arrayfun doesn't work with rmoutliers

2 vues (au cours des 30 derniers jours)
Tom K.
Tom K. le 5 Jan 2022
Modifié(e) : Tom K. le 5 Jan 2022
Goal
Remove outliers along dimension d of a 3D matrix using arrayfun and rmoutliers (or superior alternatives).
Minimal Working Example
Take the following matrix:
A = [0 2 0 0 0; 0 0 0 4 0; 3 0 0 3 0];
A(:,:,2) = A
A =
A(:,:,1) = 0 2 0 0 0 0 0 0 4 0 3 0 0 3 0 A(:,:,2) = 0 2 0 0 0 0 0 0 4 0 3 0 0 3 0
Since each 1D array along dimension d will have a different number of outliers, the output of arrayfun will have to be a cell array (by specifying "UniformOutput" = false). Calling rmoutliers through arrayfun returns the original matrix with the outliers (undesired behaviour):
arrayfun(@rmoutliers,A,"UniformOutput",false)
ans = 3×5×2 cell array
ans(:,:,1) = {[0]} {[2]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[4]} {[0]} {[3]} {[0]} {[0]} {[3]} {[0]} ans(:,:,2) = {[0]} {[2]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]} {[4]} {[0]} {[3]} {[0]} {[0]} {[3]} {[0]}
The function works as expected when called on each row by itself:
rmoutliers(A(1,:,1)) % removes the 2
ans = 1×4
0 0 0 0
rmoutliers(A(3,:,1)) % removes both 3's
ans = 1×3
0 0 0
Hopefully I'm missing something trivial.

Réponse acceptée

Voss
Voss le 5 Jan 2022
A = [0 2 0 0 0; 0 0 0 4 0; 3 0 0 3 0];
A(:,:,2) = A
A =
A(:,:,1) = 0 2 0 0 0 0 0 0 4 0 3 0 0 3 0 A(:,:,2) = 0 2 0 0 0 0 0 0 4 0 3 0 0 3 0
C = num2cell(A,2)
C = 3×1×2 cell array
C(:,:,1) = {[0 2 0 0 0]} {[0 0 0 4 0]} {[3 0 0 3 0]} C(:,:,2) = {[0 2 0 0 0]} {[0 0 0 4 0]} {[3 0 0 3 0]}
cellfun(@rmoutliers,C,"UniformOutput",false)
ans = 3×1×2 cell array
ans(:,:,1) = {[0 0 0 0]} {[0 0 0 0]} {[ 0 0 0]} ans(:,:,2) = {[0 0 0 0]} {[0 0 0 0]} {[ 0 0 0]}
  1 commentaire
Tom K.
Tom K. le 5 Jan 2022
This works.
Thank you!

Connectez-vous pour commenter.

Plus de réponses (1)

Mike Croucher
Mike Croucher le 5 Jan 2022
arrayfun operates on every element of the array. So
arrayfun(@rmoutliers,A,"UniformOutput",false)
is like doing
rmoutliers(0)
rmoutliers(0)
rmoutliers(3)
etc
  1 commentaire
Tom K.
Tom K. le 5 Jan 2022
Modifié(e) : Tom K. le 5 Jan 2022
Hi Mike,
I was able to produce the desired results using @Benjamin's solution.
Thanks!

Connectez-vous pour commenter.

Catégories

En savoir plus sur Dates and Time dans Help Center et File Exchange

Produits


Version

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by