Replace missing values in lapply

-----Original Message-----
From: Gabor Grothendieck [mailto:ggrothendieck at gmail.com] 
Sent: Wednesday, January 24, 2007 11:06 AM
To: Doran, Harold
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] Replace missing values in lapply

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


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