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 Kommentare

Geoff Hayes
Geoff Hayes am 3 Apr. 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 am 3 Apr. 2020
Thank you very much it fixed it

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 Akzeptierte Antwort

Steven Lord
Steven Lord am 3 Apr. 2020

1 Stimme

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.

Weitere Antworten (1)

David Hill
David Hill am 3 Apr. 2020
Bearbeitet: David Hill am 3 Apr. 2020

0 Stimmen

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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