Indexing from equivalent rows in two matrices

Hi all,

I am extract data from a structure based on a rule from one field, and another rule in the corresponding row of a second field. For that I have the following code:

    for k= 1:size(MouseData, 2) % for each entry in the structure
        for n= 1:numel(Behaviours1) %for each value in this cell in this array
            % Create an index with the following rules:
                % 1. If BehaviourType (first field), matches the value in the cell array
                % 2. If the value in the 4th column of the second field (in the corresponding row), matches the value "NovMouse"
            idx = find (strcmp(MouseData(k).BehaviourType, Behaviours1(n)) & ((MouseData(k).BehaviourData(:,4) == NovMouse)));   % make an idx (index) where the behaviour type == behaviour (logical array of 0/1)
            
            % Take that data (and so on......)
            temp = MouseData(k).BehaviourType(idx);
            NovMouseData(k).BehaviourType = vertcat(NovMouseData(k).BehaviourType, temp);
            temp1 = MouseData(k).BehaviourData(idx, :);
            NovMouseData(k).BehaviourData = vertcat(NovMouseData(k).BehaviourData, temp1);
        end
    end
  

I think I am going wrong in the line beginning with "idx", I don't think the code knows that I am asking it to find the corresponding rows. I know that I could create an if statement for this, but I worry that the code might take a long time to run if I do this?

Answer 1

[Raunak Gupta]

Hi,

One improvement I see in the code is you can remove

for n= 1:numel(Behaviours1) 

Because strcmp also works when comparing a string with a cell array containing multiple strings. Here since the structures are stored in vector format you may not be able to remove outermost for loop and you are using k downwards so outermost for loop is required. I am assuming that Behaviours1 is a cell array containing strings and MouseData(k).BehaviourType is also a cell array because temp is calculated with some index idx. Rest all look ok to me.