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 comentarios

Geoff Hayes
Geoff Hayes el 3 de Abr. de 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 el 3 de Abr. de 2020
Thank you very much it fixed it

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 Respuesta aceptada

Steven Lord
Steven Lord el 3 de Abr. de 2020

1 voto

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.

Más respuestas (1)

David Hill
David Hill el 3 de Abr. de 2020
Editada: David Hill el 3 de Abr. de 2020

0 votos

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;

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