FAQ
But, assuming you accept replacing each NA by the
global mean or median, the same operation can
be done avoiding the for(), which is always
better:
x<-c(NA,12,NA,14,15,17,21)
x[is.na(x)] <- median(x,na.rm=T)
x
[1] 15 12 15 14 15 17 21

Agus

Dr. Agustin Lobo
Instituto de Ciencias de la Tierra (CSIC)
Lluis Sole Sabaris s/n
08028 Barcelona SPAIN
tel 34 93409 5410
fax 34 93411 0012
alobo at ija.csic.es

On Wed, 27 Feb 2002, E. D. Isaia wrote:

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.meanúLSE))
[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

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