how to use function handles
古いコメントを表示
Instead of the following, I want to use function handles and move if-statments out of the loop:
for n=1:10
if a==0
A = function2(n,a,b);
else
A = function1(n,a,b,c,d);
end
end
function A = function1(n,a,b,c,d)
A = n+a+b+c+d; % here it can be any expression
end
function A = function2(n,a,b)
A = n*a*b; % here it can be any expression
end
moving if-statement outside of the loop and using handles:
if a==0
func = @function2;
else
func = @function1;
end
and then I can solve my problem in two ways as
for n=1:10
A = func(n,a,b,c,d);
end
function A = function1(n,a,b,c,d)
A = n+a+b+c+d;
end
function A = function2(n,a,b,~,~)
A = n*a*b;
end
or as
for n=1:10
A = func(n,a,b,c,d)
end
function A = function1(n,a,varargin)
b = varargin{1};
c = varargin{2};
d = varargin{3};
A = n+a+b+c+d;
end
function A = function2(n,a,varargin)
b = varargin{1};
A = n*a*b;
end
Is there any more elegant way to do this?
採用された回答
その他の回答 (1 件)
I think the basic idea you need is
f1 = @(x) x;
f2 = @(x) x.^2;
f = @(a,x) (a~=0)*f1(x) + (a==0)*f2(x);
f(0,2)
f(1,2)
4 件のコメント
Hm. Not finding a good way to handle this that does not potentially result in empty, Inf, or NaN when one of the functions does not have all the arguments. For example, you could pass zeros into the function when they are not needed, but it is potentially hazardous:
f1 = @(x) x;
f2 = @(x,y) x.^2 + y.^2;
f = @(a,x,y) (a~=0)*f1(x) + (a==0)*f2(x,y);
f(0,2,3)
f(1,2,0)
the cyclist
2021 年 8 月 29 日
Searching keywords such as conditional anonymous function MATLAB turns up a lot of these same ideas. I did not find a great solution.
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