How to use the fields in the structure?

My structure has 83 Fields, and each field has 1059 *3 Double. Firstly how do I access these fields from the structure and then how do I multiply a matrix(rotation matrix) on these fields?. I know that I have to use for loop to access all the values and I have found this code But I do not know how It does function.

       for i = fieldnames(markerStruct)'
       newMarkerStruct.(i{1}) = markerStruct.(i{1});
end

I understand that values from markerStruct gets copied to newmarkerStruct. But I don't understand what i{1} means and How the value is getting assigned.

Answer 1

[Adam Danz]

I've responded to your quesitons below. First I make some fake data to work with.

% Fake data
S.f1 = rand(1059, 3); 
S.f2 = rand(1059, 3); 
S.f3 = rand(1059, 3); 

"Firstly how do I access these fields from the structure?"

% Field f1 is accessed like this
S.f1
% or dynamically
fn = fieldnames(S);   %list of field names
S.(fn{1})

% add a field
S.f4 = rand(1059, 3); 

"How do I multiply a matrix(rotation matrix) on these fields?"

% rotation matrix
ry = [ cos(pi)  0  sin(pi);
    0        1    0       ;
    -sin(pi)  0  cos(pi)] ;

% Multiply the matrix by the first field . 
f1Rot = S.f1 * ry; 

"I know that I have to use for loop to access all the values "

% Actually, not true.  You can use structfun() to apply a function to all fields.  
% In this example, it's expected that all fields have 3 columns of data.
S2 = structfun(@(x) x * ry, S, 'UniformOutput', false); 

% But if you must use a loop
fn = fieldnames(S);   %list of field names
S2 = struct;          % create new structure (no fields yet)
for i = 1:length(fn)
    S2.(fn{i}) = S.(fn{i}) * ry; %store results in a new struct with same field names
end

"I have found this code But I do not know how It does function"

% This for-loop does nothing.  It's just looping through each field and re-assigning the identical value.
% It's like A = A.  
for i = fieldnames(markerStruct)'
    newMarkerStruct.(i{1}) = markerStruct.(i{1});
end

fieldnames() lists all field names in a cell array of strings.

[Praveen Kumar Pakkirisamy]

Thank you for the detailed answer. The last where you explained that it's going through each field and re-assigning the identical value, that I understood. But I did not understand why is that i{1}, What does that 1 signify?

[Walter Roberson]

fieldnames() returns a cell array of character vectors, in the form of a column vector cell. The ' at the end of the fieldnames(markerStruct) call transforms that to a cell array row vector.

When you for i = something then on each iteration, i will be assigned the next column of the values. With you having a cell array row vector, the columns are cell array scalars -- but still cell arrays. The i{1} is pulling out the content of that cell array scalar, giving you the character vector that is stored inside of it.

[Adam Danz]

The function fieldnames() lists all field names in a cell array. The loop is looping through each field name using dynamic field names which I showed in my examples above.

Here's an example

markerStruct.f1 = 'first'; 
markerStruct.f2 = 'second'; 
markerStruct.f3 = 'third'; 

% list all field names
fn = fieldnames(markerStruct)' 
%fn =
%  1×3 cell array
%    {'f1'}    {'f2'}    {'f3'}

% Look at the first field name
fn{1}
%ans =
%    'f1'

% Look at the value stored in the first field
markerStruct.(fn{1})
%ans =
%   'first'

% Loop through all fields
for i = fieldnames(markerStruct)'
   markerStruct.(i{1})
end

[Walter Roberson]

The action of for when given things that are not numeric row vectors is admittedly a bit obscure.

[Adam Danz]

Yeah, to be honest, this is my first time seeing it.

If you're going to use a loop (rather than structfun, which I showed in my examples), then this is the better approach:

fn = fieldnames(S);   %list of field names
for i = 1:length(fn)
    S.(fn{i}) * ry
end

[Praveen Kumar Pakkirisamy]

Thank you so much for the answer. Appreciate it.