Effacer les filtres
Effacer les filtres

unable to get desired results

1 vue (au cours des 30 derniers jours)
Prashant Bhagat
Prashant Bhagat le 13 Juil 2022
Commenté : Voss le 13 Juil 2022
Please find the below code.
User = zeros(1,10);
for i = 1:10
probability_of_transmission = rand(1);
if probability_of_transmission > p
j = 5;
User(i) = 1;
else
User(i) = 0;
end
end
I just want the output to be somewhat like: User = [ 0 0 0 1 1 1 1 1 0 0 ]. How can i put one more if argument in the for loop so that at the end i will get consecutive ones insetead of ones at random index.
  4 commentaires
Abderrahim. B
Abderrahim. B le 13 Juil 2022
and what does that j doing there in your loop ? With using rand to compute probability_of_transmission you can't get the User vector you want.
Prashant Bhagat
Prashant Bhagat le 13 Juil 2022
j is the number of time slots required for transmission of one packet. Then what i have to do to get the desired output

Connectez-vous pour commenter.

Réponse acceptée

Voss
Voss le 13 Juil 2022
Maybe this?
User = zeros(1,10);
p = 0.785;
for i = 1:10
probability_of_transmission = rand(1);
if probability_of_transmission > p
j = 5;
User(i+(0:j-1)) = 1;
end
end
User
User = 1×10
0 0 0 1 1 1 1 1 0 0
Note that the sequences of 5 ones may overlap each other and may also extend User to beyond 10 elements; it's not clear whether those aspects would be bugs or features for what you want to do.
  2 commentaires
Prashant Bhagat
Prashant Bhagat le 13 Juil 2022
Thanks
Voss
Voss le 13 Juil 2022
You're welcome!

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Tags

Produits


Version

R2022a

Community Treasure Hunt

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

Start Hunting!

Translated by