How do I rectify the error corresponding to the Dimensions of arrays being concatenated no being consistent
1 view (last 30 days)
Show older comments
ecg_struct = load('test.mat');
fn = fieldnames(ecg_struct);
if length(fn) ~= 1
fprintf(2, 'Warning: file contains %d different variables, arbitrarily picking the first: %s\n', length(fn), fn{1});
end
ecg_i = ecg_struct.(fn{1});
s=ecg_i;
N=size(s,2);
ECG=s;
FIR_c1=[0.0041,0.0053,0.0068,0.0080,0.0081,0.0058,-0.0000,-0.0097,-0.0226,...
-0.0370,-0.0498,-0.0577,-0.0576,-0.0477,-0.0278,0,0.0318,0.0625,0.0867,...
0.1000,0.1000,0.0867,0.0625,0.0318,0,-0.0278,-0.0477,-0.0576,-0.0577,...
-0.0498,-0.0370,-0.0226,-0.0097,-0.0000,0.0058,0.0081,0.0080,0.0068,...
0.0053,0.0041];
FIR_c2=[0.0070,0.0094,0.0162,0.0269,0.0405,0.0555,0.0703,0.0833,0.0928,...
0.0979,0.0979,0.0928,0.0833,0.0703,0.0555,0.0405,0.0269,0.0162,0.0094,...
0.0070];
l1=size(FIR_c1,2);
ECG_l=[ones(1,l1)*ECG(1) ECG ones(1,l1)*ECG(N)]; %Error here
ECG=filter(FIR_c1,1,ECG_l);
ECG=ECG((l1+1):(N+l1));
1 Comment
Stephen23
on 15 Dec 2022
Edited: Stephen23
on 15 Dec 2022
The data you uploaded has size 1200x18000, but your code:
ECG_l=[ones(1,l1)*ECG(1) ECG ones(1,l1)*ECG(N)];
is unclear, what you expect to occur. ECG(1) and ECG(N) will be scalars, but because N is the number of columns which you then use as a linear index it will select... element (80,15). Very unusual, and most likely not what you intended.
Then you concatenate horizontally, but with two row vectors: what do you expect to occur when you concatenate such arrays horizontally together?:
[1x18000, 1280x18000, 1x18000] % not possible: the number of rows are different
PS: why do you need three different names for exactly the same thing?:
ecg_i = ecg_struct.(fn{1});
s=ecg_i;
ECG=s;
Accepted Answer
Stephen23
on 15 Dec 2022
Making a wild guess about the wanted output array size:
ECG_l = ECG([1,1:end,end],:);
0 Comments
More Answers (0)
See Also
Categories
Find more on Matrices and Arrays in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!