向量化嵌套循環

Question

我寫了這段代碼，它工作正常，但我想優化它通過刪除兩個for循環如果可能的話。有沒有人有任何想法，我可能會做到這一點？非常感謝！

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 

Answer 1

所以你要vectorise此代碼：

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，除去外循環：

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，L_Sides可以用一個操作來創建：

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); 

由於th in_point（l）的長度：out_point（l）是可變的，沒有簡潔的方法來對此進行矢量化（我知道的;人有我喜歡的好辦法知道），你可以只是把它按原樣或使用本：

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{:}) 

但不會有性能優勢，這是不太明顯的是怎麼回事，所以我不會使用這段代碼。