Consider, Y is a 1000 by 100 matrix...100 columns and 1000 rows
Y1=Y(:,1);
if I run this code
S1 = numel(Y1);
xx1 = reshape(Y1(1:S1 - mod(S1, 20)), 20, []);
if I run this code
xx1 is a 20 by 50 matrix
What code can I use to do the same thing for the remaining 99 columns of Y

 Réponse acceptée

KSSV
KSSV le 14 Août 2019
Modifié(e) : KSSV le 14 Août 2019

0 votes

Y = rand(1000,100) ;
iwant = zeros(20,50,100) ;
for i = 1:100
Y1 = Y(:,i) ;
S1 = numel(Y1); % this is same always 1000
iwant(:,:,i) = reshape(Y1(1:S1 - mod(S1, 20)), 20, []);
end
With out loop
% no loop
m = 1000 ;
n = mod(m,20) ;
P = Y(1:m-n,:) ;
Q = reshape(P,20,50,[]) ;

3 commentaires

DARLINGTON ETAJE
DARLINGTON ETAJE le 14 Août 2019
Thank You the loop works...
Asking the question further:
Consider, Y is a 1000 by 100 matrix...100 columns and 1000 rows
Y1=Y(:,1);
if I run this code
S1 = numel(Y1);
xx1 = reshape(Y1(1:S1 - mod(S1, 20)), 20, []);
y1 = sum(xx1, 1).' / 20;
y1 is a 50 by 1 matrix
assuming z is also a 50 by 1 matrix
subtr=abs(y1-z)
What code can I use to do the same thing for the remaining 99 columns of Y
KSSV
KSSV le 14 Août 2019
Modifié(e) : KSSV le 14 Août 2019
If you want same to repeat for every column ..you can run a loop...but there are options to avoid looping.

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