Increasing the value of elements in a vector by a given sequence/statement - r

Question

I want to create a sample dataset for an analysis:
set.seed(123)
d <- data.frame(ID=rep(1:10, each = 8),
AGE=rep(sample(20:40, size=10),each=8),
YEAR=rep(2011:2014, 10, each = 2),
HAND=rep(c("LEFT","RIGHT"), 40),
OUTCOME=(rnorm(80)) )
> d[1:8,]
ID AGE YEAR HAND OUTCOME
1 1 26 2011 LEFT 1.7150650
2 1 26 2011 RIGHT 0.4609162
3 1 26 2012 LEFT -1.2650612
4 1 26 2012 RIGHT -0.6868529
5 1 26 2013 LEFT -0.4456620
6 1 26 2013 RIGHT 1.2240818
7 1 26 2014 LEFT 0.3598138
8 1 26 2014 RIGHT 0.4007715
ID would be subject, AGE is the age of the subject, YEAR is the year the measurement was taken, HAND is left or right hand, OUTCOME is some measure of outcome.
Now I realized that the AGE for each subject should ideally increase by one year for each YEAR the subject has been measured, i.e.:
26,26,27,27,28,28,29,29
I came up with this solution:
age <- unique(d$AGE)
AGE2=c()
for(i in 1:10){
a <- rep(age[i]+0:3, each=2)
AGE2 <- c(AGE2,a)
}
d$AGE2 <- AGE2
d[1:8,]
> d[1:8,]
ID AGE YEAR HAND OUTCOME AGE2
1 1 26 2011 LEFT 1.7150650 26
2 1 26 2011 RIGHT 0.4609162 26
3 1 26 2012 LEFT -1.2650612 27
4 1 26 2012 RIGHT -0.6868529 27
5 1 26 2013 LEFT -0.4456620 28
6 1 26 2013 RIGHT 1.2240818 28
7 1 26 2014 LEFT 0.3598138 29
8 1 26 2014 RIGHT 0.4007715 29
Question: I was wondering if there is a more efficient way to do this? For example would it be possible to add the "corrected" age right away in the data.frame() function above?

Answer 1

We can do this with duplicated
library(dplyr)
res <- d %>%
group_by(ID) %>%
mutate(AGE2 = AGE + cumsum(!duplicated(YEAR))-1)
head(res)
# ID AGE YEAR HAND OUTCOME AGE2
# <int> <int> <int> <fctr> <dbl> <dbl>
#1 1 26 2011 LEFT 1.7150650 26
#2 1 26 2011 RIGHT 0.4609162 26
#3 1 26 2012 LEFT -1.2650612 27
#4 1 26 2012 RIGHT -0.6868529 27
#5 1 26 2013 LEFT -0.4456620 28
#6 1 26 2013 RIGHT 1.2240818 28

Answer 2

Using dplyr you can simply group by ID and HAND:
d %>% group_by(ID, HAND) %>% mutate(AGE2 = AGE + (0:(length(AGE)-1)))
Source: local data frame [80 x 6]
Groups: ID, HAND [20]
ID AGE YEAR HAND OUTCOME AGE2
<int> <int> <int> <fctr> <dbl> <int>
1 1 26 2011 LEFT 1.7150650 26
2 1 26 2011 RIGHT 0.4609162 26
3 1 26 2012 LEFT -1.2650612 27
4 1 26 2012 RIGHT -0.6868529 27
5 1 26 2013 LEFT -0.4456620 28
6 1 26 2013 RIGHT 1.2240818 28
7 1 26 2014 LEFT 0.3598138 29
8 1 26 2014 RIGHT 0.4007715 29
9 2 35 2011 LEFT 0.1106827 35
10 2 35 2011 RIGHT -0.5558411 35
# ... with 70 more rows

Answer 3

With data.table, you could use rep and a little algebra.
library(data.table)
setDT(d)
d[, AGE2 := AGE + rep(0L:((.N-1)/2L), each=2), by=ID][]
ID AGE YEAR HAND OUTCOME AGE2
1: 1 26 2011 LEFT 1.715064987 26
2: 1 26 2011 RIGHT 0.460916206 26
3: 1 26 2012 LEFT -1.265061235 27
4: 1 26 2012 RIGHT -0.686852852 27
5: 1 26 2013 LEFT -0.445661970 28
6: 1 26 2013 RIGHT 1.224081797 28
7: 1 26 2014 LEFT 0.359813827 29
8: 1 26 2014 RIGHT 0.400771451 29
9: 2 35 2011 LEFT 0.110682716 35
10: 2 35 2011 RIGHT -0.555841135 35
11: 2 35 2012 LEFT 1.786913137 36
12: 2 35 2012 RIGHT 0.497850478 36
...
Here, AGE2 is constructed by adding AGE to rep(0L:((.N-1)/2L), each=2) which counts 0 through the number of observations, minus 1, divided by 2. The by statement repeats this for each ID.