# Why doesn't the for loop work?

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Emilia il 6 Gen 2023
Commentato: Vilém Frynta il 6 Gen 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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### Risposta accettata

Vilém Frynta il 6 Gen 2023
Modificato: Vilém Frynta il 6 Gen 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 il 6 Gen 2023
Since p is already a logical vector, p==0 is unnecessary. x(~p) would be a bit simpler.
Vilém Frynta il 6 Gen 2023
Yes, I thought that would be possible, thanks for additional information ÷)

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### Più risposte (1)

Bora Eryilmaz il 6 Gen 2023
Modificato: Bora Eryilmaz il 6 Gen 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 il 6 Gen 2023
It's all clear now, thank you!!!

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