Why do I get the "Array indices must be positive integers or logical values" error?
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I keep getting the "Array indices must be positive integers or logical values" error for the line of my code "fs_2(i) = 1 - exp((log((CS_scheil(i)/(k*C0)))/(k-1)));" that comes after the "else". How do I fix this? My entire entry of code is below.
C0 = 8;
k = 0.13;
FS = linspace(0,0.4079,100);
CS_scheil = k*C0*(1 - FS).^(k - 1);
for i = 0:0.01:1
if i == 1
fs_2(i) = 1 - exp((log((CS_scheil(i)/(k*C0)))/(k-1)));
a(i) = 1;
b(i) = (k*C0*((1 - fs_2(i))^(k - 1)));
else
fs_2(i) = 1 - exp((log((CS_scheil(i)/(k*C0)))/(k-1)));
a(i) = (fs_2(i) - fs_2(i-1));
b(i) = (1/2)*[(k*C0*((1 - fs_2(i))^(k - 1))) + (k*C0*((1 - fs_2(i-1))^(k - 1)))];
end
SUM(i) = a(i)*b(i);
end
for n = 1:1:100
if n == 1
SUM(n) = a(n)*b(n);
else
SUM(n) = SUM(n-1) + (a(n)*b(n));
end
end
C_avg = SUM;
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采纳的回答
FScherff
2020-10-6
Hello,
Your for-loop is running over 0:0.01:1, resulting in an index-vector [0 0.1 0.2 0.3 ... 0.99 1].That means that already in the first iteration of your foor loop, you are using the value 0 as an array index. But as the error message states, "Array indices must be positive integers or logical values" in Matlab (as you are corrctly using in the second code snippet with n=1:100). That means you should e.g. use 1:101 as index-vecotr of your foor loop, resulting in the same number of iterations, but with valid integer indices. If you really need the values of 0 to 1 in your code (although I don't see that place in the code you posted), use something like
valuesBetweenZeroAndOne = 0:0.01:1;
for i = 1:101
if i == 1
...
else
...
end
%Line where you need the value, e.g. for scaling or something like that
SUM(i) = a(i)*b(i)*valuesBetweenZeroAndOne(i);
end
I have an additional remark:
In your first for-loop you run through 101 iterations (0 to 1 in 0.01-steps), in your second for loop you only have 100 iterations (1 to 100 in 1-steps). As you are using the same vector SUM in booth loops, so I'm relativly sure this is not on purpose.
Greetings!
更多回答(2 个)
Mario Malic
2020-10-6
编辑:Mario Malic
2020-10-6
% i = 0:0.01:1 % 101 element
i = 1 : 1 : 101 % 101 element
You should index arrays/variables with integers.
Walter Roberson
2020-10-6
for i = 0:0.01:1
i is first set to 0, then to 0.01, then about 0.02, then about 0.03, and so on.
fs_2(i) = 1 - exp((log((CS_scheil(i)/(k*C0)))/(k-1)));
and there you use i as a subscript for CS_scheil and for the output variable fs_2
First you have to ask what is meant by CS_scheil(i) . It would seem to imply the value of CS_scheil that is associated with the floating point value stored in i . But look at how you created CS_scheil :
FS = linspace(0,0.4079,100);
CS_scheil = k*C0*(1 - FS).^(k - 1);
There might be some ground for saying that you want CS_scheil(i) to mean that value of CS_scheil that is associated with FS having the same value as i, but FS runs 0 to 0.4079 while i runs 0 to 1, so when i > 0.4079 then there would be no obvious meaning to CS_scheil. Furthermore, if you ask for ismembertol(0:0.01:1, FS) you will find that the only intersection between the two, to within reasonable tolerances, is 0 .
Beyond that, you need to solve the problem of indexing with i for the output variable. For that, please read https://www.mathworks.com/matlabcentral/answers/?term=tag%3A%22generalized+for+loop%22
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