# Create a matrix out of single values if a for loop?

2 vues (au cours des 30 derniers jours)
gamer le 17 Juin 2021
Commenté : gamer le 17 Juin 2021
Hello,
is it possible to create a 1xF matrix in a for loop out of single values?
n = 2;
r = 0.5;
a = 10;
b = 5;
p = [r + (a-2*r)*rand(n,1),r + (b-2*r)*rand(n,1 )];
for i = 1:n
for j = i:n
if i == j
continue
end
H = (norm(p(i,:)-p(j,:))<=2*r)
end
end
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SungJun Cho le 17 Juin 2021
You should preallocate H as a matrix and save each values into your matrix. For example,
% ...
H = zeros(n,n);
for i = 1:n
for j = i:n
% ...
H(i,j) = (norm(p(i,:)-p(j,:))<=2*r)
end
end
gamer le 17 Juin 2021
I tried it like that but it just gives me again 3 matrices back instead of one
r = 0.5; a = 0; b = 5; n = 3
p=[r + (a-2*r)*rand(n,1),r + (b-2*r)*rand(n,1)];
H = zeros(1,((n-1)*n)/2 )
for i = 1:n
for j = i:n
if i == j
continue
end
H(1,((n-1)*n)/2) = (norm(p(i,:)-p(j,:)))
end
end
I want the norm of (p(1,:) - p(2,:), p(1,:) - p(3,:) and p(2,:) - p(3,:) in one matrix. This is just an example for n = 3.

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### Réponse acceptée

KSSV le 17 Juin 2021
r = 0.5; a = 0; b = 5;
n = 3 ;
p=[r + (a-2*r)*rand(n,1),r + (b-2*r)*rand(n,1)];
H = zeros(1,[]) ;
count = 0 ;
for i = 1:n
for j = i:n
if i == j
continue
end
count = count+1 ;
H(1,count) = (norm(p(i,:)-p(j,:)))
end
end
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gamer le 17 Juin 2021
Thanks:)

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