How to improve calculation time for a huge matrix?

syms Q1 Q2 D3
EndEffectorVariables (Q1,Q2,D3) = [
    (cos(Q1)*cos(Q2))/2 + cos(Q1)*cos(Q2)*(D3 + 3/4);
    sin(Q2)/2 + sin(Q2)*(D3 + 3/4) + 1;
    - (cos(Q2)*sin(Q1))/2 - cos(Q2)*sin(Q1)*(D3 + 3/4)];


SS = 50;        %Step number. 0~360 divided into 50 different points. 
n = 1;

skip = [360];   %Skipping the values for 360 because it's equal to 0 degrees.

for angle1 = 0:(360/(SS)):360
    if ~ismember(angle1,skip)
        angle1rad = deg2rad (angle1);
        
        for angle2 = 0:(360/(SS)):360
            if ~ismember(angle2,skip)
                angle2rad = deg2rad (angle2);
                
                for distance = 0:(1/(SS-1)):1       %SS-1 to have 50 different values between 0~1 meters
                    
                    EndEffectorPosition = EndEffectorVariables (angle1rad,angle2rad,distance);
                    PositionMatrix(n,:) = EndEffectorPosition;          
                    n = n+1;
                end
            end
        end
        fprintf('Calculated for Q1=%.2f\n', angle1)
    end
end
fprintf('Already calculated for Q1=%.2f=0.00\nFinished!', angle1)

Mechatronics eng. student. Supposed to get the positions of the end effector of a RRP manipulator. Here I wrote a nested for loop to calculate all X Y Z positions regarding to the end effector position equations given in the first part. Need to get a 125000x3 matrix. Tried yesterday but this code takes hours and hours. Is there another method I can use to improve the calculation time?

Answer 1

[Voss]

The original method but with SS = 5 to reduce the time (takes ~3.5 seconds):

tic

syms Q1 Q2 D3
EndEffectorVariables (Q1,Q2,D3) = [
    (cos(Q1)*cos(Q2))/2 + cos(Q1)*cos(Q2)*(D3 + 3/4);
    sin(Q2)/2 + sin(Q2)*(D3 + 3/4) + 1;
    - (cos(Q2)*sin(Q1))/2 - cos(Q2)*sin(Q1)*(D3 + 3/4)];


SS = 5;        %Step number. 0~360 divided into 50 different points. 
n = 1;

skip = [360];   %Skipping the values for 360 because it's equal to 0 degrees.

for angle1 = 0:(360/(SS)):360
    if ~ismember(angle1,skip)
        angle1rad = deg2rad (angle1);
        
        for angle2 = 0:(360/(SS)):360
            if ~ismember(angle2,skip)
                angle2rad = deg2rad (angle2);
                
                for distance = 0:(1/(SS-1)):1       %SS-1 to have 50 different values between 0~1 meters
                    
                    EndEffectorPosition = EndEffectorVariables (angle1rad,angle2rad,distance);
                    PositionMatrix(n,:) = EndEffectorPosition;          
                    n = n+1;
                end
            end
        end
        fprintf('Calculated for Q1=%.2f\n', angle1)
    end
end

Calculated for Q1=0.00
Calculated for Q1=72.00
Calculated for Q1=144.00
Calculated for Q1=216.00
Calculated for Q1=288.00

toc

Elapsed time is 3.549845 seconds.

Another method with no loops (takes ~0.02 seconds):

tic

SS = 5;        %Step number. 0~360 divided into 50 different points. 

angle1 = deg2rad(0:(360/(SS)):360);
angle2 = deg2rad(0:(360/(SS)):360);
distance = 0:(1/(SS-1)):1;

angle1(end) = []; % skip the 360
angle2(end) = [];

[D3,Q2,Q1] = ndgrid(distance,angle2,angle1);

Q1 = Q1(:);
Q2 = Q2(:);
D3 = D3(:);

PositionMatrix_new = [
    (cos(Q1).*cos(Q2))/2 + cos(Q1).*cos(Q2).*(D3 + 3/4) ...
    sin(Q2)/2 + sin(Q2).*(D3 + 3/4) + 1 ...
    -(cos(Q2).*sin(Q1))/2 - cos(Q2).*sin(Q1).*(D3 + 3/4)];

toc

Elapsed time is 0.021408 seconds.

isequal(PositionMatrix_new,double(PositionMatrix))

ans = logical
   0

max(abs(PositionMatrix_new(:) - double(PositionMatrix(:))))

ans = 4.4409e-16

Same new method but now with SS back to 50 (takes ~0.03 seconds):

tic

SS = 50;        %Step number. 0~360 divided into 50 different points. 

angle1 = deg2rad(0:(360/(SS)):360);
angle2 = deg2rad(0:(360/(SS)):360);
distance = 0:(1/(SS-1)):1;

angle1(end) = [];
angle2(end) = [];

[D3,Q2,Q1] = ndgrid(distance,angle2,angle1);

Q1 = Q1(:);
Q2 = Q2(:);
D3 = D3(:);

PositionMatrix_new = [
    (cos(Q1).*cos(Q2))/2 + cos(Q1).*cos(Q2).*(D3 + 3/4) ...
    sin(Q2)/2 + sin(Q2).*(D3 + 3/4) + 1 ...
    -(cos(Q2).*sin(Q1))/2 - cos(Q2).*sin(Q1).*(D3 + 3/4)];

toc

Elapsed time is 0.031613 seconds.

[Ulas Dogan]

Thanks a lot Benjamin.

Answer 2

[Matt J]

When speed matters, avoid symbolic math. Also, download

https://www.mathworks.com/matlabcentral/fileexchange/74956-ndgridvecs

skip=360;
SS = 50;        %Step number. 0~360 divided into 50 different points. 

angle1=setdiff(0:(360/(SS)):360, skip);

tic;

[Q1,Q2,D3]=ndgridVecs(angle1,angle1,  0:(1/(SS-1)):1 );
e=ones(size(Q1));

PositionMatrix={
    (cosd(Q1).*cosd(Q2))/2+cosd(Q1).*cosd(Q2).*(D3 + 3/4),...
    sind(Q2)/2+sind(Q2).*(D3 + 3/4) + e,...
    -(cosd(Q2).*sind(Q1))/2-cosd(Q2).*sind(Q1).*(D3 + 3/4)};

PositionMatrix=cell2mat(cellfun( @(x) x(:),PositionMatrix,'uni',0));

toc%Elapsed time is 0.003500 seconds.

[Ulas Dogan]

Thanks a lot Matt. Yours is also working.