A simpler solution would be:
> x <- c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
> y <- rep(NA,length(x))
> y
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> z <- which(x != 0)
> l <- z - c(1,z[-length(z)])
> y[z] <- log(x[z]/(l+1))
> y
[1] 1.0986123 0.0000000 NA -1.0986123 NA -0.4054651
NA NA -1.3862944
[10] NA NA NA NA -0.4054651 -0.6931472
--- On Tue, 3/6/08, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> From: Marc Schwartz <[EMAIL PROTECTED]>
> Subject: Re: [R] NOT-SO-SIMPLE function!
> To: "T.D.Rudolph" <[EMAIL PROTECTED]>
> Cc: r-help@r-project.org
> Received: Tuesday, 3 June, 2008, 5:59 AM
> on 06/02/2008 01:30 PM T.D.Rudolph wrote:
> > I am trying to set up a function which processes my
> data according to the
> > following rules:
> >
> > 1. if (x[i]==0) NA
> > 2. if (x[i]>0) log(x[i]/(number of consecutive
> zeros immediately preceding
> > it +1))
> >
> > The data this will apply to include a variety of whole
> numbers not limited
> > to 1 & 0, a number of which may appear
> consecutively and not separated by
> > zeros. Below is an example with a detailed
> explanation of the output
> > desired:
> >
> > x <- c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
> > output desired = c(1.098, 0.69, NA, -0.69, NA, -0.41,
> NA, NA, 1.098, NA, NA,
> > NA, NA, -0.22, 0)
> >
> > the 1st element, 3, becomes log(3) = 1.098612
> > the 2nd element, 2, becomes log(2) = 0.6931472
> > the 3rd element, 0, becomes NA (cannot log zero).
> > the 4rd element, 1, becomes log(1/(1(number of
> consecutive zeros immediately
> > preceding it) + 1 (constant))) = log(1/2) =
> -0.6931472
> > the 5th element, 0, becomes NA
> > the 6th element, 2, becomes log(2/(1(number of
> consecutive zeros immediately
> > preceding it) + 1 (constant))) = log(2/3) = -0.4054651
>
>
> The above should be log(2/2) = 0
>
> There is only 1 consecutive zero preceding the 2 in the 6th
> position
>
> > the 7th and 8th elements, both zeros, become NA
> > the 9th element, 1, becomes log(1/(2(number of
> consecutive zeros immediately
> > preceding it) + 1 (constant))) = log(1/3) = 1.098612
>
> The above should be log(1/3) = -1.098612 (negative, not
> positive)
>
> > the 10-13th elements, all zeros, each become NA
> > the 14th element, 4, becomes log(4/(4(number of
> consecutive zeros
> > immediately preceding it) + 1 (constant))) = log(4/5)
> = -0.2231436
> > the 15th element, 1, becomes log(1) = 0
> >
> > This one has been in the works for some time and I
> can't quite seem to crack
> > it.
> > I would be indebted to anyone who could with success -
> it seemed so simple
> > at the offset!
> > Tyler
>
> I am presuming that you have some typos/errors above in
> your per element
> explanation of the processing of the vector. If so, then
> the following
> should work as a first pass and could probably be optimized
> further:
>
> zeroes <- function(x, i)
> {
> if (x[i] == 0) {
> NA
> } else if (i == 1) {
> log(x[i])
> } else if (x[i - 1] != 0) {
> log(x[i])
> } else {
> rz <- rle(x[1:(i-1)])
> log(x[i] / (rz$lengths[length(rz$lengths)] + 1))
> }
> }
>
>
> x <- c(3, 2, 0, 1, 0, 2, 0, 0, 1, 0, 0, 0, 0, 4, 1)
>
>
> > sapply(seq(along = x), function(i) zeroes(x, i))
> [1] 1.0986123 0.6931472 NA -0.6931472
> NA 0.0000000
> [7] NA NA -1.0986123 NA
> NA NA
> [13] NA -0.2231436 0.0000000
>
>
> See ?rle for more information on the identification of the
> sequential
> zeroes in the vector.
>
> HTH,
>
> Marc Schwartz
>
> ______________________________________________
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> and provide commented, minimal, self-contained,
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