# Why doesn't the for loop work?

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Emilia el 6 de En. de 2023
Comentada: Vilém Frynta el 6 de En. de 2023
x=1:16;
p=isprime(x);
for i=1:16
if p(i)==1
p(i)=i;
end
end
A=reshape(p,4,4);
A=A';
display(A);
I want to keep only the prime numbers in vector and to replace the others with 0s. It seems so simple yet it's not working, what's to be done? :(
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### Respuesta aceptada

Vilém Frynta el 6 de En. de 2023
Editada: Vilém Frynta el 6 de En. de 2023
x=1:16;
p=isprime(x);
x(p==0)=0; % Put zeros on the positions where p == 0 in vector X
Is this what you wanted?
This way you do not even need for loop, which is generally better.
However, if you intend to use the for loop, this is the possible way:
x=1:16;
p=isprime(x);
z = x; % create new variable, so you do not overwrite the original one
for i=1:16
if p(i)==1
z(i)=i;
else
z(i)=0;
end
end
z
z = 1×16
0 2 3 0 5 0 7 0 0 0 11 0 13 0 0 0
I'm sure there are other ways to do this as well. Hope I helped.
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Bora Eryilmaz el 6 de En. de 2023
Since p is already a logical vector, p==0 is unnecessary. x(~p) would be a bit simpler.
Vilém Frynta el 6 de En. de 2023
Yes, I thought that would be possible, thanks for additional information ÷)

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### Más respuestas (1)

Bora Eryilmaz el 6 de En. de 2023
Editada: Bora Eryilmaz el 6 de En. de 2023
p = isprime(p) gives you a logical vector, similar to
p = [true true false true]
p = 1×4 logical array
1 1 0 1
When you try to assign i to p(i) it converts the value of i to a logical value; it does not assign i to p(i). See for example
p(3) = 3
p = 1×4 logical array
1 1 1 1
p(3) did not become 3, it became the logical equivalent of 3, which is true, a.k.a. 1.
Instead you can do as simple assignment of 0 to elements of x that are not prime numbers:
x = 1:16;
p = isprime(x);
x(~p) = 0
x = 1×16
0 2 3 0 5 0 7 0 0 0 11 0 13 0 0 0
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Emilia el 6 de En. de 2023
It's all clear now, thank you!!!

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