is there a way to perform this task w/o using loops?

% this script sets up a structure p.  In this minimal version, p only contains 1 variable.
% then expands it into a 1x4 structure array q 
% and inserts values into one of the variables.  (Only the L variable is shown here.)
% and then retrieves the values
% The variable is a row vector and has name given by varName
% The values go into (and are retrieved from) the 2nd element of L because varIndex = 2;
p.varName = 'L';
p.varIndex = 2;
p.L = [3, 4, 5];
nval = 4;
q = repmat(p,1,nval);
vals = [6 7 8 9];
% Insert values
for i=1:nval
    q(i).(p.varName)(p.varIndex) = vals(i);
end
Xv = zeros(1,nval);
% Retrieve values
for i=1:nval
    Xv(i) = q(i).(q(1).varName)(q(1).varIndex);
end

[Jeffrey]

In this case, q is a 1x4 structure array, but it could be 4x1 if it makes things easier.

vals and Xv are row vectors, but they could be column vectors if it make things easier.

[Walter Roberson]

Xv = arrayfun(@(IDX) q(IDX).(q(1).varName)(q(1).varIndex), 1:nval);

The assignment is more difficult to arrange.

[Jeffrey]

Understood. Thank you.

Answer 1

[Walter Roberson]

The more general vectorized strategy would probably be to set up cell arrays of values, and then use the fact that when you struct() and pass a cell array, then the output is a struct array the size of the cell array.

[Jeffrey]

Understood. I like this answer the best because it's elegant, but will probably stick with the loop instead of introducing cell arrays into my code, to keep it simple.

Answer 2

[Matt J]

There are ways to do it without loops, but doing it without loops will be of no benefit to you. It will take more lines of code, consume more memory, and run more slowly.

p.varName = 'L';
p.varIndex = 2;
p.L = [3, 4, 5];
nval = 4;
q = repmat(p,1,nval);
vals = [6 7 8 9];
% Insert values

TMP=vertcat(q.(p.varName));
TMP(:,p.varIndex)=vals;
TMP=arrayfun(@(i)TMP(i,:), 1:nval,'uni',0);

[q.(p.varName)]=deal(TMP{:});

q.L

ans = 1×3

3 6 5

ans = 1×3

3 7 5

ans = 1×3

3 8 5

ans = 1×3

3 9 5

[Jeffrey]

Understood. Thank you!

Answer 3

[Bruno Luong]

You might consider store in table instead

NOTE: using table would be more convenient to access data, not necessary faster

p.varName = "L";
p.varIndex = 2;
nval = 4;
q = repmat(p,1,nval);
vals = [6 7 8 9];
% Insert values
for i=1:nval
    q(i).(p.varName)(p.varIndex) = vals(i);
end

T = struct2table(q)

T = 4×3 table
    varName    varIndex      L   
    _______    ________    ______

      "L"         2        0    6
      "L"         2        0    7
      "L"         2        0    8
      "L"         2        0    9

T.varName(1)

ans = "L"

T.(T.varName(1))

ans = 4×2

0 6 0 7 0 8 0 9

T.(T.varName(1))(:,T.varIndex(1))

ans = 4×1

6 7 8 9

[Jeffrey]

Very good. Thank you. I hadn't heard of tables.