All possible combination based on 2^n but with 1 and -1
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hi all, thank you for those of you who have answered my question below.
I have a slightly different question but still does not know how to achieve this.
Again, I want to create a matrix containing all possible combination. Example of this is shown below. The size of the matrix depends on the number of variable n and the total of combination should follow the . The example below is valid for and, hence, the total number of rows is 8. The value of each element is 1 and -1.
How to create this matrix automatically depending on the number of variable n?
2 commentaires
Réponses (3)
Dyuman Joshi
le 14 Mai 2021
y=dec2bin([7 4 2 1])-'0';
y(y==0)=-1;
z =[y; flipud(y)]
This only works for this particular example. If you want a generalised answer, give more examples.
2 commentaires
Dyuman Joshi
le 14 Mai 2021
Because only this combination corresponds to the desired result.
That's why I mentioned - "This only works for this particular example. If you want a generalised answer, give more examples"
Daniel Pollard
le 14 Mai 2021
You could take the answer from your previous question, subtract 0.5 and multiply by 2. Your accepted answer was
n = 3;
m = dec2bin(0:pow2(n)-1)-'0' % limited precision
which then becomes
n = 3;
m = dec2bin(0:pow2(n)-1)-'0'; % limited precision
m = 2*(m-0.5)
0 commentaires
Jan
le 14 Mai 2021
Modifié(e) : Jan
le 14 Mai 2021
dec2bin creates a CHAR vector, while -'0' converts it to a double again. This indirection costs some time. The direct approach:
n = 3;
m = rem(floor((0:2^n-1).' ./ 2 .^ (0:n-1)), 2)
pool = [1, -1]; % Arbitrary values
result = pool(m + 1) % Add 1 to use m as index
Voir également
Catégories
En savoir plus sur Resizing and Reshaping Matrices dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!