'For' loop conditions

2 visualizaciones (últimos 30 días)
Lev Mihailov
Lev Mihailov el 5 de Jul. de 2019
Comentada: Geoff Hayes el 5 de Jul. de 2019
for chek= 1:length(d)-1
if c(chek)+20>20;
nmm=MM(c(chek):c(chek)-20,chek) ;
mn=MM((c(chek)-20:c(chek)),chek) ;
C{chek}=mn;
X{chek}=nmm;
else
c(chek)+20<20
mn=0 ;
nmm=0 ;
C{chek}=mn;
X{chek}=nmm;
end
end
Help please, I have a cycle, I work with 'c', I know that some of its values ​​= 0, first add the essence of the cycle 20 to 'c', then back.
The error is that X knocks out all zeros.
With the values ​​of 'C' everything is fine
  2 comentarios
David Goodmanson
David Goodmanson el 5 de Jul. de 2019
Hi Lev,
abbreviating c(chek) as b, then
b:b-20 produces: 1×0 empty double row vector
and
b:-1:b-20
produces a vector of descending values.
Geoff Hayes
Geoff Hayes el 5 de Jul. de 2019
Lev - perhaps this
else
c(chek)+20<20
should read
elseif c(chek)+20<20
instead. Also, I'm not sure I understand why you subtract 20 from the following
nmm=MM(c(chek):c(chek)-20,chek) ;
mn=MM((c(chek)-20:c(chek)),chek) ;
Please clarify.

Iniciar sesión para comentar.

Respuestas (0)

Etiquetas

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by