How to recursively reduce the function arguments

Question

Victoria Li 2021-11-9

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582464-how-to-recursively-reduce-the-function-arguments

评论： Victoria Li 2021-11-10

Is there a way to create a function with one less argument dynamically? Basicly I'd like to find a way to do the following recursively or in a loop.

f4 = @(x1,x2,x3,x4) f5(x1, x2, x3, x4, 1);

f3 = @(x1,x2,x3) f4(x1, x2, x3, 1);

...

f1 = @(x1) f2(x1, 1);

r = f1(1)

function r = f5(x1, x2, x3, x4, x5)

r = x1 + x2 + x3 + x4 + x5

end

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David Hill 2021-11-9

Very confusing. An example would be helpful. It seems to me that r=f1=f2=f3=f4=f5=4+x1; which equals 5 in the example above (x1=1). Not sure what you are trying to do, but it is not wise to have all those variables.

Victoria Li 2021-11-9

Here is an example to calculate the double integral from 0 to 1 with composite simpson's rule. The function simpson2 takes a function argument. In the simpson2 function, I can only start the calculation when the function is a single variable. In the code below, I can transform the 2 variable function to a single variable function with given y values. I am looking for a more generic way to do this so it can handle integrals with more variables. To do this, it requires to transform the integral function with one less variable until it has only a single variable in each recursive iteration. The question above is simplified for achieving this.

f = @(x,y) (x.^2 + y.^2 );

a = 0;

b = 1;

m = 100;

integral = simpson2(a, b, m, f)

exact = integral2(f,a,b,a,b)

fprintf("Numerical Integral = %5.3f, Exact = %5.3f\n", integral, exact);

function integral = simpson2(a, b, m, fun)

% a, b: intregral boundary

% m: number of panels

% fun: function for integral

h = (b - a)/(2*m);

xi = a:h:b;

if nargin(fun) == 2

for i=1:length(xi)

fx = @(x) fun(x, xi(i)); % tranform to a single variable function with giver y varle

fx_val(i) = simpson2(a , b, m, fx);

end

elseif nargin(fun) == 1

fx_val = fun(xi);

fx_val(1);

else

fprintf("ERROR")

end

odd = double(sum(fx_val(3:2:end-1)));

even = double(sum(fx_val(2:2:end-1)));

integral = double(h/3 * (fx_val(1) + fx_val(end) + 4*(odd) + 2*(even)));

end

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Answer 1

Stephen23 2021-11-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582464-how-to-recursively-reduce-the-function-arguments#answer_827634

在 MATLAB Online 中打开

It can be done with VARARGIN:

N = 5;
C = cell(1,N);
C{N} = @f5;
for k = N-1:-1:1
    C{k} = @(varargin) C{k+1}(varargin{:},1);
end
C{1}(1)
ans = 5

function r = f5(x1, x2, x3, x4, x5)
r = x1 + x2 + x3 + x4 + x5;
end

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Victoria Li 2021-11-10

Thanks Stephen for the help. Your solution works perfectly.

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Answer 2

Jan 2021-11-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582464-how-to-recursively-reduce-the-function-arguments#answer_827529

This should work with str2func and eval.

This is such a cruel programming technique, that I recommend not to use it. Such meta-programming for creating code dynamically is hard to debug and Matlab's JIT acceleration cannot work. This can slow down the code by a factor 100 compared to some code, which avoid the dynamic creation of variables and functions.

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Adam Danz 2021-11-9

The function string could also be parsed to remove the penultimate input and then converted back to a function handle without using eval. But I still think there's a better approach.

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Answer 3

Adam Danz 2021-11-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1582464-how-to-recursively-reduce-the-function-arguments#answer_827549

在 MATLAB Online 中打开

x1 = rand(1);
x2 = rand(1);
x3 = rand(1);
x4 = rand(1);
p = [x1, x2, x3, x4]; 
f5([p,1])
ans = 3.3148
p = [x1, x2, x3];
f5([p,1])
ans = 2.4283
p = [x1, x2];
f5([p,1])
ans = 1.8867
function f5(varargin)
    sum([varargin{:}])
end