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Assigning Null / Multi-Dimensional Matrix

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Hello,
I have a matrix with the dimensions of 4x2500 and I am generating this matrix in for loop. I want this matrix to be 4x1250. Without multi dimension I just write A(1:length(X)) = [] but I am stuck in multi dimension. This logic does not work in my operation.
when I say
A = [1:1:2500];
A (1:1250) = [];
it works and that is what i want to do.
but in multi dimension,
for i = 1:1:4
A(i,:) = ????????
end
Can someone help me? Thanks.
  2 Commenti
KSSV
KSSV il 7 Apr 2022
Question is not clear. What you want to do in the loop?
tinkyminky93
tinkyminky93 il 7 Apr 2022
I have an array and I am writing this array to every row of the matrix. For example lets say I have [1:1:5]. What I do in for loop is
A(i,:) = my array
i'th row : 1 2 3 4 5
i+1'th row: 1 2 3 4 5
i+2'th row: 1 2 3 4 5
i+3'th row: 1 2 3 4 5
I want to assign null to every row of this matrix, then it will become
i'th row : 4 5
i+1'th row: 4 5
i+2'th row: 4 5
i+3'th row: 4 5

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Risposta accettata

Arif Hoq
Arif Hoq il 7 Apr 2022
vectorized solution is the most efficient and simple. But, still if you need for loop, try this
x=1:5;
A = repmat(x,4,1);
for i = 3:-1:1 % if you want to delete from Column 3 then index i will be started from 3.
A(:,i) = []
end

Più risposte (1)

KSSV
KSSV il 7 Apr 2022
A = repmat(1:5,4,1)
A = 4×5
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
A(:,1:3) = []
A = 4×2
4 5 4 5 4 5 4 5
  1 Commento
tinkyminky93
tinkyminky93 il 7 Apr 2022
Modificato: tinkyminky93 il 7 Apr 2022
Can you try it with for loop? Will it be the same? I am getting error when i try
A = repmat(1:5,4,1);
for i = 1:1:4
A(:,1:3) = [];
end
>> Matrix index is out of range for deletion

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