Deleting the row of a 3D matrix with a condition
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Ashane Fernando
am 15 Feb. 2019
Kommentiert: Ashane Fernando
am 22 Feb. 2019
I have a simple 3D matrix. Every time I do this operation on the (:,:,1) matirx it gives me an error saying "A null assignment can have only one non-colon index." I dont understand this. But in a simple 2D matrix it works. If I assign "NaN" instead of "[]" then the second row is assigned with NaN NaN values (in 3D). Can I not delete the row in a 3D matrix? Appreciate some insight to this.
A(:,:,1) = [1 2 ; 0 5 ; 7 8 ];
A(:,:,2) = [10 11 ; 1 14 ; 0 17 ];
A((A(:,1,1).*A(:,2,1))==0,:,1)=[]
1 Kommentar
KALYAN ACHARJYA
am 15 Feb. 2019
As per undestanding from your question, you have to delete the complete plane, not a row line.
Akzeptierte Antwort
Jan
am 15 Feb. 2019
Bearbeitet: Jan
am 15 Feb. 2019
Of course you cannot delete one row in a 3D array: The resulting array must have the same length for all subvectors. Removing a row from a 3D array is equivalent to removing one element from a matrix: The result would not be rectangular anymore, but matrices must have the same number of elements for all rows, and all columns, respectively.
A(:,:,1) = [1 2 ; 0 5 ; 7 8 ];
A(:,:,2) = [10 11 ; 1 14 ; 0 17 ];
B = A;
B(:, :, 1) = [] % Working
A(1, :, 1) = [] % Error: output cannot be a valid array
Therefore I cannot guess, what you expect as output and cannot suggest a solution. Assigning NaN to a row works, because this does not change the number of elements.
5 Kommentare
Jan
am 15 Feb. 2019
Sorry, I cannot follow you. I asked you what happens if you have only 1 zero in the input, such that you cannot remove a single subvector without destroying the rectangular shape of the array. Now your answer is: If the array contains one 0 only, then it actually contains two? What does this mean?
If all submatrices contain the same number of zeros, than you can remove the corresponding rows or columns or what you want. But you did not clarify yet, if this is guaranteed.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!