[R] Seeking help with an apparently simple recoding problem

[Marc Schwartz (via MN)]

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