Borrar filtros
Borrar filtros

regroupe the results in a matrix

1 visualización (últimos 30 días)
Dija
Dija el 30 de Nov. de 2014
Comentada: dpb el 1 de Dic. de 2014
hii i have this
f= 1 0
0 1
0 2
1 3
0 4
1 5
1 6
0 7
this indicate the numbers from 0 to 7 in decimal i want to regroupe the 1 in a matrix and at the same time tell which number it is and do the same thing with 0 i have a mistake somewhere here's the program
N=8;
r=3;
A=zeros(N,r);
Q=zeros(N,1);
for k=0:1:N-1
A(k+1,:)=dec2bin(k,r)-'0';
Q(k+1,1)=xor(xor(A(k+1,1),A(k+1,2)),A(k+1,3));
f=xor(Q,1);
end
for k=0:1:N-1
if f(k+1)==1
se=k
S(k+1,1)=se
else
ss=k
C(k+1,1)=ss
end
end
S =
0
0
0
0
3
5
0
6
C =
0
1
2
0
4
0
0
7
i don't know how to do to delete the 0 i just want 0356 and 1247 to appears
i hope you understand me

Respuesta aceptada

Thorsten
Thorsten el 1 de Dic. de 2014
Editada: Thorsten el 1 de Dic. de 2014
Ok, here is my 2nd solution. There's also a faster way to compute f, I think:
f = 1 - (mod(sum(A'), 2) == 1);
x = 0:7;
s = x(~logical(f));
c = x(logical(f));
  3 comentarios
Thorsten
Thorsten el 1 de Dic. de 2014
Sure, but please start a new question to that others can help you, too.
Dija
Dija el 1 de Dic. de 2014
it is about this program too and i guess that you can understand me more than others it is okey if you can't help me but i will try with you first :)
how can i discard any bianry digit i want in A for example if i discard the first one in A
A =
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
i want to get this
A =
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1

Iniciar sesión para comentar.

Más respuestas (3)

dpb
dpb el 30 de Nov. de 2014
Editada: dpb el 30 de Nov. de 2014
>> s=find(f(:,1))
s =
1
4
6
7
>> c=f(s,2)
c =
0
3
5
6
>>
Looks like your desired S above is from some other sample dataset...f(2,1)==0
ADDENDUM
OK, following your comment/plea; the complementary is by the above
s=find(f(:,1);
c=f(s,2);
s=f(find(f(:,1)==0,2));
This is a little convoluted; I did it that way because as noted I thought you were looking for the position in the original vector, not the two results.
More straightforward in the latter case is to use logical addressing instead of find --
ix=f(:,1)==1); % the '1' location vector
s=f(ix,2);
c=f(~ix,2);
  4 comentarios
Dija
Dija el 1 de Dic. de 2014
it doesnt work too
dpb
dpb el 1 de Dic. de 2014
Does too...
>> f
f =
1 0
0 1
0 2
1 3
0 4
1 5
1 6
0 7
>> ix=f(:,1)==1;
>> s=f(ix,2)
ans =
0
3
5
6
>> c=f(~ix,2)
c =
1
2
4
7
>>

Iniciar sesión para comentar.


Thorsten
Thorsten el 1 de Dic. de 2014
x = f(:,2);
x(logical(f(:,1)))
ans =
0
3
5
6
x(~logical(f(:,1)))
ans =
1
2
4
7
  9 comentarios
Dija
Dija el 1 de Dic. de 2014
thank you
dpb
dpb el 1 de Dic. de 2014
Well, again we're mis-communicating on what you mean by "which number it is" -- whether it's the location in the array or the numeric value. Since you weren't happy with the latter in the demonstrated solution I thought you meant the position again which find returns.
If it's actually the numeric value then the logic array solution is the more straightforward as demonstrated with the inputs in a 2-column array. If you don't put them in the array, then use whatever is the storage for each column in place of the array.

Iniciar sesión para comentar.


Andrei Bobrov
Andrei Bobrov el 1 de Dic. de 2014
S = num2str(f(f(:,1)>0,2))';
C = num2str(f(~f(:,1),2))';
  1 comentario
Dija
Dija el 1 de Dic. de 2014
the same error :( Index exceeds matrix dimensions.
Error in Untitled2 (line 12) S = num2str(f(f(:,1)>0,2))'

Iniciar sesión para comentar.

Categorías

Más información sobre Operating on Diagonal Matrices en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by