Hello ,
I want a better speed in operating my code ,if i did enter a large number of points like 10000 it will operate in 10 mins or more and my processor is good enough ,this code is plotting points in a uniform shape like rectangular and randomly with different x and y , thanks for attention.
%%
RN=input('Enter number of points =');
U1=[];U2=[];
for i=1:RN
xa=-L/2+L*rand(1,1)+0.2*rand(1,1)-.4;
ya=-N/2+N*rand(1,1)+0.2*rand(1,1)-.4;
xra(i)=xa;yra(i)=ya; z=[xra(i) yra(i)];
sub1=z(1);sub2=z(2);U1=[U1;sub1];U2=[U2;sub2];
USERS=[U1 U2];
figure(1)
plot(xra(i),yra(i),'k.');hold on
end

 Akzeptierte Antwort

Daniel Shub
Daniel Shub am 25 Apr. 2012

0 Stimmen

The best thing to do would be to move the plotting outside the loop:
...
end
figure(1);
plot(xra,yra,'k.');

Weitere Antworten (2)

Junaid
Junaid am 25 Apr. 2012

1 Stimme

Though your code requires cleaning but this code will generate exact output you are required. In your code there are many variables which are not required.
RN=input('Enter number of points =');
L=input('Enter the length x =');
N=input('Enter the length y =');
U1=[];U2=[];
xa = -L/2+L*rand(RN,1)+0.2*rand(RN,1)-.4;
ya = -N/2+N*rand(RN,1)+0.2*rand(RN,1)-.4;
xra=xa;yra=ya; z=[xra yra];
sub1=z(:,1);sub2=z(:,2);U1=[U1;sub1];U2=[U2;sub2];
USERS=[U1 U2];
figure(1)
plot(xra,yra,'k.');
Junaid
Junaid am 25 Apr. 2012

0 Stimmen

I guess this can be done in one line without using for loop, and second line for plot. Could you tell us about L, N and other varibales you are using in side loop

3 Kommentare

Ahmed Hassaan
Ahmed Hassaan am 25 Apr. 2012
L=input('Enter the length x =');
N=input('Enter the length y =');
Ahmed Hassaan
Ahmed Hassaan am 25 Apr. 2012
L=input('Enter the length x =');
N=input('Enter the length y =');
Daniel Shub
Daniel Shub am 25 Apr. 2012
Without having timed it, my guess is the time improvement from complete vectorization will be small compared to the use of a single plot.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Startup and Shutdown finden Sie in Hilfe-Center und File Exchange

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by