Thanks for all the suggestions. The last one is certainly minimalist!
On Aug 16, 1:20 pm, Dieter Menne <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck gmail.com> writes:
>
> > is.na(DF) <- is.na(DF)
>
> I knew you would win the shortest competition. I missed you at the useR
> meeting.
>
> Diete
Gabor Grothendieck gmail.com> writes:
> is.na(DF) <- is.na(DF)
I knew you would win the shortest competition. I missed you at the useR meeting.
Dieter
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PLEASE do read th
Try this:
is.na(DF) <- is.na(DF)
or this which returns a transformed data frame without
overwriting DF:
replace(DF, is.na(DF), NA)
On Fri, Aug 15, 2008 at 10:27 PM, Peck, Jon <[EMAIL PROTECTED]> wrote:
> I am looking for the most efficient way to replace all occurrences of NaN in
> a data fram
riday, August 15, 2008 10:28 PM
> To: r-help@r-project.org
> Subject: [R] Dealing with NaN's in data frames
>
> I am looking for the most efficient way to replace all occurrences of
> NaN in a data frame with NA. I can do this with a double loop, but it
> seems that there shoul
Peck, Jon spss.com> writes:
>
> I am looking for the most efficient way to replace all occurrences of NaN in a
data frame with NA. I can do this
> with a double loop, but it seems that there should be a higher level and more
efficient way. With is.na, I
> could use ifelse, but if.nan seems not
try this:
dat <- data.frame(x = rnorm(10), y = rnorm(10),
z = rnorm(10), g = gl(5,2))
dat$x[sample(10, 3)] <- NaN
dat$y[sample(10, 3)] <- NaN
dat$z[sample(10, 3)] <- NaN
dat[] <- lapply(dat, function(x){
x[is.nan(x)] <- NA
x
})
I hope it helps.
Best,
Dimitris
Peck, Jon wrote:
I
I am looking for the most efficient way to replace all occurrences of NaN in a
data frame with NA. I can do this with a double loop, but it seems that there
should be a higher level and more efficient way. With is.na, I could use
ifelse, but if.nan seems not to have similar capabilities.
T
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