[R] NOT-SO-SIMPLE function!

[Gabor Grothendieck]

In the following cs increases by 1 for each nonzero
entry in x.  Thus the zeros plus the preceding nonzero
form runs in cs.  seq(cs)-match(cs,cs) assigns 0
to the first element of each such run, 1 to the next,
2 to the next and so on.  Thus shifting it forward and
adding 1 gives the number of preceding zeros, num0p1.

cs <- cumsum(x != 0)
num0p1 <- head(c(0, seq(cs) - match(cs, cs)), -1) + 1
ifelse(x == 0, NA, log(x/num0p1))

On Mon, Jun 2, 2008 at 2:30 PM, T.D.Rudolph <prairie.picker at gmail.com> 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 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 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
> --
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