[R] help with simple function

[T.D.Rudolph]

I'm trying to build on Jim's approach to change the parameters in the 
function, with new rules: 

1. if (x[i]==0) NA 
2. if (x[i]>0) log(x[i]/(number of consecutive zeros preceding it +1)) 

x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1) 
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38, 
NA, NA, NA, NA, -1.61) 
y <- rle(x) 
# attempting to modify Jim's function: 
result <- lapply(seq_along(y$lengths), function(.indx){ 
     if (y$values[.indx-1] == 0) 
     log(y$values[.indx]/seq(y$lengths[.indx-1]+1, by=-1, 
     length=y$lengths[.indx])) 
     else rep(log(y$values[.indx]), y$lengths[.indx]) 
}) 
# but I am clearly missing something! 

Does it not work because I haven't addressed what to do with the zeros and 
log(0)=-Inf? 
I've tried adding another "ifelse" but I still get the same result. 
Can someone find the error in my ways?   
Tyler 


T.D.Rudolph wrote:
> 
> 
> I have a matrix of frequency counts from 0-160.
> x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1))
> 
> I would like to apply a function creating a new column (x[,2])containing
> values equal to:
> a) log(x[m,1]) if x[m,1] > 0; and
> b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values
> +1)
> 
> for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472
> whereas x[3,2] should equal log(x[5,1]/3) = log (1/3) = -1.098612
> 
> I will be applying this to nrow(x)=~70,000 so I would prefer to not do it
> by hand!
> Tyler
> 
> 
> 
> 

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