'For' loop conditions

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Lev Mihailov
Lev Mihailov 2019-7-5
评论： Geoff Hayes 2019-7-5
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 个评论
David Goodmanson
David Goodmanson 2019-7-5
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 2019-7-5
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.

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