Using a cell array as an argument of a function handle
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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
b = @(y) y(1) + y(2)
c = b(a)
I would like to get c = a(1) + a(2) or c = @(x) x(1) + 2 + x(2) + 3;
Thanks!
回答（1 个）
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)
my_input(1) + 2 + my_input(2) + 3 % to test
3 个评论
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 = double(subs(gy,u,[0 0 0 0 0 0]))
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
gy = gx(U)
And class(gy) = 'function_handle', so I could evaluate Gy = gy([0 0 0 0 0 0] = 0.0066
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)
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)
y = zeros(1, 6);
nablaGy = double(subs(nablagy,u,y))
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)
nablagx = @(x) gradient(gx(x))
nablaGx = nablagx(U)
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2023-7-19
2023-7-20
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