Error using function handles in for loop

Question

Andrew Poissant 2017-4-26

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https://ww2.mathworks.cn/matlabcentral/answers/337318-error-using-function-handles-in-for-loop

评论： KSSV 2017-4-27

I have a function, pi_t, which is defined using a function handle but also is a function i in the for loop. When I try to incorporate my function handle inside the for loop it gives an error for pi_t saying "Nonscalar arrays of function handles are not allowed; use cell arrays instead." What I want to do is to have Np amount of pi_t values with the function handle but with the dpi(i) value correctly substituted in the equation for pi_t. Any ideas how to effectively use a function handle in a for loop?

a_pi20 = 1.50;
b_pi20 = 2.25; 
c_pi20 = 3;
dist_pi20 = makedist('Triangular', a_pi20, b_pi20, c_pi20);
pi08 = 8.50;
Np = 1000;
for i = 1:Np
    pi20 = random(dist_pi20);
    dpi(i) = (pi20 - pi08)/(12*pi08);
    pi_t(i) = @(x)abs(-dpi(i)*x(1)*pi08 - pi08);
end

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Answer 1

Walter Roberson 2017-4-26

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https://ww2.mathworks.cn/matlabcentral/answers/337318-error-using-function-handles-in-for-loop#answer_264527

在 MATLAB Online 中打开

pi_t{i} = @(x)abs(-dpi(i)*x(1)*pi08 - pi08);

and to call it, you would need something like

pi_t{17}(37.8)

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Andrew Poissant 2017-4-26

Yes but I need the -dpi(i) value to substitute into pi_t(i) every time. I cannot call pi_t as you did because I am solving for x later on using optimization. What I need is that every iteration dpi has a new value based on its statistical distribution and I need that to go into the pi_t term. When I type in pi_t(1) after using your suggestion it gives me @(x)(-dpi(i)*x(1)*pi08-pi08). What I want to do is have it give me for example dpi(1) = 0.06, so pi_t(1) = @(x)(-0.06*x(1)*pi08-pi08). Any ideas how to do this?

Walter Roberson 2017-4-26

在 MATLAB Online 中打开

fzero( pi_t{17}, x0 )

"When I type in pi_t(1) after using your suggestion it gives me @(x)(-dpi(i)*x(1)*pi08-pi08)"

p{1} would give that, and it is fine. It has "captured" the value that dpi(i) had at the time the function handle was created. It will not display as @(x)(-0.06*x(1)*pi08-pi08) but it will execute that way. You can use

temp = functions(pi_t{1});
temp2 = temp.workspace{1};
temp2.dpi
temp2.i

to see that it has copied dpi and i; those are what it will use at execution time.

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Answer 2

KSSV 2017-4-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/337318-error-using-function-handles-in-for-loop#answer_264526

在 MATLAB Online 中打开

You can follow with out suing function as below. (Note that x is not defined, I have taken some random value).

a_pi20 = 1.50;
b_pi20 = 2.25; 
c_pi20 = 3;
dist_pi20 = makedist('Triangular', a_pi20, b_pi20, c_pi20);
pi08 = 8.50;
Np = 1000;
pi_t = zeros(1,Np) ;
dpi = zeros(1,Np) ;
x = rand ;
for i = 1:Np
    pi20 = random(dist_pi20);
    dpi(i) = (pi20 - pi08)/(12*pi08);
    pi_t(i) = abs(-dpi(i)*x*pi08 - pi08);
end

If you want to go by function. You have to define the function first and then call the function in the loop. An example code given below would help you to understand.

f = @(x) x+2 ;
y = zeros(1,10) ;
for i = 1:10
    y(i) = f(i) ;
end

Also note that, your entire code can be vectorised with out using loop. As a beginner it is always good to know loops and then vectorize it.

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Andrew Poissant 2017-4-26

I see why you had to make x = rand but the issue is that I solve for x later on in an optimization so I can't define it earlier like you did. Is there a way to do this without defining what x is initially?