slow drawing of a set of points on a plane

Dear Sir

please let me have a help

this is a simple set of statements that:

-define figure 1

-clear the figure

-define a matrix 2 x 82 ,where each column represent the coordinate of a point on a plane

these set of points represent a circle and a cross , centered inside

-then do a loop that draw each point at a time, as red dot

the "hold on" is needed because at each loop i wish to maintain the points already drawed

the purpose of this code is to visualize this figure first undeformed, and after, with some deformation gradient F applied to the set of points, (and) with another (more external) loop , to visualize the advance of the deformation making the gradient F varying

(F not present in the code below)

----------------------------------------

clear

figure(1),clf

p=[[cos(0:.1:1.9*pi) (-1:.2:1) 0*(-1:.2:1) ]' [sin(0:.1:1.9*pi) 0*(-1:.2:1) (-1:.2:1) ]']'

for j=1:max(size(p)) % visualize

figure(1),plot(p(1,j),p(2,j),'r.'),grid on, axis([-2 2 -2 2]),axis('square')

hold on

end

----------------------------------------

now , to draw this simple figure it take roughly 1 second , so i ask were i'm wrong , and how can I speed up the execution of this code

the specification of the pc is as follow

Device name s.

Processor Intel(R) Pentium(R) CPU G860 @ 3.00GHz 3.00 GHz

Installed RAM 4.00 GB

Device ID ....8C39

Product ID ...9

System type 64-bit operating system, x64-based processor

Pen and touch No pen or touch input is available for this display

Edition Windows 10 Home

Version 22H2

Installed on ‎11/‎7/‎2020

OS build 19045.4355

Experience Windows Feature Experience Pack 1000.19056.1000.0

Matlab :

ver

-----------------------------------------------------------------------------------------------------

MATLAB Version: 24.1.0.2689473 (R2024a) Update 6

MATLAB License Number: STUDENT

Operating System: Microsoft Windows 10 Home Version 10.0 (Build 19045)

Java Version: Java 1.8.0_202-b08 with Oracle Corporation Java HotSpot(TM) 64-Bit Server VM mixed mode

-----------------------------------------------------------------------------------------------------

MATLAB Version 24.1 (R2024a)

Symbolic Math Toolbox Version 24.1 (R2024a)

>>

Thankyou very much

Kind Regards

manu1965

Answer 1

[Aquatris]

In essence for loop are always slower. So try to vectorize your code whereever possible for efficiency, here is a good read. In your case vectorizing the code makes it ~10 times faster

p=[[cos(0:.1:1.9*pi) (-1:.2:1) 0*(-1:.2:1) ]' [sin(0:.1:1.9*pi) 0*(-1:.2:1) (-1:.2:1)  ]']';

% for loop ploting
tic
figure(1)
for j=1:max(size(p))      % visualize
figure(1),plot(p(1,j),p(2,j),'r.'),grid on, axis([-2 2 -2 2]),axis('square')
hold on
end

timeViaForLoop = toc

timeViaForLoop = 0.6282

% vector plotting
tic
figure(2)
plot(p(1,:),p(2,:),'r.'),grid on, axis([-2 2 -2 2]),axis('square')

timeViaVectorizeCode = toc

timeViaVectorizeCode = 0.0679

[M A]

Good evening

thanks to all for the answers

Dear Aquatris, thanks for the detailed answer, I read the article on vectorization, I imagined that treating a vector as a single block instead of a set of scalars could be faster, but not 10 times faster!

anyway I would like to ask you if there is a utility that analyzes a script and tells me what improvements can be made for the execution time

Kind regards

manu1965

[Voss]

https://www.mathworks.com/help/matlab/ref/profile.html

[Aquatris]

There is matlab profiler that might come in handy, here is the link

Answer 2

[Voss]

p = [ cos(0:.1:1.9*pi), -1:.2:1, 0*(-1:.2:1) ; sin(0:.1:1.9*pi), 0*(-1:.2:1), -1:.2:1 ]

p = 2x82

1.0000 0.9950 0.9801 0.9553 0.9211 0.8776 0.8253 0.7648 0.6967 0.6216 0.5403 0.4536 0.3624 0.2675 0.1700 0.0707 -0.0292 -0.1288 -0.2272 -0.3233 -0.4161 -0.5048 -0.5885 -0.6663 -0.7374 -0.8011 -0.8569 -0.9041 -0.9422 -0.9710 0 0.0998 0.1987 0.2955 0.3894 0.4794 0.5646 0.6442 0.7174 0.7833 0.8415 0.8912 0.9320 0.9636 0.9854 0.9975 0.9996 0.9917 0.9738 0.9463 0.9093 0.8632 0.8085 0.7457 0.6755 0.5985 0.5155 0.4274 0.3350 0.2392

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figure
plot(p(1,:),p(2,:),'r.')
grid on
axis([-2 2 -2 2])
axis('square')