[R] Odp: replace NA with 9999 in zoo object

[Petr PIKAL]

Hi

not sure about zoo but data frame or matrix can be indexed

mat[is.na(mat)] <- 9999

Regards
Petr
petr.pikal at precheza.cz

r-help-bounces at r-project.org napsal dne 06.03.2008 14:09:16:

> This is the same set of data that I have been working with for those
> in the know.  it is a matrix of ~174 columns and ~70,000 rows.  I have
> it as a zoo object, but I could read it in as just a matrix as long as
> the date time stamp won't be corrupted.
> 
> here is an example of what a column would look like:
> 1/1/06 12:00, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5,6 ,7, NA
> #read in with the following thanks to Gabor
>  # chron
> > library(chron)
> > fmt.chron <- function(x) {
> +    chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
> + }
> > z1 <- read.zoo("all.csv", sep = ",", header = TRUE, FUN = fmt.chron)
> #this part works just fine and I can plot and analyze data till my
> hearts content
> 
> I need to replace NA (~700,000+) with the numeric value 9999 for a
> beam forming exercise that we are conducting with a geophysicist in
> matlab.  I can't get it into excel in the present form because it is
> too big.  I was wondering if there was an easy way to do such a thing
> in R?
> 
> I tried the following:
> 
> dat <- sapply(z1, function(x) {x[is.na(x)] <- 9999; x})
> 
> and I got the following error:
> 
> Error in array(unlist(answer, recursive = FALSE), dim = c(common.len,  :
>   'dim' specifies too large an array
> 
> thanks for the help
> 
> Stephen
> 
> -- 
> Let's not spend our time and resources thinking about things that are
> so little or so large that all they really do for us is puff us up and
> make us feel like gods.  We are mammals, and have not exhausted the
> annoying little problems of being mammals.
> 
>                         -K. Mullis
> 
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