[R] Some difficulties to use the apply command on an array
Laurent Rhelp
L@urentRHe|p @end|ng |rom |ree@|r
Mon Jul 5 17:33:57 CEST 2021
Thank you for the explanations. Indeed I made a mistake for the MARGIN
understanding. But when I use the command I have an error because there
is no simplify argument in the apply command ?
Le 05/07/2021 à 17:05, Andrew Simmons a écrit :
> Hello,
>
>
> It seems as though you may be using `MARGIN` incorrectly. `MARGIN` is
> the indices you are looping through, so it should be MARGIN = 1, not
> MARGIN = c(2, 3). Additionally, you probably want to set argument
> `simplify` to FALSE. Something like
>
> apply(x, 1, FUN = function(x) diag( svd(x)$d ), simplify = FALSE)
>
> seems to work well for me. I hope this helps!
>
> On Mon, Jul 5, 2021 at 10:56 AM Laurent Rhelp <LaurentRHelp using free.fr
> <mailto:LaurentRHelp using 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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