Using a cell array as an argument of a function handle

Question

Geovane Gomes 2023-7-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1998243-using-a-cell-array-as-an-argument-of-a-function-handle

评论： Geovane Gomes 2023-7-20

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Hi all,

Is it possible to use a cell array as arguments in a function handle?

For example:

a(1) = {@(x) x(1) + 2};
a(2) = {@(x) x(2) + 3};
a
a = 1×2 cell array
    {@(x)x(1)+2}    {@(x)x(2)+3}
b = @(y) y(1) + y(2)
g = function_handle with value:
    @(y)y(1)+y(2)
c = b(a)
Operator '+' is not supported for operands of type 'cell'.

Error in solution>@(y)y(1)+y(2) (line 4)
g = @(y) y(1) + y(2)

I would like to get c = a(1) + a(2) or c = @(x) x(1) + 2 + x(2) + 3;

Thanks!

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Answer 1

Angelo Yeo 2023-7-19

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https://ww2.mathworks.cn/matlabcentral/answers/1998243-using-a-cell-array-as-an-argument-of-a-function-handle#answer_1275848

在 MATLAB Online 中打开

a(1) = {@(x) x(1) + 2};
a(2) = {@(x) x(2) + 3};
b = @(x) a{1}(x) + a{2}(x); % merging function handles
my_input = [5, 8]; % for example
c = b(my_input)
c = 18
my_input(1) + 2 + my_input(2) + 3 % to test
ans = 18

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Geovane Gomes 2023-7-20

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Thanks Angelo!

In this simple example your answer works good, but my problem is a more complex one.

I implemented the FORM algorithm using symbolic math and it works perfectly, however I want to use the same code with function handles.

In one part of the original code I have the following lines:

mu = [2e4 12 0.04 2e10 9.82e-4 1e11]; % Mean

sigma = [1.4e3 0.12 4.8e-3 1.2e9 5.9852e-5 6e9]; % Standard deviation

dist = ["normal", "normal", "normal", "normal", "normal","normal"]; % Distribution

numVar = length(mu);

% Using symbolic

x = sym("x", [1 numVar]);

u = sym("u",[1 numVar]);

gx = 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation

for i = 1:numVar

U(i) = mu(i) + sigma(i) * u(i);

end

gy = subs(gx,x,U)

gy =

Gy = double(subs(gy,u,[0 0 0 0 0 0]))

Gy = 0.0066

I would like to have the same result, but with gx as a function handle. Something like this:

clear
mu = [2e4 12 0.04 2e10 9.82e-4 1e11]; % Mean
sigma = [1.4e3 0.12 4.8e-3 1.2e9 5.9852e-5 6e9]; % Standard deviation
dist = ["normal", "normal", "normal", "normal", "normal","normal"]; % Distribution
numVar = length(mu);
% Using function handle
gx = @(x) 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation
for i = 1:numVar
   U(i) = {@(u) mu(i) + sigma(i) * u(i)};
end
U
U = 1×6 cell array
    {@(u)mu(i)+sigma(i)*u(i)}    {@(u)mu(i)+sigma(i)*u(i)}    {@(u)mu(i)+sigma(i)*u(i)}    {@(u)mu(i)+sigma(i)*u(i)}    {@(u)mu(i)+sigma(i)*u(i)}    {@(u)mu(i)+sigma(i)*u(i)}
gy = gx(U)
Operator '.^' is not supported for operands of type 'cell'.

Error in ^ (line 22)
    Z = X.^Y;

Error in solution>@(x)0.03-(x(1)*x(2)^2)/2*(3.81/(x(3)*x(4))+1.13/(x(5)*x(6))) (line 22)
gx = @(x) 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation

And class(gy) = 'function_handle', so I could evaluate Gy = gy([0 0 0 0 0 0] = 0.0066

Angelo Yeo 2023-7-20

编辑：Angelo Yeo 2023-7-20

在 MATLAB Online 中打开

I don't actually understand your intention. Why do you want to make two function handles to input "u"? The code below would work with the same result. This uses function handle but only once. You don't have to put "u" as a input for a function handle. You can use element-wise multiplication (.*).

mu = [2e4 12 0.04 2e10 9.82e-4 1e11]; % Mean
sigma = [1.4e3 0.12 4.8e-3 1.2e9 5.9852e-5 6e9]; % Standard deviation
dist = ["normal", "normal", "normal", "normal", "normal","normal"]; % Distribution
numVar = length(mu);
u = zeros(1, 6);
U = mu + sigma.*u;
gx = @(x) 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation
gx(U)
ans = 0.0066

Geovane Gomes 2023-7-20

在 MATLAB Online 中打开

Actually I need to evaluate the gradient of this:

mu = [2e4 12 0.04 2e10 9.82e-4 1e11]; % Mean

sigma = [1.4e3 0.12 4.8e-3 1.2e9 5.9852e-5 6e9]; % Standard deviation

dist = ["normal", "normal", "normal", "normal", "normal","normal"]; % Distribution

numVar = length(mu);

% Using symbolic

x = sym("x", [1 numVar]);

u = sym("u",[1 numVar]);

gx = 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation

for i = 1:numVar

U(i) = mu(i) + sigma(i) * u(i);

end

gy = subs(gx,x,U);

nablagy = gradient(gy)

nablagy =

y = zeros(1, 6);

nablaGy = double(subs(nablagy,u,y))

nablaGy = 6×1

-0.0016 -0.0005 0.0008 0.0004 0.0010 0.0010

clear

mu = [2e4 12 0.04 2e10 9.82e-4 1e11]; % Mean

sigma = [1.4e3 0.12 4.8e-3 1.2e9 5.9852e-5 6e9]; % Standard deviation

dist = ["normal", "normal", "normal", "normal", "normal","normal"]; % Distribution

numVar = length(mu);

u = zeros(1, 6);

U = mu + sigma.*u;

gx = @(x) 0.03 - (x(1) * x(2) ^ 2) / 2 * (3.81 / (x(3) * x(4)) + 1.13 / (x(5) * x(6))); % Limit state equation

gx(U)

ans = 0.0066

nablagx = @(x) gradient(gx(x))

nablagx = function_handle with value:

@(x)gradient(gx(x))

nablaGx = nablagx(U)

nablaGx = 0

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Using a cell array as an argument of a function handle

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