R help - Sep 2011 - last observation carried forward +1

Hi R-helpers

I'm looking for a vectorised function which does missing value replacement
as in last observation carried forward in the zoo package but instead of a
locf, I would like the locf function to add +1 to each time a missing value
occurred. See below for an example.
> require(zoo)
> x <- 5:15
> x[4:7] <- NA
> coredata(na.locf(zoo(x)))
 [1]  5  6  7  7  7  7  7 12 13 14 15
But what I need is
5  6  7  7+1  7+1+1  7+1+1+1  7+1+1+1+1 12 13 14 15
to obtain
[1]  5  6  7  8  9 10 11 12 13 14 15
I could program this in C but if anyone has already done this I would be
interested in seeing their vectorized solution.

thanks,
Jan

-- 
groeten/kind regards,
Jan

Jan Wijffels
Statistical Data Miner
www.bnosac.be  | +32 486 611708

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On Fri, Sep 30, 2011 at 5:43 AM, Jan Wijffels <jwijffels at bnosac.be>
wrote:
> Hi R-helpers
>
> I'm looking for a vectorised function which does missing value
replacement
> as in last observation carried forward in the zoo package but instead of a
> locf, I would like the locf function to add +1 to each time a missing value
> occurred. See below for an example.
>
>> require(zoo)
>> x <- 5:15
>> x[4:7] <- NA
>> coredata(na.locf(zoo(x)))
> ?[1] ?5 ?6 ?7 ?7 ?7 ?7 ?7 12 13 14 15
> But what I need is
> 5 ?6 ?7 ?7+1 ?7+1+1 ?7+1+1+1 ?7+1+1+1+1 12 13 14 15
> to obtain
> [1] ?5 ?6 ?7 ?8 ?9 10 11 12 13 14 15
> I could program this in C but if anyone has already done this I would be
> interested in seeing their vectorized solution.
>
Try this (where x is from the question):
> ix <- cumsum(is.na(x))
> na.locf(x) + ix - cummax(ix * !is.na(x))
 [1]  5  6  7  8  9 10 11 12 13 14 15

Because of the way x was constructed in this test case the result is
actually the same as linear interpolation (though not in general) so
if that were what is wanted then it would just be:
> na.approx(x)
 [1]  5  6  7  8  9 10 11 12 13 14 15

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