On Tue, 2005-08-23 at 10:12 -0500, Greg Blevins wrote:
> Hello,
>
> I have struggled, for longer than I care to admit, with this seemingly
> simple problem, but I cannot find a solution other than the use of
> long drawn out ifelse statements. I know there has to be a better
> way. Here is stripped down version of the situation:
>
> I start with:
> a <- c(1,0,1,0,0,0,0)
> b <- c(1,1,1,1,0,0,0)
> c <- c(1,1,0,1,0,0,0)
>
> rbind(a,b,c)
> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
> a 1 0 1 0 0 0 0
> b 1 1 1 1 0 0 0
> c 1 1 0 1 0 0 0
>
> I refer to column 3 as the target column, which at the end of the day
> will be NA in all instances.
>
> The logic involved:
>
> 1) If columns 2, 4 thru 7 do NOT include at least one '1', then recode
> columns 2 thru 7 to NA and recode column 1 to code 2.
>
> 2) If columns 2, 4 thru 7 contain at least one '1', then recode column
> 3 to NA.
>
> Desired recoding of the above three rows:
> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
> a 2 NA NA NA NA NA NA
> b 1 1 NA 1 0 0 0
> c 1 1 NA 1 0 0 0
>
> Thanks you.
You left out one key detail in the explanation, which is that the
recoding appears to be done on a row by row basis, not overall.
The following gets the job done, though there may be a more efficient
approach:
> a <- c(1,0,1,0,0,0,0)
> b <- c(1,1,1,1,0,0,0)
> c <- c(1,1,0,1,0,0,0)
> d <- rbind(a, b, c)
> d
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
a 1 0 1 0 0 0 0
b 1 1 1 1 0 0 0
c 1 1 0 1 0 0 0
> mod.row <- function(x)
{
if (all(x[c(2, 4:7)] == 0))
{
x[2:7] <- NA
x[1] <- 2
} else {
x[3] <- NA
}
x
}
> y <- t(apply(d, 1, mod.row))
> y
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
a 2 NA NA NA NA NA NA
b 1 1 NA 1 0 0 0
c 1 1 NA 1 0 0 0
HTH,
Marc Schwartz
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