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 ??

 Akzeptierte Antwort

Youssef  Khmou
Youssef Khmou am 24 Mai 2013

1 Stimme

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 Kommentare

Alex
Alex am 24 Mai 2013
Bearbeitet: Alex am 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 am 31 Mär. 2021
For loops are too complex for me. I hate it. Nobody explains it to me in a logical way.
Image Analyst
Image Analyst am 31 Mär. 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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Weitere Antworten (6)

Image Analyst
Image Analyst am 24 Mai 2013

0 Stimmen

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 Kommentar

Alex
Alex am 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 am 24 Mai 2013
Bearbeitet: Matt Kindig am 24 Mai 2013

0 Stimmen

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 am 25 Mai 2013

0 Stimmen

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 am 25 Mai 2013

0 Stimmen

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 am 25 Mai 2013

0 Stimmen

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 am 31 Mär. 2021

0 Stimmen

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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Gefragt:

am 24 Mai 2013

Beantwortet:

am 31 Mär. 2021

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