Question

在用一些例子来压倒你之前，我试图尽可能简单地说明问题：

如果f11 , ... , fnm是n*m实值的功能，我希望通过n个步骤一次评估m，通过一些更高阶函数b，即

v = []
f1 = @(x) [f11(x) f12(x) ... f1m(x)]
v = [v b(f1)]
f2 = @(x) [f21(x) f22(x) ... f2m(x)]
v = [v b(f2)]

我如何通过迭代解决这个问题？就像这样：

f = @(x) [f11(x) ... f1m(x) ; ... ; fn1(x) ... fnm(x)];
% now iterate over the rows of f
for i=1:n
    v = [v b(f(i,:)) ]
end

以下是我所拥有的一个例子（为了不错过我实际现实问题的任何细节而增长，但我试图让它尽可能小）：

% 4 functions that take a 1x2 real valued vector as argument
% and return a real value
f11 = @(x) x(1) + x(2);
f12 = @(x) x(1) - x(2);
f21 = @(x) x(2) - x(1);
f22 = @(x) x(1) * x(2);

% we'll run function b for 2 steps, then 4 steps
steps = [2 4];

% start value
x = [1 2];
% vector to hold results
v = []

% get the result of passing the 1:st 2 functions to b with steps(1)
f1 = @(x) [f11(x) f12(x)];
v = [v ;b(x, f1, steps(1))]
% update x
x = v(end,:)

% add the result of passing the 2:nd 2 functions to b with steps(2)
f2 = @(x) [f21(x) f22(x)];
v = [v ;b(x, f2, steps(2))];
% update x
x = v(end,:)

其中b是定义如下的函数：

function [ X ] = b( x, f, n )
%   @param:
%   x = an 1x2 real valued vector
%   f = a real valued function returning
%       a 1x2 real valued vector
%   n = an integer defining the rows of return matrix
%
%   @return:
%   X = an nx2 real valued matrix defined as below 

    X = zeros(n,2);
    for i=1:n
        % apply the functions on x
        a = f(x+1);
        b = f(x+2);
        % update x
        x = a+b
        % add x to return matrix
        X(i,:) = x;
    end
end

上述代码可以概括为：

% n*m functions that take a 1xm real valued vector as argument
% and return a real value
f11 = @(x) ... ;
f12 = @(x) ... ;
.
.
.
fnm = @(x) ... ;

% we'll run function b for a1 steps, then a2 steps, ... , then an steps
steps = [a1 a2 ... an];

% start value
x = [1 2 ... m];
% vector to hold results
v = []

% get the result of passing the 1:st m functions to b with steps(1)
f1 = @(x) [f11(x) ... f1m(x)];
v = [v ;b(x, f1, steps(1))]
% update x
x = v(end,:)

% add the result of passing the 2:nd m functions to b with steps(2)
f2 = @(x) [f21(x) ... f2m(x)];
v = [v ;b(x, f2, steps(2))];
% update x
x = v(end,:)

.
.
.

% add the result of passing the n:ed m functions to b with steps(n)
fn = @(x) [fn1(x) ... fnm(x)];
v = [v ;b(x, fn, steps(n))];
% update x
x = v(end,:)

其中b是返回步骤（i）x m矩阵的任何函数。

我想知道小的具体例子和一般例子是否应该通过一般迭代来解决，如下所示：

% let f hold all the functions as a matrix
f = @(x) [f11(x) ... f1m(x) ; ... ; fn1(x) ... fnm(x)];
% now iterate over the rows of f
for i=1:n
    v = [v ; b(x, f(i,:), steps(i)) ]
end

Answer 1

所以诀窍在于将函数定义为单元矩阵，然后使用一些矢量化来解决问题。这是我提出的代码：

%First define your functions in a cell matrix
fn_mat = {@(x) x(1) + x(2), @(x) x(1) - x(2); ...
          @(x) x(2) - x(1), @(x) x(1) * x(2)};
%Store the sixe of this matrix in two variables
[n, m] = size(fn_mat);
%Number of steps
steps = [2, 4];
% start value
x = [1 2];
% vector to hold results
v = [];
%This will run the required code for n iterations
for ii = 1:n
    %This is the tricky part. What I have done is used arrayfun to run over
    %all the functions in the row defined by ii and pass x as an argument 
    %each time. The rest is same as before
    fn = @(x) arrayfun(@(a, b) fn_mat{ii, a}(b{:}), 1:m, repmat({x}, 1, m));
    v = [v; b(x, fn, steps(ii))];
    x = v(ii, :);
end

对于当前值，输出为：

v =
          12          -2
          26          28
         -28         -13
          30         610
        1160       38525
       74730    89497060

for循环足以适应任何大小的fn_mat。

从函数矩阵中获取函数行

1 个答案: