Dear list members,
Berwin gave me the right hint, it works!
Look at this:
## R Code
a <- (1:5)
a[2] <- NA
b <- (6:10)
b[3] <- NA
a
1 NA 3 4 5
b
6 7 NA 9 10
matrix <- array(1:5*2, dim=c(5,2))
matrix[,1] <- a
matrix[,2] <- b
matrix[ !apply(is.na(matrix), 1, any),]
[,1] [,2]
[1,] 1 6
[2,] 4 9
[3,] 5 10
## /R Code
Cheers,
Fabian
Berwin A Turlach wrote:
> G'day Fabian,
>
>
>>>>>>"FL" == Fabian Lienert <[EMAIL PROTECTED]> writes:
>
>
> FL> ## R Code
> FL> awithoutan <- a[(1:length(a))[(!is.na(a))&(!is.na(b))]]
> a[(!is.na(a))&(!is.na(b))] ## works too
>
> FL> bwithoutan <- b[(1:length(b))[(!is.na(a))&(!is.na(b))]]
> b[(!is.na(a))&(!is.na(b))] ## ditto
>
> FL> How can I do the same matching in a matrix and get a matrix without NA
> FL> elements (and less colums of course)?
>
> FL> matrix <- array(1:5*2, dim=c(5,2))
> Not a good name, since matrix is also a function, could lead to
> confusion in more complicated situations. :)
>
> FL> matrix[,1] <- a
> FL> matrix[,2] <- b
> FL> matrix[,][(!is.na(matrix[,]))]
>
>>matrix[ !apply(is.na(matrix), 1, any), ]
>
> [,1] [,2]
> [1,] 1 6
> [2,] 4 9
> [3,] 5 10
>
> Hope this helps.
>
> Cheers,
>
> Berwin
>
--
Beste Grüsse,
Fabian Lienert
--
pgp public key: http://lienert.org/keys/lienert.asc
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