convert Vector into matrix ?

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 ??

 Accepted Answer

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 Comments

Alex
Alex on 24 May 2013
Edited: Alex on 24 May 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
For loops are too complex for me. I hate it. Nobody explains it to me in a logical way.
@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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More Answers (6)

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 Comment

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 on 24 May 2013
Edited: Matt Kindig on 24 May 2013
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))
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
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 on 25 May 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);
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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Asked:

on 24 May 2013

Answered:

on 31 Mar 2021

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