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Mean-replacing NAs in a 3d array

2 messages · T.R. Marshall, Eik Vettorazzi

T.R. Marshall
# 
Hi all

I have a 3d array containing missing values.
, , 1

     [,1] [,2]
[1,]    1    3
[2,]   NA    4

, , 2

     [,1] [,2]
[1,]    5    7
[2,]   NA   NA

, , 3

     [,1] [,2]
[1,]    9   11
[2,]   10   12

I want to replace the missing values with the mean, but the mean of each
'page' in the array (wrong terminology I'm sure). So - for the array
above - [2,1,2] and [2,2,2] should both read '6', and [2,1,1] should be
'2.66667'.

I can obtain a vector of those means by:
[1]  2.666667  6.000000 10.500000

But what to do next?

Help would be appreciated. Also I'm a noob to this mailing list so
apologies if I've not given enough detail or failed to find the answer
somewhere obvious.

Best
Tom Marshall, Universiteit van Amsterdam
Eik Vettorazzi
# 
Hi Tom,
it would have been nice (and it is in fact requested by the posting 
guide) to give a running example instead of letting us construct your data.
Anyway, it wasn't too hard with
 Xa<-array(1:12,dim=c(2,2,3))
 Xa[c(2,6,8)]<-NA

#so next, create a vector for the colMeans with length 2*2*3, repeating 
each mean 4 times
cM<-rep(colMeans(Xa,na.rm=TRUE,dims=2),each=dim(Xa)[1]*dim(Xa)[2])

#... and use the fact, that most r-objects have also an one-dimensional 
index (which was used in the construction above as well)
Xa[is.na(Xa)]<-cM[is.na(Xa)]

hth.

T.R. Marshall schrieb: