how can i get a matrix from an array like this:

Starting with
a = [12 21 32];
I would like to get this:
M = [12 21 32; 21 32 12; 32 12 21];
with size and length fun. only
Is it some kind of loop???

 Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 2 Jan. 2013
Bearbeitet: Azzi Abdelmalek am 2 Jan. 2013

0 Stimmen

a = [12 21 32 42];
n=length(a)
M=a
for k=1:n-1
c=a(1)
a(1:end-1)=a(2:end)
a(end)=c
M=[M;a]
end

2 Kommentare

Zaza
Zaza am 2 Jan. 2013
thanks a lot Azzi
ןt was a bit tricky for me
This does not satisfy your stated requirements, as assignment and subscripting and [] are all functions in MATLAB: in particular, those operations call upon subsasgn(), subsref(), and vertcat (and horzcat).

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (5)

Roger Stafford
Roger Stafford am 2 Jan. 2013

1 Stimme

M = hankel(a,fliplr(a));
or if you can't use the 'hankel' function do this
n = length(a);
M = reshape(a(mod(floor(((n+1)*(0:n^2-1))/n),n)+1),n,n);
Azzi Abdelmalek
Azzi Abdelmalek am 2 Jan. 2013
Bearbeitet: Azzi Abdelmalek am 2 Jan. 2013

0 Stimmen

M=[a ;circshift(a,[0 -1]); circshift(a,[0 -2])]
%or more general
a=[12 21 32 42 52]
M=cell2mat(arrayfun(@(x) circshift(a,[0 -x]),(0:numel(a)-1)','un',0))

6 Kommentare

Zaza
Zaza am 2 Jan. 2013
Bearbeitet: Zaza am 2 Jan. 2013
can i get this using for or while? because in HW i can use length and size fun. only
Azzi Abdelmalek
Azzi Abdelmalek am 2 Jan. 2013
Bearbeitet: Azzi Abdelmalek am 2 Jan. 2013
Yes you can, try for loop
Zaza
Zaza am 2 Jan. 2013
for i=1:length(a)
M(i,:) = a(i);
i get the 1st column of M but i can't manipulate M to get the correct form of M...
Use circshift function in the loop
Zaza
Zaza am 2 Jan. 2013
i can't use this fun.
but i can't do it with size and length only
oh i must take a deep breath and try later...
I will give you a hint
a=[12 21 32]
b=a([2 3 1])

Melden Sie sich an, um zu kommentieren.

Walter Roberson
Walter Roberson am 2 Jan. 2013

0 Stimmen

You can look at the code for toeplitz.m
Image Analyst
Image Analyst am 2 Jan. 2013

0 Stimmen

I don't know what "with size and length fun. only" means, but the simplest solution is
M = [a(1,1), a(1,2), a(1,3); a(1,2), a(1,3), a(1,1); a(1,3), a(1,1), a(1,2)];
Since you haven't given any other requirements, like it needs to work for other sizes of matrices, this will work exactly as you have said. No functions like toeplitz() or circshift() needed. If you needed other requirements and flexibility, I assume you would have said so (ok, that's probably a bad assumption given the types of posts people make).

4 Kommentare

Zaza
Zaza am 2 Jan. 2013
i meant that i can use only "size" and "length" functions
and i forget to mention that this has to work on any size of given array...
Image Analyst
Image Analyst am 2 Jan. 2013
Bearbeitet: Image Analyst am 2 Jan. 2013
Did you also forget to mention that it is homework? Because I can think of no other reason why you would be prohibited from using any function at your disposal to get the job done. Add that tag to your post if it's homework. We won't just give you the answer to your homework outright, but may give hints.
Zaza
Zaza am 2 Jan. 2013
i did mention that
tips are welcome
Image Analyst
Image Analyst am 2 Jan. 2013
Bearbeitet: Image Analyst am 2 Jan. 2013
OK - I didn't know the acronym HW. Others probably didn't recognize that as homework either, because it looks like three people have already done your homework for you and given you an answer that should work. I'll add the "homework" tag for you. Next time you can add it yourself so you won't be caught copying.

Melden Sie sich an, um zu kommentieren.

Sean de Wolski
Sean de Wolski am 2 Jan. 2013

0 Stimmen

M = a(sortrows(bsxfun(@(x,y)mod(x+y,numel(a))+1,1:numel(a),(1:numel(a))')))

Kategorien

Mehr zu Programming finden Sie in Hilfe-Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by