Some difficulties to use the apply command on an array

apply _always_ returns a matrix of vector results. That is, the result of each call to FUN may be a matrix, but a matrix is merely a vector with dimensions that are in this case ignored. So restore the dimensions on the result:

array(apply(x, 1, FUN = function(x) diag( svd(x)$d )), dim=c(3,3,5))


On July 5, 2021 8:15:16 AM PDT, Eric Berger <ericjberger at gmail.com> wrote:

Hi Laurent,
I am not sure how to get apply()  to work but the following uses
lapply()
and returns the matrices in a list.

lapply(1:3, FUN = function(i) diag( svd(x[i,,])$d ))

HTH,
Eric

On Mon, Jul 5, 2021 at 5:56 PM Laurent Rhelp <LaurentRHelp at free.fr>
wrote:

Dear R-Help,

I have an array x made up of three matrices of 5 rows and 3 columns

of

complex numbers (the complex numbers are not the problem)

## my array
x <- structure(c(5.6196790161893828+0i, 5.7565523942393364+0i,
8.5242834298729342+0i,
                10.304766710160479+0i, 11.412967010108229+0i,
-2.6952197604866495-5.7324226520260237e-18i,
                -3.8053698833683476-1.5535793240880411i,
-4.440850591952894-0.21277190153473785i,
                -4.9049897326853316+0.22487294128201613i,
-4.6982778900868931-0.35986943359186585i,
                -6.7722948827866034-1.8252881225149525e-16i,
-5.6864255243941733-0.82149125972147463i,
                -2.5321753317962115-3.4309121438578525i,
0.85377696343414344-4.3720244641138883i,
                6.3363344379954638-2.23654533694011i,
-2.6952197604866495+5.7324226520260237e-18i,
                -3.8053698833683476+1.5535793240880411i,
-4.440850591952894+0.21277190153473785i,
                -4.9049897326853316-0.22487294128201613i,
-4.6982778900868931+0.35986943359186585i,
                3.1683154084671417+0i, 5.397392162921844+0i,
8.3085887222115922+0i,
                11.150391891976289+0i, 13.173434845070231+0i,
3.7978840709983359+5.0735539531499429e-17i,
                3.0599215005995717-2.3669737981128267i,
0.5951827523696811-3.2116907301709845i,
                -1.7735244105472532-3.8366652813316566i,
-5.0558348297797169-2.6645747173403049i,
                -6.7722948827866034+1.8252881225149525e-16i,
-5.6864255243941733+0.82149125972147463i,
                -2.5321753317962115+3.4309121438578525i,
0.85377696343414344+4.3720244641138883i,
                6.3363344379954638+2.23654533694011i,
3.7978840709983359-5.0735539531499429e-17i,
                3.0599215005995717+2.3669737981128267i,
0.5951827523696811+3.2116907301709845i,
                -1.7735244105472532+3.8366652813316566i,
-5.0558348297797169+2.6645747173403049i,
                8.5581082281979697+0i, 8.6908832216086331+0i,
12.394094469562258+0i,
                16.106350896659897+0i, 19.569513600539693+0i), .Dim =
c(5L, 3L,
3L))

str(x)
#  cplx [1:5, 1:3, 1:3] 5.62+0i 5.76+0i 8.52+0i ...

The dimensions (i,j,k) of my array are 5, 3, 3. For every value of i,

I

am interested in the squared matrices made up of the j columns for

all

the k indice. For a given value i I have to keep the row i for every

k.

For example if i=1, I am interested by the matrix x[1,,]
##
x[1,,]

##              [,1]         [,2]         [,3]
## [1,]  5.619679+0i -2.695220+0i -6.772295+0i
## [2,] -2.695220-0i  3.168315+0i  3.797884-0i
## [3,] -6.772295-0i  3.797884+0i  8.558108+0i

Now, for every value of i, I want to calculate the SVD decomposition

and

get the diagonal matrix created by the singular values. When I set

the i

value it works:
diag( svd(x[1,,])$d )

# [,1]     [,2]      [,3]
# [1,] 15.73833 0.000000 0.0000000
# [2,]  0.00000 1.502771 0.0000000
# [3,]  0.00000 0.000000 0.1049992


But when I try to do a loop on the i value it doesn't work:

apply(x, c(2,3), FUN = function(x) diag( svd(x)$d ))


# [,1]        [,2]        [,3]
# [1,] Numeric,361 Numeric,81  Numeric,144
# [2,] Numeric,81  Numeric,400 Numeric,81
# [3,] Numeric,144 Numeric,81  Numeric,900

I do not understand how to do the apply on an array in order to get

my

diagonals matrices. I think that I miss something about the
understanding of the array object.

Thank you for your help

Laurent




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