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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