Matrix index is out of range for deletion.

Hello!

I cannot figure out why this is breaking. It should not be out of range as we are just comparing values. Could you please help me get to the bottom of this?

Function Call in main script:

%% Regional Channel Break down:

% each channel

% each sub-region

% included in the function call

R1 = [1,2,3,17,5,18,6,11]; % right ventrolateral prefrontal cortex

R2 = [4, 9, 10]; % right ventrolateral prefrontal cortex

R3 = [7, 8, 21,22, 12, 13,25, 26]; % right frontopolar prefrontal cortex

R4 = [14, 15, 16, 29, 30]; % right oribitofrontal cortex

L1 = [19,20,33,34,35, 38,39,43]; % left ventrolateral prefrontal cortex

L2 = [40, 44, 45]; % left ventrolateral prefrontal cortex

L3 = [23, 24, 27, 28, 36, 37, 41, 42]; % left frontopolar prefrontal cortex

L4 = [31, 32, 46, 47, 48]; % left oribitofrontal cortex

regions = input(['Which Region? Enter one of the following: ' ...

'R1, R2, R3, R4, L1, L2, L3, L4: '])

%Looping through a struct:

epochMatrixRegions = zeros(24,4);

fn=fieldnames(dataStruct); %datastruct contains the timestamps

fn=fn(2:end);

i = 1;

%loop through the fields

for i = 1:numel(fn)

%access the data

stamps=dataStruct.(fn{i});

oxyhb= Participants.(fn{i});

%call function

[epochMatrixRegions(i,1), epochMatrixRegions(i,2),...

epochMatrixRegions(i,3), epochMatrixRegions(i,4)] = ...

thinkAloudAUC_Regions(stamps,oxyhb, regions);

end

Part inside the fcn that is breaking:

%% Regions

R1 = [1,2,3,17,5,18,6,11]; % right ventrolateral prefrontal cortex

R2 = [4, 9, 10]; % right ventrolateral prefrontal cortex

R3 = [7, 8, 21,22, 12, 13,25, 26]; % right frontopolar prefrontal cortex

R4 = [14, 15, 16, 29, 30]; % right oribitofrontal cortex

L1 = [19,20,33,34,35, 38,39,43]; % left ventrolateral prefrontal cortex

L2 = [40, 44, 45]; % left ventrolateral prefrontal cortex

L3 = [23, 24, 27, 28, 36, 37, 41, 42]; % left frontopolar prefrontal cortex

L4 = [31, 32, 46, 47, 48]; % left oribitofrontal cortex

%% Load Participant Data:

% The odd columns are HbO, the even are HbR

% for now, we only want to look at oxygenated data

HbO = oxyhb.data(:,2:2:end);

%filter the HbO data for the desired region:

channels = 1:1:48;

for j = 1:length(channels)

if ismember(j,regions) == false

HbO(:,j) = [];

end

end

The error I get is:

Matrix index is out of range for deletion.

Error in thinkAloudAUC_Regions (line 45)

HbO(:,j) = [];

Error in ThinkAloudProcessing (line 265)

thinkAloudAUC_Regions(stamps,oxyhb, regions);

Answer 1

[Steven Lord]

What happens if you try to delete the 11th element of a vector with 10 elements?

try
    x = 1:10;
    x(11) = [];
catch ME
    fprintf("This code threw error '%s'", ME.message)
end

This code threw error 'Matrix index is out of range for deletion.'

What happens if you have a vector with 10 elements, delete one of those elements, then try to delete the 10th element of the result?

try
    x = 1:10;
    for k = 1:10
        if k == 5 | k == 10
            fprintf('Deleting element %d of a %d element long vector.\n', k, length(x))
            x(k) = [];
        end
    end
catch ME
    fprintf("This code threw error '%s'", ME.message)
end

Deleting element 5 of a 10 element long vector.
Deleting element 10 of a 9 element long vector.

This code threw error 'Matrix index is out of range for deletion.'

One common solution to this is to work your way backwards, starting with the last element working your way towards the first. Another solution would be to create a vector of elements to delete or to keep inside the loop and only perform the actual pruning at the end.

x = 1:10;
todelete = false(size(x));
for k = 1:10
    if k == 5 | k == 10
        fprintf('Deleting element %d of a %d element long vector.\n', k, length(x))
        todelete(k) = true;
    end
end

Deleting element 5 of a 10 element long vector.
Deleting element 10 of a 10 element long vector.

fprintf("Original x\n")

Original x

disp(x)

     1     2     3     4     5     6     7     8     9    10

x(todelete) = [];
fprintf("x after deletion\n")

x after deletion

disp(x)

     1     2     3     4     6     7     8     9

[Emma Walker]

I do not follow. Here is another explanation maybe:

It seems like a simple question.

Example:

regions = [ 2 5 8 12 36 44] 6 channels

hbo is 7008x48 double

i want hbo to be 7008x6, containing the 7008 pieces of data in columns 2,5,8,12,36,44.

How would I do that?

[Stephen23]

"How would I do that?"

out = hbo(:,regions);

Your approach (using FOR loop and removing individual columns one at a time) suffers from the exact problem that Steven Lord explained. Using a loop is a red-herring.

[Steven Lord]

@Stephen23 is correct. This is a case where instead of a list of columns to delete you have a list of columns to keep.

[Emma Walker]

Thank you!