accessing single function from an array of functions

2022 3 4
2 个回答
回答已采纳
更新时间：2022 3 4
6 次查看（30 天）

0 个投票

I have defined a matrix of functions:
a=@(t) (1+0.8*sin(2*pi*t));
b=@(t) .8*(1-0.8*sin(2*pi*t));
z = @(t) 0*t;
%T = matlabFunction(fmat);
T=@(t) [-3*a(t) a(t) a(t) a(t) z(t) z(t);...
a(t) -5*a(t) a(t) a(t) a(t) a(t); ...
a(t) a(t) -5*a(t) a(t) a(t) a(t); ...
b(t) b(t) b(t) -5*b(t) b(t) b(t); ...
b(t) b(t) b(t) b(t) -5*b(t) b(t); ...
a(t) a(t) a(t) b(t) 5*b(t) -3*a(t)-6*b(t)];
If I want to access just a single function from the matrix, how do I do it? I have tried several approaches and keep getting: Error: Invalid array indexing.

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Walter Roberson
Walter Roberson 2022-3-4
See https://www.mathworks.com/matlabcentral/answers/539642-is-it-possible-to-have-an-array-of-function-handles#comment_2020739

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Torsten
Torsten 2022-3-4
编辑：Torsten 2022-3-4

0 个投票

f = {@(t) t, @(t) exp(t), @exp} % the second and third are equivalent
SecondFunction = f{2};
ValueOfSecondFunction = SecondFunction(1)
Does that help ?

6 个评论
显示 4更早的评论 隐藏 4更早的评论

I have also tried defining the function in this way:
Tcell = {@(t) -3*a(t), a, a, a, z, z;...
a, @(t) -5*a(t), a, a, a, a; ...
a, a, @(t) -5*a(t), a, a, a; ...
b, b, b, @(t) -5*b(t), b, b; ...
b, b, b, b, @(t) -5*b(t), b; ...
a, a, a, b, @(t) 5*b(t), @(t)-3*a(t)-6*b(t)};
The trouble is that I want to exponentiate the matrix for fixed values of t and define a new function in terms of that. Something like
myLT = @(s)eye(mysize);
% temporary definition***
C=eye(mysize);
for k=0:n-1
tobeexp = T(k*deltat)*deltat;
myLT = @(s) myLT(s)*expm(tobeexp-s*C);
end
This does work, but I can't figure out how to get to a particular function within the matrix of functions.
Torsten
Torsten 2022-3-4
编辑：Torsten 2022-3-4
myLT = @(s)eye(mysize);
Why does eye(mysize) depend on a parameter s ?
This does work, but I can't figure out how to get to a particular function within the matrix of functions.
Within which matrix of functions ?
The trouble is that I want to exponentiate the matrix for fixed values of t and define a new function in terms of that.
Why trouble ?
a=@(t) (1+0.8*sin(2*pi*t));
b=@(t) .8*(1-0.8*sin(2*pi*t));
z = @(t) 0*t;
Tcell = @(t){ -3*a(t), a(t), a(t), a(t), z(t), z(t);...
a(t), -5*a(t), a(t), a(t), a(t), a(t); ...
a(t), a(t), -5*a(t), a(t), a(t), a(t); ...
b(t), b(t), b(t), -5*b(t), b(t), b(t); ...
b(t), b(t), b(t), b(t), -5*b(t), b(t); ...
a(t), a(t), a(t), b(t), 5*b(t), -3*a(t)-6*b(t)};
Tcell_45_at_t_eq_2 = Tcell(2)(4,5);
Please include a piece of code and explain comprehensively what you are trying to do.
Sorry, it is a code snippet. mysize is just 6. It does not depend on s. It is so the loop definition of the function will work.
a = @(t)(1+0.8*sin(2*pi*t));
b = @(t)0.8*(1-0.8*sin(2*pi*t));
z = @(t)0*t;
n = 10;
%T = matlabFunction(fmat);
% T is a function of t which returns a 6-by-6 matrix.
% Each element of T is derived from the outputs of functions a, b, and z,
% which are also functions of t.
T = @(t)[ ...
-3*a(t) a(t) a(t) a(t) z(t) z(t);...
a(t) -5*a(t) a(t) a(t) a(t) a(t); ...
a(t) a(t) -5*a(t) a(t) a(t) a(t); ...
b(t) b(t) b(t) -5*b(t) b(t) b(t); ...
b(t) b(t) b(t) b(t) -5*b(t) b(t); ...
a(t) a(t) a(t) b(t) 5*b(t) -3*a(t)-6*b(t); ...
];
mysize = max(size(T(0)));
myLT = @(s)eye(mysize);
C=diag([1 2 3 -4 -3 0]);
deltat = 1/n;
mysize = max(size(T(0)));
myLT = @(s)eye(mysize);
C=diag([1 2 3 -4 -3 0]);
for k=0:n-1
tobeexp = T(k*deltat)*deltat;
myLT = @(s) myLT(s)*expm(tobeexp-s*C);
end
myLT = @(s)det(eye(mysize)-myLT(s));
myLT(.4)
28.0296
h=0:.01:.99;
plot(myLT(h));
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix
matches the number of rows in the second matrix. To perform elementwise multiplication, use '.*'.
Torsten
Torsten 2022-3-4
编辑：Torsten 2022-3-4
h=0:.01:.99;
myLTh = arrayfun(myLT,h);
plot(h,myLTh)
But to be honest: I can't decipher what you are doing in your code.

请先登录，再进行评论。

更多回答（1 个）

Voss
Voss 2022-3-4

0 个投票

a = @(t)(1+0.8*sin(2*pi*t));
b = @(t)0.8*(1-0.8*sin(2*pi*t));
z = @(t)0*t;
%T = matlabFunction(fmat);
% T is a function of t which returns a 6-by-6 matrix.
% Each element of T is derived from the outputs of functions a, b, and z,
% which are also functions of t.
T = @(t)[ ...
-3*a(t) a(t) a(t) a(t) z(t) z(t);...
a(t) -5*a(t) a(t) a(t) a(t) a(t); ...
a(t) a(t) -5*a(t) a(t) a(t) a(t); ...
b(t) b(t) b(t) -5*b(t) b(t) b(t); ...
b(t) b(t) b(t) b(t) -5*b(t) b(t); ...
a(t) a(t) a(t) b(t) 5*b(t) -3*a(t)-6*b(t); ...
];
% the 6-by-6 matrix returned from the function T when it is called with
% input 1:
result = T(1)
% get some element of the result (the result is the matrix returned from
% calling the function T with input 1):
result(4,5)
% trying to do the same with the output of the function T called with
% input 1 directly doesn't work in MATLAB (the output from T should be
% stored in a variable first, i.e., 'result' here):
try
T(1)(4,5);
catch ME
disp(ME.message);
% but you can index T(1) without storing in a variable, if you really
% want to:
subsref(T(1),substruct('()',{4,5}))
end

类别

在 帮助中心 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

产品

版本

R2021b

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by