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 个评论
KSSV
2022-6-21
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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