Try this.  It does a small sample with 15 weeks and only 10 patients, but
the algorithm is easily changed for 52.

require(zoo)  # need the 'zoo' library for 'na.locf'
# create test data
visits <- data.frame(id=rep(1:10, each=10),value=runif(100),
week=sample(1:15,100,TRUE))
# removed duplicated values
visits <- visits[!duplicated(paste(visits$id, visits$week)),]
# now do the algorithm
full <- by(visits, visits$id, function(x){
    # fill to 15 weeks and add a baseline (0)
    x <- rbind(x, data.frame(id=x$id[1], value=runif(1), week=0))
    # determine which weeks are missing
    missing <- setdiff(1:15, x$week)  # change to 52 for your case
    # add missing weeks
    x <- rbind(x, data.frame(id=x$id[1], value=NA, week=missing))
    # sort by week
    x <- x[order(x$week),]
    # replace 'value' with previous good value
    x$value <- na.locf(x$value)
    x
})
# create new data frame
(full <- do.call(rbind, full))



On 1/18/07, Niccolò Bassani <biostatistica@gmail.com> wrote:
>
> I have a dataset like this:
>
> ID        value     week of visit
> 1          4.2              0  (baseline)
> 1          2.5              6
> 1          4.1              10
> 1          3.0              12
> .
> .
> .
>
> And so on for more than 11000 records (more than 1000 subjects). I need to
> add rows in order to have 52 records per patients  (corresponding to the
> 52
> weeks of the year); that is that weeks 1-2-3-4-5 will have the same
> content
> of baseline one, except for the column "week of visit". The fact is that
> each of the rows can have a very different timing, so the only thing I can
> do is creating a vector giving the number of replications needed for each
> row, but how can I practically replicate them increasing the index ok the
> week?
> thanks
> niccolò
>
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>
>
>
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>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

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