[R] Replace missing values in lapply

Gabor Grothendieck ggrothendieck at gmail.com
Wed Jan 24 17:06:16 CET 2007

```I wonder if a list of matrices is the best representation?
Do your matrices all have the same dimension as in:

TP <- list(matrix(c(1:3, NA), 2), matrix(c(NA, 1:3), 2))

# Then you could consider representing them as an array:

TPa <- array(unlist(TP), c(2,2,2))

# in which case its just

TPa[is.na(TPa)] <- 0
TPa

On 1/24/07, Doran, Harold <HDoran at air.org> wrote:
> I have some matrices stored as elements in a list that I am working
> with. On example is provided below as TP[[18]]
>
> > TP[[18]]
>      level2
> level1  1  2  3  4
>     1 79  0  0  0
>     2  0  0  0  0
>     3  0  0  0  0
>     4  0  0  0  0
>
> Now, using prop.table on this gives
>
> > prop.table(TP[[18]],1)
>      level2
> level1   1   2   3   4
>     1   1   0   0   0
>     2
>     3
>     4
>
> It is important for the zero's to retain their position as this matrix
> will subsequently be used in some matrix multiplication and hence, must
> be of dimension 4 by 4 so that is it conformable for multiplcation with
> another matrix.
>
> In looking at the structure of the object resulting from prop.table I
> see NaNs, and so I can do this
>
> > rr <- TP[[18]]
> > rr[is.na(rr)] <- 0
> > rr
>      level2
> level1  1  2  3  4
>     1 79  0  0  0
>     2  0  0  0  0
>     3  0  0  0  0
>     4  0  0  0  0
>
> This is exactly what I want for each matrix. But, I have multiple
> matrices stored within the list that need to be changed and so I am
> trying to resolve this via lapply, but something is awry (namely the
> user), but I could use a little help.
>
> I was thinking the following function should work, but it doesn't. It
> reduces each matrix within the list to a 0.
>
> PP <- lapply(TP, function(x) x[is.na(x)] <- 0)
>
> Am I missing something obvious?
>
> Harold
>
>
>        [[alternative HTML version deleted]]
>
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