[R] Replace zeroes in vector with nearest non-zero value

[William Dunlap]

approx() almost does what you want, but you have to patch
up its output to account for a possible initial run of 0's in the
input:
> x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0)
> y <- approx(seq_along(x)[x!=0], x[x!=0], xout=seq_along(x),
method="const", f=0, rule=2)$y
> y
 [1] -1 -1 -1 -1 -1 -1  1 -1  1  1

Doing tricks with cumsum will do it more directly:

f<-function(x){
     i<-cumsum(x!=0) # indices to last non-zero value in x
     c(x[i==0], x[x!=0][i]) # x[i==0] is initial run of 0's, x[x!=0][i]
gets the rest
}
> rbind(x,f(x))
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
x    0   -1   -1   -1    0    0    1   -1    1     0
     0   -1   -1   -1   -1   -1    1   -1    1     1

If there are NA's in x you will have to add an NA test to
the x!=0 and x==0 tests or replace the NA's by 0's before
doing this.

Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com  

> -----Original Message-----
> From: r-help-bounces at r-project.org 
> [mailto:r-help-bounces at r-project.org] On Behalf Of 
> Murali.MENON at fortisinvestments.com
> Sent: Thursday, June 18, 2009 9:48 AM
> To: r-help at r-project.org
> Subject: [R] Replace zeroes in vector with nearest non-zero value
> 
> Folks,
>  
> If I have a vector such as the following:
>  
> x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0)
>  
> and I want to replace the zeroes by the nearest non-zero number to the
> left, is there a more elegant way to do this than the following loop?
>  
> y <- x
> for (i in 2 : length(x))
> {
>     if (y[i] == 0) {
>         y[i] <- y[i - 1]
>     }
> }
>  
> > y
> [1]  0 -1 -1 -1 -1 -1  1 -1  1  1
>  
> You can see the first zero is left as is, the next two zeroes 
> become -1,
> which is the closest non-zero to the left of them, and the last zero
> becomes 1.
>  
> Cheers,
> Murali
> 
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