How to update struct array fields with mutiple values

I am trying to update a field value in a struct array. For example if I have 1*10 struct of A with a field in it called B, I want to replace the following loop with another method:

for iLoop=1:10
  A(iLoop).B = iLoop;
end

I tried:

[A.B] = deal(1:10);

And also:

A = setfield(A,num2cell(1:10),'B',num2cell(1:10),(1:10));

But none of them worked (the first method assigns the whole (1:10) vector to each 'B' field in the struct array. The second one crashes). Does anyone know how to make it work?

Answer 1

[Walter Roberson]

t = num2cell(1:10);
[A.B] = t{:};

See comments for the case where the struct does not already exist.

[Mohammad Tabesh]

I feel stupid for not having tried this one :D

Simple and Works Perfectly!

[Bradley Stiritz]

Thank you Walter, but doesn't your solution force all the field contents to be cell values? This isn't convenient for all contexts.

[Stephen23]

"doesn't your solution force all the field contents to be cell values?"

No, it doesn't. The LHS has nothing to do with any cell arrays. Both the LHS and RHS use comma-separated lists, which happen to be defined using a non-scalar structure and a non-scalar cell array respectively. No field/element/value of the LHS is "forced" to be a cell array, which you can easily check yourself:

>> V = 1:10;
>> C = num2cell(V);
>> [A.B] = C{:};
>> A(1).B
ans =  1
>> class(A(1).B)
ans = double

Do you see any cell array anywhere in A or its fields? I don't.

You can learn about comma-separated lists here:

https://www.mathworks.com/matlabcentral/answers/320713-how-to-operate-on-comma-separated-lists

Your comment and five-star rating to this FEX submission

https://www.mathworks.com/matlabcentral/fileexchange/33866-disperse

is interesting, as it does exactly the same thing as Walter Roberson's answer. Lets try it:

>> [S.B] = disperse(V);
>> S(1).B
ans =  1
>> class(S(1).B)
ans = double

Exactly the same as Walter Roberson's answer above. There is absolutely no difference.

[Walter Roberson]

Note: better is

[A(1:length(t)).B] = t{:};

[breathi]

Walter,

please edit your answer and insert the solution you posted in the comments.

It was driving me nuts to find out, that only an existing struct array can be filled with your current solution, just to find out a few angry debug steps later that you posted a comment and that this solution could just create a new struct array by explicitly using length(t) as the left side input.

Thanks.

[Walter Roberson]

For the case where the struct does not already exist, there is a different method that can be easier:

A = struct('B', t);

where t is the cell array from above. This also permits you to store to multiple fields, and to leave some fields empty, and to put in non-scalar values.

A = struct('B', num2cell(1:10), 'C', [], 'D', num2cell(rand(2,10),1));

Answer 2

[Andrew Newell]

What timing! It happens that the File Exchange Pick of the Week is the function disperse. If you download disperse and put it on your path, you can use the following command:

[A.B] = disperse(1:10);

[Mohammad Tabesh]

Thank you for your help but Walter's suggestion works perfectly and does not need any functions.

[Bradley Stiritz]

Thank you Andrew! disperse() is just what I needed.

[Stephen23]

"disperse() is just what I needed."

Actually Walter Roberson's answer does exactly the same thing, without requiring third-party functions.

Answer 3

[Tom]

This works perfectly, until you have a field in between.

[A.B] = t{:}; %Great!

[A.B(1).C] = t{:}; % doesn't work

Anyone have a suggestion on how to make it work this way?

[Mohammad Tabesh]

If your question structure is that A has 10 rows, each have a field B which in turn has a field C that you want to update, I suggest creating an intermediate structure array similar to B, assign the values there and then assign that to field B:

D = A.B;
[D.C] = t{:};
[A.B] = deal(D);

Answer 4

[Mitja M]

I would highly appreciate the solution to the following problem, which I believe is highly related to the previous, but I somehow don't find the appropriate solution.

I have the following variables:

AA, 1×3 struct array with fields: bb

bb, nx6 double array

cc, nx3 double array

n for all three bb and cc arrays is equal to 100 right now

Now I would like to change the fifth column of all three bb arrays to corresponding column in cc array. It can be correctly done using the following for loop:

for i=1:3
AA(i).bb(:,5)=cc(:,i);
end

Is it possible to achieve this without the for loop.

Thank you

[Mohammad Tabesh]

Hi Mitja, I suggest a similar approach as I suggested to Tom.

BB=cat(3,AA.bb);
BB(:,5,:)=cc;
BBB=num2cell(BB,[1 2]);
[AA.bb]=deal(BBB{:});