Thompson, David (MNR <David.John.Thompson <at> ontario.ca> writes:
> Thank you Jim Holtman and Mark Leeds for your help.
>
> Original question:
> >How do I do the following more concisely?
> > Bout[is.na(Bout$bd.n), 'bd.n'] <- 0
> > Bout[is.na(Bout$ht.n), 'ht.n'] <- 0
> > Bout[is.na(Bout$dbh.n), 'dbh.n'] <- 0
> >. . .
>
> Solution:
> for (i in c('bd.n', 'ht.n', 'dbh.n')) Bout[is.na(Bout[[i]]), i] <- 0
ABove solution is completely OK, but can be cumbersome. I wrote functions
NAToUnknown() and unknownToNA() with exactly the same problem in mind.
Take a look in gdata package. I also described the function in RNews
G. Gorjanc. Working with unknown values: the
gdata package. R News, 7(1):24–26, 2007.
http://CRAN.R-project.org/doc/Rnews/Rnews_2007-1.pdf.
For your example try the following:
library(gdata)
df.0 <- as.data.frame( cbind(
c1=c(NA, NA, 10, NA, 15, 11, 12, 14, 14, 11),
c2=c(13, NA, 16, 16, NA, 12, 14, 19, 18, NA),
c3=c(NA, NA, 11, 19, 17, NA, 11, 16, 20, 13),
c4=c(20, NA, 15, 11, NA, 15, NA, 13, 14, 15),
c5=c(14, NA, 13, 16, 17, 17, 16, NA, 15, NA),
c6=c(NA, NA, 13, 11, NA, 16, 15, 12, NA, 20)) )
df.0
c1 c2 c3 c4 c5 c6
1 NA 13 NA 20 14 NA
2 NA NA NA NA NA NA
3 10 16 11 15 13 13
4 NA 16 19 11 16 11
5 15 NA 17 NA 17 NA
6 11 12 NA 15 17 16
7 12 14 11 NA 16 15
8 14 19 16 13 NA 12
9 14 18 20 14 15 NA
10 11 NA 13 15 NA 20
NAToUnknown(df.0, unknown=0)
c1 c2 c3 c4 c5 c6
1 0 13 0 20 14 0
2 0 0 0 0 0 0
3 10 16 11 15 13 13
4 0 16 19 11 16 11
5 15 0 17 0 17 0
6 11 12 0 15 17 16
7 12 14 11 0 16 15
8 14 19 16 13 0 12
9 14 18 20 14 15 0
10 11 0 13 15 0 20
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