Hello, i need to convert a vector into matrix like this:
A = [1 2 3 4 5]
into
B=[1 2 3; 2 3 4; 3 4 5]
Any ideas ??

 Réponse acceptée

Youssef  Khmou
Youssef Khmou le 24 Mai 2013

1 vote

hi Alex
This is not conversion but you are adding new elements to the original vector :
If you have a vector of size Mx1 you can convert it to matrix using function reshape to get a matrix of size PxN such M=P*N.
r=randn(100,1);
y=reshape(r,10,10);
In your case , try :
A=1:5;
t=1;
r=3;
for n=1:3
B(n,:)=t:n+2;
t=t+1;
r=r+1;
end

3 commentaires

Alex
Alex le 24 Mai 2013
Modifié(e) : Alex le 24 Mai 2013
thank you ! I was trying to avoid the for, but i think is impossible, My final code is
N=32;
A=0:2*N-2;
t=A(1);
r=N-1;
for n=0:r
B(n+1,:)=t:N-1+n;
t=t+1;
end
Mohammed Ghouse Mohiuddin
Mohammed Ghouse Mohiuddin le 31 Mar 2021
For loops are too complex for me. I hate it. Nobody explains it to me in a logical way.
Image Analyst
Image Analyst le 31 Mar 2021
@Mohammed Ghouse Mohiuddin, uncommented code (like Alex's) can be hard to understand. Everyone should use comments. If you can't figure it out, post a new question with the code you are trying to adapt and an explanation of what you really want to do (if you could adapt it). But read this link first.

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Plus de réponses (6)

Image Analyst
Image Analyst le 24 Mai 2013

0 votes

I know it seem really really obvious, but if that is all we have to go on (i.e. no indication that it needs to be generalized in any way), then why not just do this:
B = [A(1:3);A(2:4);A(3:5)]

1 commentaire

Alex
Alex le 24 Mai 2013
i am sorry, my mistake, A is big, it was just an example, exist other way to make this automatic ??

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Matt Kindig
Matt Kindig le 24 Mai 2013
Modifié(e) : Matt Kindig le 24 Mai 2013

0 votes

It might not be very efficient, but I think it should work:
nc = 3; %number of columns in B
nc = nc-1;
c = 1:(length(A)-nc);
B = cell2mat(cellfun(@(n) A(n:(n+nc)), num2cell(c(:)), 'uni', false))
Youssef  Khmou
Youssef Khmou le 25 Mai 2013

0 votes

hi, there are other alternatives , :
try :
N=32;
A=1:N;
A2=0:N-1;
A2=A2';
B=repmat(A,N,1);
for x=1:N
B(:,x)=B(:,x)+A2;
end
Youssef  Khmou
Youssef Khmou le 25 Mai 2013

0 votes

hi,
here is the best solution without using loops :
N=32;
A=1:N;
B=repmat(A,N,1);
A2=(0:N-1)';
B2=repmat(A2,1,N);
C=B+B2;
anukriti dureha
anukriti dureha le 25 Mai 2013

0 votes

hi alex, you can do this:
i=1;
ind=0;
num=3;
while num <=5
ind=ind+1;
z{i}=A(ind:num);
num=num+1;
i=i+1;
end
z=cell2mat(x);b=vec2mat(z,3);
Stephen23
Stephen23 le 31 Mar 2021
A = [1,2,3,4,5];
B = hankel(A(1:3),A(3:5))
B = 3×3
1 2 3 2 3 4 3 4 5

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