this analysis added to it. The overall mean time trend is quadratic in time.
set.seed(11)
< 40
n < lapply(1 : n,
w function(i) { npre < sample(1:5, 1)
< sample(1:5, 1)
npost < sort(c(sample(21 : 0, npre),
day sample(1 : 180, npost)))
< sample(40:80, 1)
age < round(day + (day  90)^2 / 20 + 30 * rnorm(npre + npost))
y data.table(id=i, age, day, y=y)
} )head(w)
< rbindlist(w)
u
u# Show distribution of number of points per subject
with(u, table(table(id)))
:= factor(id)]
u[, id ggplot(u, aes(x=day, y=y)) +
geom_line(aes(col=id, alpha=I(0.4)), data=u) +
geom_point(aes(col=id), data=u) +
guides(color='none')
 1

separately for integers 1, 2, 3, …, n runs a function on that integer with the function producing a
list
that is stored in an element of a largerlist
; responsey
is normal with mean being a quadratic function ofday
 2

argument
i
is the subject number; for that subject simulate the number of observations then simulate data using that number  3

stack all the little data tables into one tall data table
u
 4

ggplot2
needs a variable mapped to thecolor
aesthetic to be factor so it treats subjectid
as categorical and doesn’t try to create a continuous color gradient
[[1]]
id age day y
1: 1 51 6 427
2: 1 51 5 486
3: 1 51 37 196
4: 1 51 60 104
[[2]]
id age day y
1: 2 42 20 572
2: 2 42 15 511
3: 2 42 14 538
4: 2 42 9 443
5: 2 42 5 399
6: 2 42 6 335
7: 2 42 39 181
8: 2 42 47 172
[[3]]
id age day y
1: 3 66 15 508
2: 3 66 126 211
3: 3 66 133 178
4: 3 66 156 348
5: 3 66 165 461
[[4]]
id age day y
1: 4 77 16 573
2: 4 77 6 429
3: 4 77 1 440
4: 4 77 0 395
5: 4 77 6 293
6: 4 77 14 329
7: 4 77 21 281
[[5]]
id age day y
1: 5 42 11 544
2: 5 42 10 492
3: 5 42 5 421
4: 5 42 1 483
5: 5 42 8 341
6: 5 42 18 219
7: 5 42 161 429
[[6]]
id age day y
1: 6 58 20 659
2: 6 58 112 141
3: 6 58 136 257
4: 6 58 138 255
id age day y
1: 1 51 6 427
2: 1 51 5 486
3: 1 51 37 196
4: 1 51 60 104
5: 2 42 20 572

218: 39 42 20 589
219: 39 42 18 613
220: 39 42 139 222
221: 40 48 2 414
222: 40 48 66 92
2 3 4 5 6 7 8 9
4 3 4 9 5 9 3 3
For one subject write a function to compute all of the three summary measures for which the data needed are available. Don’t try to estimate the slope if some times are not at least 3 days apart. By returning a list
the g
function causes data.table
to create three variables.
< function(x, y) {
g < x <= 0 & ! is.na(y)
j < x[j]; y < y[j]
x < length(x)
n if(n == 0) return(list(y0 = NA_real_, ym8 = NA_real_, slope = NA_real_))
< x >= 7
pre0 < x < 7
prem8 list(y0 = if(any(pre0)) mean(y[pre0]) else NA_real_,
ym8 = if(any(prem8)) mean(y[prem8]) else NA_real_,
slope = if(length(x) > 1 && diff(range(x)) >= 3)
unname(coef(lsfit(x, y))[2]) else NA_real_)
}
 1

Analyze observations with nonmissing
y
that are measured preintervention  2

data.table
requires all variables being created to have the same type for every observation; this includes the type ofNA
 3

range(x)
computes a 2vector with the min and maxx
, anddiff
subtracts the min from the max. The&&
and operator causesdiff(...)
to not even be evaluated unless there are at least two points.lsfit
is a simple function to fit a linear regression.coef
extracts the vector of regression coefficients from the fit, and we keep the second coefficient, which is the slope. The slope carries an element name ofX
(created bylsfit
) which is removed byunname()
.
Check the function.
g(numeric(0), numeric(0))
$y0
[1] NA
$ym8
[1] NA
$slope
[1] NA
g(10, 1)
$y0
[1] NA
$ym8
[1] 1
$slope
[1] NA
g(c(10, 9), c(1, 2))
$y0
[1] NA
$ym8
[1] 1.5
$slope
[1] NA
g(c(10, 3), c(1, 2))
$y0
[1] 2
$ym8
[1] 1
$slope
[1] 0.1428571
Now run the function separately on each subject. Then drop the preintervention records from the main data table u
and merge the new baseline variables with the followup records.
< u[, g(day, y), by = .(id)]
base head(base)
id y0 ym8 slope
1: 1 456.5000 NA NA
2: 2 399.0000 516 11.900901
3: 3 NA 508 NA
4: 4 421.3333 573 10.073095
5: 5 452.0000 518 6.131274
6: 6 NA 659 NA
dim(u)
[1] 222 4
< u[day > 0] # Keep only postintervention (followup) records
u dim(u)
[1] 115 4
< Merge(base, u, id = ~ id) # Merge is in Hmisc u
Vars Obs Unique IDs IDs in #1 IDs not in #1
base 4 40 40 NA NA
u 4 115 40 40 0
Merged 7 115 40 40 0
Number of unique IDs in any data frame : 40
Number of unique IDs in all data frames: 40
u
id y0 ym8 slope age day y
1: 1 456.5 NA NA 51 37 196
2: 1 456.5 NA NA 51 60 104
3: 2 399.0 516 11.90090 42 6 335
4: 2 399.0 516 11.90090 42 39 181
5: 2 399.0 516 11.90090 42 47 172

111: 38 407.0 498 10.05983 57 62 69
112: 38 407.0 498 10.05983 57 67 114
113: 38 407.0 498 10.05983 57 96 42
114: 39 NA 601 NA 42 139 222
115: 40 414.0 NA NA 48 66 92
13.2.2 Interpolation/Extrapolation to a Specific Time
Suppose that subjects have varying numbers of records, with some having only one measurement. Our goal is to summarize each subject with a single measurement targeted to estimate the subject’s response at 90 days. Our strategy is as follows:
 If the subject has a measurement at exactly 90 days, use it.
 If the subject has only one measurement, use that measurement to estimate that subject’s intercept (vertical shift), and use the timeresponse curve estimated from all subjects to extrapolate an estimate at 90 days.
 If the subject has two or more measurements, use the same approach but average all vertical shifts from the overall
loess
curve to adjust theloess
“intercept” in estimatingy
at the target time.
Use the data table u
created above.
< u[! is.na(day + y), approxfun(lowess(day, y))]
timeresp < data.table(day=1:180, y=timeresp(1:180))
w ggplot(u, aes(x=day, y=y)) +
geom_line(aes(col=id, alpha=I(0.4)), data=u) +
geom_point(aes(col=id), data=u) +
geom_line(aes(x=day, y, size=I(1.5)), data=w) +
guides(color='none')
 1

lowess
runs theloess
nonparametric smoother and creates a list with vectorsx
andy
;approxfun
translates this list into a function that linearly interpolates while doing the table lookup into thelowess
result.timeresp
is then a function to estimate the allsubjectscombined timeresponse curve.lowess
doesn’t handleNA
s properly so we excluded points that are missing on either variable.  2
 evaluates the whole timeresponse curve for day 1, 2, 3, …, 180
As done previously, we follow the good practice of perfecting a function that summarizes the data for one subject, then let data.table
run that function separately by
subject ID. The method adjusts the loess curve so that it perfectly fits one observed point or best fits all observed points on the average, then we use that adjusted curve to estimate the response at day 90. This is an empirical approach that trusts the raw data to provide the intercept, resulting in interpolated or extrapolated estimates that are as noisy as the real data. Function f
also returns the number of days from 90 to the closest day used for estimation of day 90’s response.
< function(x, y, id=character(0), target=90) {
f < ! is.na(x + y)
j if(! any(j)) {
cat('no nonmissing x, y pairs for subject', id, '\n')
return(list(day = target, distance=NA_real_, yest=NA_real_))
}< x[! is.na(x)]; y < y[! is.na(y)]
x < abs(x  target)
distance < which(distance == 0)
i if(length(i)) return(list(day=target, distance=0., yest=mean(y[i])))
# estimate mean y at observed x  mean loess at observed x to adjust
< timeresp(c(target, x))
z list(day = target,
distance = min(distance),
yest = z[1] + mean(y)  mean(z[1]) )
}
Test the function under a variety of situations.
# n=1, y is on the loess curve so projected y at x=90 should be also
f(25, 253.0339)
$day
[1] 90
$distance
[1] 65
$yest
[1] 115.017
timeresp(c(25, 90))
[1] 260.3086 122.2918
# n=1, x=90 exactly
f(90, 111.11)
$day
[1] 90
$distance
[1] 0
$yest
[1] 111.11
# n=3, two x=90
f(c(1, 2, 90, 90), c(37, 3, 33, 34))
$day
[1] 90
$distance
[1] 0
$yest
[1] 33.5
# n=1, not on loess curve
f(25, timeresp(25)  100)
$day
[1] 90
$distance
[1] 65
$yest
[1] 22.29177
timeresp(90)  100
[1] 22.29177
# n=2, interpolate to 90 guided by loess
f(c(80, 100), c(100, 200))
$day
[1] 90
$distance
[1] 10
$yest
[1] 146.486
# n=2, extrapolate to 90
f(c(70, 80), c(100, 200))
$day
[1] 90
$distance
[1] 10
$yest
[1] 146.9803
Now apply the function to each subject. Add the new points to the original plot. Keep the age
variable in the one subject per record data table z
.
< u[, c(list(age=age), f(day, y, id)), by=id] # just use u[, f(day, y, id), ...] if don't need age
z # Show distribution of distance between 90 and closest observed day
table(distance)] z[,
distance
1 2 3 5 6 8 9 10 12 13 14 15 17 18 22 24 26 30 32 36 40 43 47 49 52 53
3 4 1 10 5 10 5 10 8 4 3 4 5 2 3 1 3 2 5 4 1 5 1 1 5 1
54 58 69 70 71
1 1 3 1 3
ggplot(u, aes(x=day, y=y)) +
geom_line(aes(col=id), data=u) +
geom_point(aes(col=id), data=u) +
geom_line(aes(x=day, y, size=I(1.5)), data=w) +
geom_point(aes(x=day, y=yest, col=id, shape=I('x'), size=I(3)), data=z) +
guides(color='none')
13.2.3 Linear Interpolation to a Vector of Times
Suppose that 10 subjects have varying numbers of records, corresponding to assessment times t ranging from 0 to 2 years. We want to estimate the values of variable y over a regular sequence of values 0.25 years apart, for all subjects having at least two records at distinct times. Use linear interpolation, or linear extrapolation if one of the target times is outside the range of a subject’s data. It is advisable to specify the time sequence as consisting of points interior to [0, 2] so that the estimates at the end have a good chance to be symmetrically interpolated/extrapolated. For any value of t in the grid, set the calculated interpolated/extrapolated value to NA
if there is no measurement within 0.25 years of t.
Create the test dataset and a function to do the calculations on a single subject.
set.seed(15)
< 10
n < lapply(1 : n,
w function(i) { m < sample(1 : 20, 1)
< sort(runif(m, 0, 2))
t < abs(t  1) + 0.5 * rnorm(m)
y data.table(id=as.character(i), t, y)})
< rbindlist(w)
u # Show distribution of number of points per subject
with(u, table(table(id)))
ggplot(u, aes(x=t, y=y)) +
geom_line(aes(col=id, alpha=I(0.4)), data=u) +
geom_point(aes(col=id), data=u) +
guides(color='none')
 1
 Number of assessments for a subject is a random number between 1 and 20
 2
 Uniformly distributed assessment times in ascending order
 3
 True responses have means that are Vshaped in time
 4

Setting
id
to a character variable will keepggplot2
from trying to treatid
as a continuous variable  5
 Using a transparency of 0.4 makes it easier to read spaghetti plots
 6

Prevent
ggplot2
from creating a legend forid
1 5 6 8 9 12 16 17
1 1 1 1 1 2 1 2
# Set target times to regularize to
< seq(0.25, 1.75, by=0.25)
times < function(t, y) {
g < ! is.na(y)
i if(any(! i)) {
< t[i]
t < y[i]
y
}if(uniqueN(t) < 2) return(list(t=times, y=rep(NA_real_, length(times))))
< approxExtrap(t, y, xout=times)$y
z # For each target time compute the number of measurements within 0.25
# years of the target
< sapply(times, function(x) sum(abs(t  x) <= 0.25))
nclose < 1] < NA_real_
z[nclose list(t=times, y=z)
}
 1

g
is a function of two vectors (all values for one subject) that returns alist
, which makesdata.table
place the output into two variables with names given inlist()
(t and y)  2

uniqueN
is indata.table
and computes the number of distinct values of its argument. Remember that when you needNA
placeholders in building adata.table
you need to declare the storage mode for theNA
. Herereal
means double precision floating point (16 digits of precision, the standard R noninteger numeric).  3

approxExtrap
is in theHmisc
package. It extends the builtin Rapprox
function to do linear extrapolation.
Test the function.
g(1, 1)
$t
[1] 0.25 0.50 0.75 1.00 1.25 1.50 1.75
$y
[1] NA NA NA NA NA NA NA
g(1:2, 11:12)
$t
[1] 0.25 0.50 0.75 1.00 1.25 1.50 1.75
$y
[1] NA NA 10.75 11.00 11.25 NA 11.75
Create a data table with the new variables, and plot the derived estimates along with the raw data. X’s are regularized yvalues. X’s that are not touching lines represent extrapolations, and those on lines are interpolations.
< u[, g(t, y), by=id]
z ggplot(u, aes(x=t, y=y)) +
geom_line(aes(col=id), data=u) +
geom_point(aes(col=id), data=u) +
geom_point(aes(x=t, y=y, col=id, shape=I('x'), size=I(4)), data=z) +
guides(color='none')
Look at raw data for the two Xs on the plot that are not touching lines at t=0.25 (id=7) and 0.5 (id=4). For id=4 the X marks an extrapolation to the left of a steep drop over the first two points. For id=7 there is also extrapolation to the left.
%in% c(4,7)] u[id
id t y
1: 4 0.5470506 0.651103767
2: 4 0.6435926 0.008838958
3: 4 0.6618349 0.664723930
4: 4 0.8096265 0.804353411
5: 4 1.0037501 0.490364169
6: 4 1.3605564 1.030634292
7: 4 1.3618222 0.891685917
8: 4 1.3620511 0.780965231
9: 4 1.4502774 0.676070873
10: 4 1.5483572 0.117616105
11: 4 1.6356089 0.081489276
12: 4 1.6660358 0.372573268
13: 4 1.6981980 0.201474172
14: 4 1.8238231 1.344406932
15: 4 1.8530522 0.211624201
16: 4 1.8928997 0.920978736
17: 4 1.9928758 1.671926932
18: 7 0.3260752 0.938380892
19: 7 1.0025397 0.189611500
20: 7 1.1225979 0.635890259
21: 7 1.1586452 0.946132952
22: 7 1.3950600 0.753206290
23: 7 1.5650036 1.063503814
24: 7 1.8467943 1.307851168
25: 7 1.9586770 0.756841672
26: 7 1.9804584 0.825144644
id t y
Let’s summarize the regularized data by computing the interpolated y at t=1, Gini’s mean difference, and mean absolute consecutive y difference for each subject. Also compute the number of nonmissing regularized values.
< function(y) {
g < sum(! is.na(y))
n if(n < 2) return(list(n=n, y1=NA_real_, gmd=NA_real_, consec=NA_real_))
list(n = n,
y1 = y[abs(times  1.) < 0.001],
gmd = GiniMd(y, na.rm=TRUE),
consec = mean(abs(y  Lag(y)), na.rm=TRUE))
}g(c(1, 2, NA, 5, 8, 9, 10))
< z[, g(y), by=id]
w w
 1

Checking against integers such as times = 1 will not have a problem with code such as
y[times == 1]
but to work in general when values being checked for equality may not be exactly represented in base 2 it’s best to allow for a tolerance such as 0.001.  2

Lag
is in theHmisc
package; by default it shifts the y vector back one observation
$n
[1] 6
$y1
[1] 5
$gmd
[1] 4.6
$consec
[1] 1.5
id n y1 gmd consec
1: 1 6 NA 0.6785255 0.4052871
2: 2 7 0.4282241 0.1502275 0.1544205
3: 3 0 NA NA NA
4: 4 6 0.4964298 0.5336070 0.6330032
5: 5 7 0.5053917 0.6432752 0.4290084
6: 6 7 0.4744770 0.3438133 0.2189778
7: 7 6 0.1924227 0.4040584 0.3270544
8: 8 7 0.1416566 0.5904388 0.5696528
9: 9 7 0.1374026 0.4416730 0.3067097
10: 10 7 0.1504149 0.6846669 0.5107083
13.3 Overlap Joins and Nonequi Merges
Section 12.2 covered simple inexact matching/merging. Now consider more complex tasks.
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.
< data.table(w =.q(a, a, b, b, b),
d1 start = c( 5, 10, 1, 25, 50),
end = c(11, 20, 4, 52, 60))
< data.table(w =.q(a, a, b),
d2 start = c(1, 15, 1),
end = c(4, 18, 55),
name = .q(dog, cat, giraffe),
key = .q(w, start, end))
< foverlaps(d1, d2, type="any")
f < c('no a overlap with d1 511 & d2 interval',
ann 'a 1020 overlaps with a 1618',
'b 14 overlaps with b 155',
'b 2562 overlaps with b 155',
'b 5060 overlaps with b 155')
:= ann]
f[, annotation f
w start end name i.start i.end annotation
1: a NA NA <NA> 5 11 no a overlap with d1 511 & d2 interval
2: a 15 18 cat 10 20 a 1020 overlaps with a 1618
3: b 1 55 giraffe 1 4 b 14 overlaps with b 155
4: b 1 55 giraffe 25 52 b 2562 overlaps with b 155
5: b 1 55 giraffe 50 60 b 5060 overlaps with b 155
# Don't include record for nonmatch
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.
< data.table(id = .q(a,b,c), low=10, hi=20)
base < data.table(id = .q(a,b,b,b,k),
events 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)
< foverlaps(base, events,
w by.x = .q(id, low, hi),
by.y = .q(id, start, end ),
type = 'any', mult='all', nomatch=NA)
:= pmin(end, hi) + 1  pmax(start, low)]
w[, elapsed total.dose = sum(dose * elapsed, na.rm=TRUE)), by=id] w[, .(
id total.dose
1: a 0
2: b 103
3: c 0
Similar things are can be done with nonequi 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("20030325","20060427","20081205",
"20040103","20050918")),
stop = as.Date(c("20060402","20120203","20110503",
"20050630","20100712")),
key = 'ID')
medication
ID medication start stop
1: 1 a 20030325 20060402
2: 1 b 20060427 20120203
3: 2 a 20081205 20110503
4: 3 a 20040103 20050630
5: 3 b 20050918 20100712
set.seed(123)
< data.table(
visit ID = rep(1:3, 4),
date = sample(seq(as.Date('20030101'), as.Date('20130101'), 1), 12),
sbp = round(rnorm(12, 120, 15)),
key = c('ID', 'date'))
visit
ID date sbp
1: 1 20040609 113
2: 1 20060606 147
3: 1 20080116 126
4: 1 20090928 127
5: 2 20030714 138
6: 2 20060215 122
7: 2 20060621 127
8: 2 20091115 101
9: 3 20051103 91
10: 3 20090204 110
11: 3 20110305 125
12: 3 20120324 112
# Variables named in inequalities need to have variables in
# medication listed first
< medication[visit, on = .(ID, start <= date, stop > date)]
m m
ID medication start stop sbp
1: 1 a 20040609 20040609 113
2: 1 b 20060606 20060606 147
3: 1 b 20080116 20080116 126
4: 1 b 20090928 20090928 127
5: 2 <NA> 20030714 20030714 138
6: 2 <NA> 20060215 20060215 122
7: 2 <NA> 20060621 20060621 127
8: 2 a 20091115 20091115 101
9: 3 b 20051103 20051103 91
10: 3 b 20090204 20090204 110
11: 3 <NA> 20110305 20110305 125
12: 3 <NA> 20120324 20120324 112
# start and stop dates are replaced with actual date of visit
# drop one of them and rename the other
:= NULL]
m[, stop setnames(m, 'start', 'date')
m
ID medication date sbp
1: 1 a 20040609 113
2: 1 b 20060606 147
3: 1 b 20080116 126
4: 1 b 20090928 127
5: 2 <NA> 20030714 138
6: 2 <NA> 20060215 122
7: 2 <NA> 20060621 127
8: 2 a 20091115 101
9: 3 b 20051103 91
10: 3 b 20090204 110
11: 3 <NA> 20110305 125
12: 3 <NA> 20120324 112