vector notation with data from struct array

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I have the following problem: I want to operate with variables from my workspace which I saved in a struct array. This will be the argument for a self defined function which you can see in the following:
t_abtast = 2;
data = load('GMapsRoute02182022095722SimulationRun.mat')
v = data.GMapsRoute02182022095722SimulationRun.velocity_kmh(1:10:end)';
a = data.GMapsRoute02182022095722SimulationRun.acc(1:10:end)';
t = data.GMapsRoute02182022095722SimulationRun.t(1:10:end);
E = data.GMapsRoute02182022095722SimulationRun.energy(1:10:end)';
v = v(~isnan(v));
a = a(~isnan(a));
E = E(~isnan(E));
v = smooth(v);
a = smooth(a);
E = smooth(E);
v = v(1:t_abtast:end);
a = a(1:t_abtast:end);
t = t(1:t_abtast:end);
E = E(1:t_abtast:end);
t_delta = t(2)-t(1);
P = [0, diff(E)/t_delta];
adot = [0, diff(a)/t_delta];
[A,B,X,Y,y,res,x_vec,x_pred,u_pred] = DMDc1(5,10,4,v,a,adot,P);
...
I get this error message:
Error using horzcat
Dimensions of arrays
being concatenated are
not consistent.
Error in DMD_Prediction
(line 42)
P = [0
diff(E)/t_delta];
Before I loaded the variables with the evalin function and there was not problem.
load('GMapsRoute02182022095722SimulationRun.mat')
v = evalin('base','GMapsRoute02182022095722SimulationRun.velocity_kmh');
a = evalin('base','GMapsRoute02182022095722SimulationRun.acc');
t = evalin('base','GMapsRoute02182022095722SimulationRun.t');
E = evalin('base','GMapsRoute02182022095722SimulationRun.energy');
When I change the notation from "," to ";", I receive the following error:
Unable to perform assignment because the size of the left side is 4-by-239
and the size of the right side is 972-by-0.
Error in DMDc1 (line 71)
X((ii-1)*states+1:ii*states,:) = y(:,ii:end-n+ii);
Error in DMD_Prediction (line 47)
[A,B,X,Y,y,res,x_vec,x_pred,u_pred] = DMDc1(5,10,4,v,a,adot,P);
So, this rewriting leads to problem later in the code.

Accepted Answer

Matt J
Matt J on 22 Mar 2022
Error in DMD_Prediction (line 42)
P = [0
diff(E)/t_delta];
In all likelihood diff(E) is not a row vector. THat should be easy to check.
  2 Comments
Matt J
Matt J on 22 Mar 2022
Edited: Matt J on 22 Mar 2022
You're welcome, but please Accept-click the answer if it resolved your question.

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