Why I get ans = logical 1

23 vues (au cours des 30 derniers jours)
Mohamed Mahir
Mohamed Mahir le 3 Avr 2020
Commenté : Mohamed Mahir le 3 Avr 2020
for i=1:100
x(i)=normrnd(0,1);
end;
A=x(:);
for i=1:100
if (A(i)<0.9557 & A(i) > 0)
Aq(i)=0.497;
elseif (A(i)>0.9957)
Aq(i)=1.493;
elseif (A(i)<-0.9957)
Aq(i)=-1.493;
else
(A(i)>-0.9957 & A(i) <0)
Aq(i)=-0.497;
end;
end;
sum=0;
for i=1:100
sum = (A(i)-Aq(i))^2+sum;
end;
Avg = sum/100
  2 commentaires
Geoff Hayes
Geoff Hayes le 3 Avr 2020
Mohamed - look closely at your else
else
(A(i)>-0.9957 & A(i) <0)
Aq(i)=-0.497;
end;
Should this be an elseif instead where (A(i)>-0.9957 & A(i) <0) is the condition?
Mohamed Mahir
Mohamed Mahir le 3 Avr 2020
Thank you very much it fixed it

Connectez-vous pour commenter.

Réponse acceptée

Steven Lord
Steven Lord le 3 Avr 2020
Another approach uses discretize.
>> A = randn(10, 1);
>> edges = [-Inf -0.9957 0 0.9957 Inf];
>> values = [-1.493, -0.497, 0.497, 1.493];
>> Aq = discretize(A, edges, values);
>> results = table(A, Aq)
If an element of A falls between (for example) edges(2) and edges(3) the corresponding element in Aq will be values(2).
I put the results in a table array so you can easily check that each element of Aq matches what it should be for the corresponding element of A.
  1 commentaire
Mohamed Mahir
Mohamed Mahir le 3 Avr 2020
Thank you that really helped me

Connectez-vous pour commenter.

Plus de réponses (1)

David Hill
David Hill le 3 Avr 2020
Modifié(e) : David Hill le 3 Avr 2020
Much easier way:
A=normrnd(0,1,[100,1]);
[~,~,a]=histcounts(A,[-10,-.9957,0,.9957,10]);
b=[-1.493;-0.497;0.497;1.493];
Aq=b(a);
s=sum((A-Aq).^2);
Avg = s/100;
  1 commentaire
Mohamed Mahir
Mohamed Mahir le 3 Avr 2020
Thank you

Connectez-vous pour commenter.

Catégories

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

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by