replace NAs

Dear R community:

it is possible to replace NA?s in a data frame with zeroes?
what should I do?

Thanks in advance

Juan Pablo

Juan,

Replacing NAs is relatively easy. Nonetheless, I wrote some simple
code to make it even easier for me.  You might find it useful.  The
first replaces NAs with the mean, the second replaces NAs with a value
you specify.  This works for a single vector, you can also use the
apply() function to use it across multiple columns in a data frame or
matrix.


If anyone is interested, I also have code to produce least squares
means by levels of a factor given any number of covariates and code
for classical item analysis (reliability and item characteristics).
replace.na.m<-

  function (x){
    X<-mean(x,na.rm=TRUE)
    ifelse ( is.na(x)=="TRUE",X,x)
}


replace.na.x<-
  function(x, value){
    ifelse (is.na(x)=="TRUE", value , x)
}


juan pablo perez wrote:

Dear R community:

it is possible to replace NA?s in a data frame with zeroes?
what should I do?

Thanks in advance

Juan Pablo

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

r-help mailing list -- Read
http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html

Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._

(x<-c(NA,12,NA,14,15,17,21))

(rep.na(x))

(rep.na(x,my.mean=FALSE))

Dear R community:

it is possible to replace NA?s in a data frame with zeroes?
what should I do?

Thanks in advance

Juan Pablo

x<-c(NA,12,NA,14,15,17,21)
x[is.na(x)] <- median(x,na.rm=T)
x

With this function you may replace the NA with the mean or median of the non
missing values

##   replace NA
##
rep.na<-function(x, my.mean=TRUE)
{
    if (!my.mean){valore<-median(x[!is.na(x)])}
    else {valore<-mean(x[!is.na(x)])}
    for (i in (1:length(x))){if (is.na(x[i])==TRUE) {x[i]<-valore}}
    x<<-x
}
##
##
i.e.

(x<-c(NA,12,NA,14,15,17,21))

[1] NA 12 NA 14 15 17 21

(rep.na(x))

[1] 15.8 12.0 15.8 14.0 15.0 17.0 21.0

(rep.na(x,my.mean=FALSE))

[1] 15 12 15 14 15 17 21

Good job, isaia.



juan pablo perez wrote:

Dear R community:

it is possible to replace NA´s in a data frame with zeroes?
what should I do?

Thanks in advance

Juan Pablo

--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~ Ennio D. Isaia
~ Dep. of Statistics & Mathematics, University of Torino
~ Piazza Arbarello, 8 - 10128 Torino (Italy)
~ Phone: +39 011 670 62 51 ~~ Fax: +39 011 670 62 39
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._