Matrix non-exceeding condition

1 visualización (últimos 30 días)
Lev Mihailov
Lev Mihailov el 6 de Ag. de 2019
Comentada: Guillaume el 6 de Ag. de 2019
Hello! I have a 1x15 matrix, the code is presented
[xdata,x]=min(Data); % Data 45x350 x=[ 16 16 16 31 16 0 16....]
a=[15 15 15 15 15....] % 1x350
for i = 1:length(x)-1
if x(i)+a(i)>a(i)
ax=Data((x(i)-a(i):x(i)),i) ;
ay=Data((x(i):x(i)+a(i),i) ;
A{i}=ax;
B{i}=ay;
else x(i)+a(i)<a(i) ;
ax=0 ;
ay=0 ;
A{i}=ax;
B{i}=ay;
end
end
%%% x(i)+a(i) 31+15=46
It so happened that in one value x (i) + a (i) exceeds the size of the matrix, can anyone tell me the condition how can I get around this?
  4 comentarios
Lev Mihailov
Lev Mihailov el 6 de Ag. de 2019
Editada: Lev Mihailov el 6 de Ag. de 2019
if x(i)+a(i)>size(Data) % in my case it's 45
ax=0; % it was just written that it was zero
else x(i)+a(i)<a(i)
ax=0;
Something is needed
Guillaume
Guillaume el 6 de Ag. de 2019
Something is needed
Yes, some much better explanation of what you're trying to do.
Once again,
else x(i)+a(i)<a(i)
ax=0;
is the same as
else
x(i)+a(i)<a(i) %useless statement that will print 0 in the command window
ax=0;
Perhaps, you meant something entirely different:
elseif x(i)+a(i)<a(i)
ax=0;
but until you explain in words what you're trying to do, we can't know for sure.

Iniciar sesión para comentar.

Respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Etiquetas

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by