R help - Jun 2009 - 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

use na.locf in zoo:
> x
 [1]  0 -1 -1 -1  0  0  1 -1  1  0
> # replace 0 with NA so na.locf works
> is.na(x) <- x == 0
> x
 [1] NA -1 -1 -1 NA NA  1 -1  1 NA
> na.locf(x, na.rm=FALSE)
 [1] NA -1 -1 -1 -1 -1  1 -1  1  1
>
You can then go back and replace NAs with 0


On Thu, Jun 18, 2009 at 12:47 PM,
<Murali.MENON@fortisinvestments.com>wrote:
> 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
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
>
http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

	[[alternative HTML version deleted]]

The zoo package has na.locf which replaces NAs with the last non-NA.
So, first replace 0's with NA's, apply na.locf and then replace NAs with
0's.
> library(zoo)
> x.na <- replace(x, x == 0, NA)
> x0 <- na.locf(x.na, na.rm = FALSE)
> replace(x0, is.na(x0), 0)
 [1]  0 -1 -1 -1 -1 -1  1  1  1  1

On Thu, Jun 18, 2009 at 12:47 PM, <Murali.MENON at fortisinvestments.com>
wrote:
> 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
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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
> 
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide 
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>