Subscript indices must either be real positive integers or logicals.

Question

Ahmed Saleem 2020-12-31

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/705998-subscript-indices-must-either-be-real-positive-integers-or-logicals

评论： Rik 2020-12-31

Error in line 21

Plz help me with this. Whenever I sign them arrays it gives me error but without array it doesnt. Im using arrays to plot it.

clc
% Part a
% ps1=exp (16.59158-(3643.31/(t-33.424)))
% ps2=exp (14.25326-(2665.54/(t-53.424)))
% T = 318.15 K and x1 = 0.25
t=318.15;
p=zeros(1,21);
y1=zeros(1,21);
y2=zeros(1,21);
for i=1:0.05:2
    syms x2 ps1 ps2 p y1 y2
    x1=i-1;
    x2=1-x1;
    k=((i-1+0.05)/0.05);
    ps1 = exp (16.59158-(3643.31/(t-33.424)));
    ps2 = exp (14.25326-(2665.54/(t-53.424)));
    % eqn1= p== (x1*ps1)+(x2*ps2);
    p(1,k)= double(solve(p== (x1*ps1)+(x2*ps2)));
    % eqn2= y1== (x1*ps1)/p;
    y1(1,k)= double(solve(y1== (x1*ps1)/p(1,k)))
    % eqn3= y2== (x2*ps2)/p;
    y2(1,k)= double(solve(y2== (x2*ps2)/p(1,k)))
end

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Answer 1

Rik 2020-12-31

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/705998-subscript-indices-must-either-be-real-positive-integers-or-logicals#answer_588193

在 MATLAB Online 中打开

Welcome to the wonderful world of floating point numbers. Just like 1/3 cannot be represented exactly with decimal values (meaning 3*(1/3) will be 0.9999), Matlab can't always store the numbers with an exact representation. This causes the rounding errors you see:

for i=1:0.05:2
    k=((i-1+0.05)/0.05);
    fprintf('k=%.20f\n',k)
end
k=1.00000000000000000000
k=2.00000000000000088818
k=3.00000000000000133227
k=3.99999999999999777955
k=4.99999999999999822364
k=5.99999999999999911182
k=7.00000000000000000000
k=8.00000000000000177636
k=8.99999999999999822364
k=9.99999999999999822364
k=11.00000000000000000000
k=12.00000000000000177636
k=13.00000000000000177636
k=13.99999999999999822364
k=15.00000000000000355271
k=16.00000000000000000000
k=17.00000000000000000000
k=18.00000000000000000000
k=18.99999999999999644729
k=20.00000000000000355271
k=21.00000000000000000000