My code taking too much time

Here is my code. How can I reduce that time without changing the result?
clear all;clc;close all;
L=1000;
c = randn(1,1000000);
cont3 = 1;
while cont3 < length(c)+1
if(abs(c(cont3)) < 1 || abs(c(cont3)) > 10)
c(cont3) = randn(1);
cont3 = 0;
end
cont3 = cont3+1;
end
c;

 Risposta accettata

Jon
Jon il 16 Nov 2020
Modificato: Jon il 16 Nov 2020
You can operate on the entire vector using for example
a(a<1) = randn
So you could make a loop something like
while any(a<1) || any(a>10)
a(a<1) = randn(sum(a<1),1)
a(a>10) = randn(sum(a>10),1)
end

5 Commenti

Mr.Chandler
Mr.Chandler il 16 Nov 2020
New randn value(n) must provide the condition ( 1 < abs(n) < 10 ) too.
Jon
Jon il 16 Nov 2020
Modificato: Jon il 16 Nov 2020
while any(a<1) || any(abs(a)>10)
condition1 = a < 1;
a(condition1) = randn(sum(condition1),1);
condition2 = abs(a) > 10;
a(condition2) = randn(sum(condition2),1);
end
Mr.Chandler
Mr.Chandler il 16 Nov 2020
I tried the last updated version of your suggestion. However, the values should be in the interval [-10,-1]U[1,10]. In your code, condition = a<1 contains all the values which are less then 1. But it is not okay with the condition. Because, while <1 it contains also [-10,-1] interval but it doesn't need to be changed.
Sorry I didn't look at your specification carefully. Anyhow you should be able to modify
while any(abs(a)<1) || any(abs(a)>10)
condition1 = abs(a) < 1;
a(condition1) = randn(sum(condition1),1);
condition2 = abs(a) > 10;
a(condition2) = randn(sum(condition2),1);
end
Mr.Chandler
Mr.Chandler il 16 Nov 2020
Thanks a lot. This solved my problem exactly.

Accedi per commentare.

Più risposte (0)

Prodotti

Release

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by