How do I code for this random situation?

1 vue (au cours des 30 derniers jours)
NAA
NAA le 20 Mar 2023
Commenté : Image Analyst le 20 Mar 2023
On a r*c size grid, 55% of the sites are randomly filled with X, 2% randomly filled with Y, and the rest are empty
  3 commentaires
NAA
NAA le 20 Mar 2023
There was no previous question like this, and I saw your comment on the other question, you didn't answer. Instead you just left a rude comment like the one here
Image Analyst
Image Analyst le 20 Mar 2023
@NAA, OK try it this way:
r = 10;
c = 20;
output = nan(r, c);
numX = round(0.55 * r * c) % Number of elements to place an X into.
numY = round(0.02 * r * c) % Number of elements to place a Y into.
X = 1;
Y = 2;
output(1 : numX) = X;
output(numX + 1 : numX + numY) = Y;
randomIndexes = randperm(numel(output));
output = reshape(output(randomIndexes), [r, c])
There are other ways that would work also.

Connectez-vous pour commenter.

Réponse acceptée

Image Analyst
Image Analyst le 20 Mar 2023
This looks like a homework problem. If you have any questions ask your instructor or read the link below to get started:
Obviously we can't give you the full solution because you're not allowed to turn in our code as your own.
Hint:
output = nan(r, c);
numX = round(0.55 * r * c); % Number of elements to place an X into.
numY = round(0.02 * r * c); % Number of elements to place a Y into.
Two ways to do it:
  1. You can do it vectorized using randperm (assign the X to the first numX elements, then assign Y to the next numY elements, then use randperm to scramble the order), or
  2. you can do it brute force using a for loop to place the x value and another loop to place the Y value, but only in the location if the value is a nan (not an X).
  1 commentaire
NAA
NAA le 20 Mar 2023
it's not the hw problem, this is just part of an assumption I am making to get started, but thank you!

Connectez-vous pour commenter.

Plus de réponses (0)

Produits


Version

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by