Re: [R] Replace missing values in lapply

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Wed 24 Jan 2007 - 16:06:16 GMT

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@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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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Thu Jan 25 13:03:18 2007

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