1

我编写了这段代码，它运行良好，但如果可能，我想通过删除两个 for 循环来优化它。有人知道我如何实现这一目标吗？非常感谢！

chosen_runs = [2:5];

% Configurations for each test in order
XY = [0 1; 0 1; 0 9; 0 1; 0 2; 0 3; 0 4; 0 5; 11 12; 11 12];

% Inductance Matrix
LMat = [0.0045  0.0045  0.0045  0.0044  0.0044  0.0044  0.0044  0.0044  0.0043  0.0043;
        0.0045  0.0046  0.0046  0.0045  0.0045  0.0045  0.0045  0.0044  0.0044  0.0044;
        0.0045  0.0046  0.0046  0.0046  0.0046  0.0046  0.0045  0.0045  0.0045  0.0045;
        0.0044  0.0045  0.0046  0.0047  0.0047  0.0047  0.0046  0.0046  0.0046  0.0046;
        0.0044  0.0045  0.0046  0.0047  0.0048  0.0048  0.0047  0.0047  0.0047  0.0046;
        0.0044  0.0045  0.0046  0.0047  0.0048  0.0048  0.0048  0.0048  0.0048  0.0047;
        0.0044  0.0045  0.0045  0.0046  0.0047  0.0048  0.0049  0.0049  0.0049  0.0048;
        0.0044  0.0044  0.0045  0.0046  0.0047  0.0048  0.0049  0.0050  0.0049  0.0049;
        0.0043  0.0044  0.0045  0.0046  0.0047  0.0048  0.0049  0.0049  0.0050  0.0050;
        0.0043  0.0044  0.0045  0.0046  0.0046  0.0047  0.0048  0.0049  0.0050  0.0051];

% Declaration of Variables
runs = chosen_runs;
num = length(runs);
in_point = zeros(num,1);
out_point = zeros(num,1);
L_Mid = zeros(10,num);
L_Sides = zeros(10,num);

%%%%%%%%%%%%%%%%%%%%%%%%%%

in_point = XY(runs,1);    % Creates a column vector each row of which is the in_point for a chosen run
out_point = XY(runs,2);   % Creates a column vector each row of which is the out_point for a chosen run

in_point
out_point

for k = 1:10
    for l = 1:num

        L_Mid(k,l) = sum(LMat(k,1+in_point(l):out_point(l)));     % Creates a matrix, each column of which is the inductance (in between current leads) for a chosen run, each row is a different layer in the solenoid.
        L_Sides(k,l) = sum(LMat(k,:))-L_Mid(k,l);    % Creates a matrix, each column of which is the inductance (either side of the current leads) for a chosen run, each row is a different layer in the solenoid.

    end
end

L_Mid
L_Sides
4

1 回答 1

3

所以你想对这段代码进行矢量化：

for k = 1:10
    for l = 1:num
        L_Mid(k,l) = sum(LMat(k,1+in_point(l):out_point(l)));  
        L_Sides(k,l) = sum(LMat(k,:))-L_Mid(k,l); 
    end
end

第一步，去掉外循环：

for l=1:num
    L_Mid(:,l)=sum(LMat(:,1+in_point(l):out_point(l)),2); % Using the dim input to sum
    L_Sides(:,l) = bsxfun(@minus,sum(LMat,2),L_Mid(:,l)); % Using bsxfun to subtract
end

下一步，可以使用单个操作创建 L_Sides：

for l=1:num
    L_Mid(:,l)=sum(LMat(:,1+in_point(l):out_point(l)),2); % Using the dim input to sum
end

L_Sides = bsxfun(@minus,sum(LMat,2),L_Mid);

由于 in_point(l):out_point(l) 的长度是可变的，因此没有整洁的方法来对其进行矢量化（据我所知；任何人都有一个我很想知道的好方法！），你可以离开它按原样，或使用这个：

L_Mid2 = arrayfun(@(x) ...
    sum(LMat(:,1+in_point(x):out_point(x)),2), 1:length(in_point),'uniformoutput',false);
L_Mid2=cat(2,L_Mid2{:})

但是不会有任何性能优势，而且发生的事情不太明显，所以我不会使用这段代码。

于 2013-07-22T11:06:13.537 回答