R help - Aug 2008 - 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
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.

 

TIA,

Jon Peck

 

Jon K. Peck

jkpeck@aya.yale.edu

 


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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 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.
> 
>  
> 
> TIA,
> 
> Jon Peck
> 
>  
> 
> Jon K. Peck
> 
> jkpeck at aya.yale.edu
> 
>  
> 
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> 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.
> 
> 
-- 
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://perswww.kuleuven.be/dimitris_rizopoulos/



Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

Peck, Jon <peck <at> 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 to have similar capabilities.
 

dt[sample(12,3)] = NaN
df = data.frame(matrix(dt,nrow=3))
df[sapply(df,is.na)] = NA

Dieter

Hi Jon,

Here's one way.
> x <- c(1,2,3,4,NaN)
> y <- c(1,2,NaN,4,5)
> 
> myDF <- data.frame(x,y)
> myDF
    x   y
1   1   1
2   2   2
3   3 NaN
4   4   4
5 NaN   5
> 
> myDF[ is.na(myDF) ] <- NA
> myDF
   x  y
1  1  1
2  2  2
3  3 NA
4  4  4
5 NA  5

Cheers,
Bob

========================================================Bob Muenchen (pronounced
Min'-chen), Manager
Statistical Consulting Center
U of TN Office of Information Technology
200 Stokely Management Center, Knoxville, TN 37996-0520
Voice: (865) 974-5230 
FAX: (865) 974-4810
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News: http://listserv.utk.edu/archives/statnews.html
========================================================
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Peck, Jon
> Sent: Friday, August 15, 2008 10:28 PM
> To: r-help at 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 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.
> 
> 
> 
> TIA,
> 
> Jon Peck
> 
> 
> 
> Jon K. Peck
> 
> jkpeck at aya.yale.edu
> 
> 
> 
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> 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.

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 <peck at spss.com>
wrote:
> 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.
>
>
>
> TIA,
>
> Jon Peck
>
>
>
> Jon K. Peck
>
> jkpeck at aya.yale.edu
>
>
>
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
>