Plotting real data and prediction
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I have a real data in excel, I want to make a future prediction.I have the following code but it only produce figure from the current data.
clc clear
%% % Reading the data from .xlsx file path (path,'C:\Users\User\Desktop\MSC PRO%');
[num, txt, raw] = xlsread('Book1.xlsx','data pull'); TN = xlsread('Book1.xlsx','data pull');
%% % Initializing the variable ID = (TN(:,1)); N = 200000000; I = (TN(:,3)); E = 4.*I; R = (TN(:,5)); D = (TN(:,7)); X = (E + I + R + D); S = N - X;
s = S/N; e = E/N; i = I/N; r = R/N; d = D/N;
%% %plotting the current data figure(1); plot(ID, I,'r',ID,R,'b',ID, D,'y'); xlabel('Days'); ylabel('population'); legend('Confirmed', 'Recovered','Death'); %% % finding apha1,apha3, apha4 and apha5 alpha1 = -1*(1/5.*e + diff(s) + diff(d) + diff(r) + diff(r))./(s.*i); alpha1(isnan(alpha1)) = []; alpha1(isinf(alpha1)) = []; alpha1 = mean(alpha1);
alpha3 = (1/5.*e - diff(d) - diff(i))./(i); alpha3(isnan(alpha3)) = []; alpha3(isinf(alpha3)) = []; alpha3 = mean(alpha3);
alpha4 = diff(d)./(i); alpha4(isnan(alpha4)) = []; alpha4(isinf(alpha4)) = []; alpha4 = mean(alpha4);
alpha5 = (1/5.*e - diff(d) - diff(i) - diff(r))./(r); alpha5(isnan(alpha5)) = []; alpha5(isinf(alpha5)) = []; alpha5 = mean(alpha5);
%% %Slving the equations t0 = 0; tf = 756; y0 = [s(end) e(end) i(end) r(end) d(end)]; [t y] = ode45('ypseird',[t0 tf],y0);
%% %plotting the prediction figure(2) plot(t,y(:,1),t,y(:,2),t,y(:,3),t,y(:,4),t,y(:,5)) title('SEIRD Model'); xlable('Times') ylable('fraction of the Population') legend('Susciple','Exposed', 'Infected', 'Recovered', 'Death')
The following are the errors. 1. Error using ==> plus Matrix dimensions must agree. Error in ==> seird at 40 alpha1 = -1*(1/5.*e + diff(s) + diff(d) + diff(r) + diff(r))./(s.*i);
2. Error in ==> seird at 43 [t y] = ode45('ypseird',[t0 tf],y0);
Please I need assistant.
2 Comments
KSSV
on 21 Jun 2022
Error is clear, you need to check the dimensions of the arrays which you are using. We cannot check as the input files are not with us.
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