This Matlab code is saying Index in position 1 is invalid. Array indices must be positive integers or logical values.

Question

CONNOR 2023-11-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2053997-this-matlab-code-is-saying-index-in-position-1-is-invalid-array-indices-must-be-positive-integers-o

回答： Image Analyst 2023-11-30

EnvironmentalForcing.mat

when I try to run this code it says "Index in position 1 is invalid. Array indices must be positive integers or logical values.", but both i and n are integers so I am assuming that it has a problem with how I use the cell arrays.

close all, clear, clc
load('EnvironmentalForcing.mat')
Bmax = 1;
uL_min = 6;
uI = 10;
e = 0.001;
Ap = 5000;
Pi = 930.27249;
Si = Pi/Ap;
Li = 0.01*Si;
Ii = 0;
Ri = uI*Ii;
Bi = 1;
Pb = 1;
for i = 1:length(T)
    if T(i)>0 && T(i)<35
        Tb(i) = (0.000241*(T(i)^2.06737))*((35-T(i))^0.72859);
    end
end
j = 1;
for i=1:length(t)
    uL(i) = sum(Tb(j:i));
    while uL(i) > uL_min
        j = j+1;
        uL(i) = sum(Tb(j:i));
    end
end
B = Bmax*Tb;
p = [B, uL, uI, e, Te, Pb];
y0 = [Pi, Si, Li, Ii, Ri, Bi];
odeFunc = SLIRmodel(tspan, y0, p);
[t,y] = rk4(odeFunc, tspan, y0);
%%      Functions
function [dydt] = SLIRmodel(t,y0,p)
%assign parameters
beta = p(1);
uL = p(2);
uI = p(3);
e = p(4);
Te = p(5);
Pb = p(6);
%assign variables
P = y0(1);
S = y0(2);
L = y0(3);
I = y0(4);
R = y0(5);
dPbdt = (0.1724*Pb - 0.0000212*Pb^2)*Te;
dPldt = (1.33*t)*Te;
dPdt = dPbdt + dPldt;
dSdt = (-beta*S*I)+dPdt;
dLdt = (beta*S*I)-((uL^-1)*L)+e;
dIdt = ((uL^-1)*L)-((uI^-1)*I);
dRdt = (uI^-1)*I;
dydt = {dPbdt, dPldt, dPdt, dSdt, dLdt, dIdt, dRdt};
end
function [t, y] = rk4(odeFunc, tspan, y0)
N = length(tspan);
t = tspan;
y = y0;
for n=2:N
    for i=1:length(odeFunc)
        k1 = odeFunc{i}(t(n), y(n));
        k2 = odeFunc{i}(t(n) + 0.5*h, y(n) + 0.5*h*k1);
        k3 = odeFunc{i}(t(n) + 0.5*h, y(n) + 0.5*h*k2);
        k4 = odeFunc{i}(t(n) + h, y(n) + h*k3);
        
        for j = 1:q
            y(j, n+1) = y(j, n) + h*(k1(j) + 2*k2(j) + 2*k3(j) + k4(j))/6;
        end
        t(n+1) = t(n) + h;
    end
end
end