how insert array in field struct

Z is matrix number

for i=1:width(A)

A(i).b=z(:,i)

end

it's possibile to avoid loop?

Answer 1

[Vishnu]

Hi @Luca Re,

Yes, it is possible to avoid the loop and achieve the same result using matrix operations in MATLAB using the following snippet

% Assuming A is a structure array and z is a matrix
z = [1 2 3; 4 5 6; 7 8 9];  % Example matrix

% Convert z to a cell array of column vectors
zCell = mat2cell(z, size(z, 1), ones(1, size(z, 2)));

% Assign the columns of z to the field b of A using comma-separated lists
[A.b] = zCell{:};

A.b

ans = 3×1

1 4 7

[Stephen23]

!!! WARNING !!!

If A does not exist in the workspace or if it is e.g. a scalar structure, then this code will make data disappear without any warning. This is shown in the author's own example: note how the 2nd and 3rd columns are not present in the expansion of A.b

[Paul]

@Stephen23

Is this a better way, at least for the case where A does not already exist. I think it will also work if A already exists.

z = [1 2 3; 4 5 6; 7 8 9];  % Example matrix
zCell = mat2cell(z, size(z, 1), ones(1, size(z, 2)));  % or use num2cell(z,1)
[A(1:size(z,2)).b] = zCell{:};
A.b

ans = 3×1

1 4 7

ans = 3×1

2 5 8

ans = 3×1

3 6 9

[Stephen23]

"Is this a better way, at least for the case where A does not already exist."

In that case STRUCT would be simpler and makes the intent much clearer.

"I think it will also work if A already exists."

In that case fiddling around with linear/subscript indexing on the LHS is not required.

Answer 2

[Stephen23]

"it's possibile to avoid loop?"

Does A already exist or not? Your question does not make this clear... here are both cases:

Z = [1,2,3; 4,5,6; 7,8,9]

Z = 3×3

1 2 3 4 5 6 7 8 9

C = num2cell(Z,1);
A = struct('a',C) % structure does not exist

A = 1×3 struct array with fields:
    a

[A.b] = C{:} % structure already exists

A = 1×3 struct array with fields:
    a
    b

Checking:

A.b

ans = 3×1

1 4 7

ans = 3×1

2 5 8

ans = 3×1

3 6 9

Read more:

https://www.mathworks.com/help/matlab/matlab_prog/comma-separated-lists.html

https://www.mathworks.com/matlabcentral/answers/1656435-tutorial-comma-separated-lists-and-how-to-use-them

Answer 3

[Rahul]

Hi Luca,

This is possible. You can try out the following code for same.

A = struct('b', cell(1, width(A)));  % Preallocate struct array

% Assign columns of z to the field 'b' of each struct element in A
[A.b] = deal(z);

Here deal(z) is used to assign each column of z to the corresponding field b in each struct element of A. This way, you can avoid the need for a loop and achieve the desired assignment efficiently.

Make sure that the number of columns in z matches the number of elements in A (i.e., width(A) or numel(A)).

Hope this helps.

Thanks.

[Luca Re]

deal is not correct....i get 82x82..

instead I want to get 82x1

[Image Analyst]

It's not clear if you want

1. an array of structures (82 A's) with each structure having one field "b" which is a column vector of 82 elements, OR
2. if you want a single structure (one A, not an array of 82 of them) and that one structure has a single field which is an 82 element column vector.

Your initial code seems to indicate 1 -- an array of structures, but is that really true?

Why do you want to avoid the loop anyway? With a microscopic data size like 82, a for loop may well be faster than messing around with inefficient cell arrays. Either way will be so fast you won't be able to perceive a difference between them anyway.

[Luca Re]

the solution of stephen23 is correct!

C = num2cella(Z,1);

[Ab] = C{:} % la struttura esiste già