What part of this for loop does not want to work with anything else than positive integers?

Question

Camille Godin 2023-3-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1926600-what-part-of-this-for-loop-does-not-want-to-work-with-anything-else-than-positive-integers

编辑： Jan 2023-3-10

Hello there,

I'm really struggling to fix this for loop. I have provided all the variables that are in the code, and the for loop as well. So far, I managed to run it properly when dt = a positive integer, which means that t ranges from 0 to Lt without decimals. However, I need my loop to work on decimals! In other words, when t = [1,2,3,4,5,6,7,8,9,10], it works. When t = [0, 0.1, 0.3, ... ,10] it does not work.

I thought that wouldve been fixed by creating the i loop that is made of positive integers only, and using i to index in t which is made of decimals. But that is not working.

This is the error code I get when I run the for loop step by step:

Array indices must be positive integers or logical values.

Error in WC_sim (line 35)

drE = (-(LE)*rE(t(i)) + F(wEE .* rE(t(i)) - wEI .* rI(t(i)) + I_ext_E(t(i))));

*Note that the F function is:

function drE = F(x) %fonction Sin
  drE = 1 ./ exp(-x)
end

Here is the code:

%Weight values
wEE = 8; 
wII = 10;
wEI = 1;
wIE = 1;
 
%Time range on axis
Lt = 10;
dt = 0.1;
t = 0.1:dt:Lt;
 
%External stim
I_ext_E = 3;
I_ext_I = 1;
I_ones = ones(1,length(t));
I_ext_E = I_ext_E .* I_ones;
I_ext_I = I_ext_E .* I_ones;
LE = 0.5;
n = length(t);
rE = zeros(1,n);
rI = zeros(1,n);

for i = 1:length(t);

t = 0:dt:Lt;

% Calculate the derivative of the E population

drE = (-rE(t(i)) + F(wEE .* rE(t(i)) - wEI .* rI(t(i)) + I_ext_E(t(i))));

i

% Calculate the derivative of the I population

drI = (-rI(t(i)) + F(wIE .* rE(t(i)) - wII .* rI(t(i)) + I_ext_I(t(i))));

% Update using Euler's method

rE(i + 1) = rE(t(i)) + (dt .* drE);

% rI(i + 1) = rI(t(i)) + (dt .* drI);

end

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Jan 2023-3-10

Please use the tools to format code and add a copy of the complete error message. Then the reader do not have to search, inwhich line the error occurs.

Camille Godin 2023-3-10

Hi Jan, I think I did it properly but feel free to let me know if you want me to add it differently/add something else.

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Answer 1

Jan 2023-3-10

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1926600-what-part-of-this-for-loop-does-not-want-to-work-with-anything-else-than-positive-integers#answer_1190490

编辑：Jan 2023-3-10

在 MATLAB Online 中打开

rE is a vector:

rE = zeros(1,n);

You cann address its elements with positive integers as indices: rE(1), rE(2), ...

t contains nonintegers and zero also:

t = 0:dt:Lt;

Here indices are valid also: t(1), t(2), ... t(1) is 0 in this case.

Now you try to use t(i) as index:

drE = (-rE(t(i)) + F(wEE .* rE(t(i)) - wEI .* rI(t(i)) + I_ext_E(t(i))));
%      ^^^^^^^^^            ^^^^^^^           ^^^^^^^^^  ^^^^^^^^^^^^^^

For i=1 this is rE(0). For i=2: rE(dt). This is not valid.

Why do you use t at all?

%Weight values
wEE = 8; 
wII = 10;
wEI = 1;
wIE = 1;
 
%Time range on axis
Lt = 10;
dt = 0.1;
t = 0:dt:Lt;
n  = length(t);
 
%External stim
I_ext_E = 3;
I_ext_I = 1;
I_ones = ones(1, n);
I_ext_E = I_ext_E .* I_ones;
I_ext_I = I_ext_E .* I_ones;
 
rE = zeros(1,n);
rI = zeros(1,n);
for i = 1:n - 1  % Not length(t)
    drE = -rE(i) + F(wEE .* rE(i) - wEI .* rI(i) + I_ext_E(i));
    % Not used: drI = -rI(i) + F(wIE .* rE(i) - wII .* rI(i) + I_ext_I(i));
    rE(i + 1) = rE(i) + dt .* drE;
end
function drE = F(x) %fonction Sin
  drE = 1 ./ exp(-x);
end

rE contains NaNs after a valid start, but this is another problem.

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Torsten 2023-3-10

编辑：Torsten 2023-3-10

在 MATLAB Online 中打开

x = 0:0.1:1

x = 1×11

0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000

y = x.^2

y = 1×11

0 0.0100 0.0400 0.0900 0.1600 0.2500 0.3600 0.4900 0.6400 0.8100 1.0000

plot(x,y)

Here, y(i) is the function f(x) = x^2 at x = x(i). y(i) is not defined as y(x(i)), but simply as y(i).

Do the same in your code to avoid errors.

Jan 2023-3-10

编辑：Jan 2023-3-10

在 MATLAB Online 中打开

For i=1 this is rE(0). For i=2: rE(dt). This is not valid. Answer:

Why is this not valid?

Because 0 is no valid index in Matlab. Indices must be > 0.

"t is my x axis. I need it to be divided with decimals (so not only positive integers such as 1)."

You are consuing the values of the vector with the indices. See this example:

t = 0:0.01:2*pi;
x = zeros(size(t));
for k = 1:numel(t)
    x(k) = sin(t(k));
end
plot(t, x);

Do you see it? Your t can have any value, but the indices are strictly from 1 to numel(t). Assigning x(k) is working, but x(t(k)) is not. Of course, mathematically you construct "x(t)", but this is not the correct notation in Matlab, but the meaning of the contents.

The for loop does work with fractional steps:

t = zeros(1, 629);
x = zeros(1, 629);
k = 0;
for T = 0:0.01:2*pi
    k    = k + 1;  % Integers > 0
    t(k) = T;
    x(k) = sin(T);
end

With this approach you have to care for another counter to create the indices.

Clearer now?

By the way, the clean Matlab way omits the loop:

t = 0:0.01:2*pi;
x = sin(t);

Less troubles, because no indices appear anywhere.

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What part of this for loop does not want to work with anything else than positive integers?

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What part of this for loop does not want to work with anything else than positive integers?

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