[R] Compare rows of two matrices

[Adaikalavan Ramasamy]

Here is an another way

   count <- is.na(x) + is.na(y)
   which( count == 1, arr.ind=TRUE )

'count' gives you the number of missing values at for each row and
column. Then you can find out how many occurances of both missing, none
missing and one missing. 
 


On Mon, 2005-02-21 at 15:48 +0100, TEMPL Matthias wrote:
> Hello,
> 
> #I have two matrices, eg.:
> 
> y <-  matrix( c(20,  NA,  NA,  45,  50,  19,  32, 101,  10,  22,  NA,  NA,  80,  49,  61, 190), ncol=4 )
> x <-  matrix( c(20,  NA,  NA,  NA,  50,  19,  32, 101,  10,  22,  NA,  NA,  80,  49,  61, 190), ncol=4 )
> 
> #Whereas x contains all NA´s from y plus some additional NA´s.
> #I want to find the index of these additional NA´s. I think, there must be a very easy way to do this.
> 
> #Here are the indices of NA´s in x and y:
> l1 <- which(is.na(x), arr.ind=TRUE)
> l2 <- which(is.na(y), arr.ind=TRUE)
> 
> #> l1
> #     [,1] [,2]
> #[1,]    2    1
> #[2,]    3    1
> #[3,]    4    1
> #[4,]    3    3
> #[5,]    4    3
> 
> #> l2
> #     row col
> #[1,]   2   1
> #[2,]   3   1
> #[3,]   3   3
> #[4,]   4   3
> 
> #Now I want to find a matrix, which includes the values of l1, without the rows of l2, 
> #which has equal entities (the index of the additional NA´S).
> #In this example the result should be row 3 of l1 with the values 4 and 1.
> #The following code works, but I think there must be a much more elegant way to do this.
> 
> l3 <- l1
> l3 <- cbind( l1, rep(0, nrow(l1)) )
> num <- 1
>    
> for( i in 1:nrow(l1) ){
>   for( j in 1:nrow(l2) ){
>     if( l1[i,1] == l2[j,1] & l1[i,2] == l2[j,2]){
>       l3[i,3] <- 1
>     }
>   }
> }
> 
> l4 <- l3[l3[,3]==0, c(1,2)]
> 
> #> l4
> #row col 
> #  4   1  
> 
> I have often such problems like this and I assume, that other people have similar tasks.
> My question is: Does anybody know a function in one package, which compares rows of two matrices like this or have anybody an idea to do this in a much more elegant way"? 
> 
> Thank you very much,
> Matthias
> 
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