i have an array whose size is 467*320 = 149440 i mean,
Y = [y1 y2 ... y149440] and i'd like to do that :
Y_1 =[y1+y2+y3...+y320 y321+y322+y321+...y640 .....]
y11=y1+y2+...+y320
y12=y321+y322...+y640
:
:
that is to say; the first element of Y_1 is sum of the first 320 elements and the second element of that is sum of the second 320 elements and going on like this and the size of Y_1 will be 1x467(1x149440/320)...
And i will have Y_2 like Y_1
for Y_2 : Y_2=[y1+y2+y3+...+y320 ..... .... ](1x467)
y21=y1+y2+y3+...+y320
:
:
at the end i create X-array that is
X = [y11+y21 y12+y22 y13+y23 ......]
this is my question :)
i thank you for your helps...

 Respuesta aceptada

Matt Fig
Matt Fig el 23 de Mzo. de 2011

1 voto

I am a little confused, do you mean like this:
Y = reshape(1:18,6,3) % A manageable dimension.
Y_1 = sum(Y,2)
So in your case it would be
Y_1 = sum(Y,2).';
.
.
Or do you mean, more like this:
Y = 1:18
Y_1 = sum(reshape(Y,3,6))
So in your case it would be
Y_1 = sum(reshape(Y,320,467))

4 comentarios

osman yurdakul
osman yurdakul el 23 de Mzo. de 2011
let me explain step-by-step:
first of all i have an array called Y whose dimension is 1x149440
Y =[y1 y2 y3 .......... y149440];
and i want to make Y_1 array whose dimension is 1x467
Y_1 = [y1+y2+y3+...+y320 y321+y322+y323+...+y640];
i mean the first element of Y_1-array is sum of the first 320(1:320) elements of Y-array and the second element of Y_1-array is sum of the second 320(321:640) of Y-array
osman yurdakul
osman yurdakul el 23 de Mzo. de 2011
Y_1 = [y1+y2+y3+...+y320 y321+y322+y323+...+y640 ...........];
i mean the first element of Y_1-array is sum of the first 320(1:320) elements of Y-array and the second element of Y_1-array is sum of the second 320(321:640) of Y-array and going on like this..dimension of Y_1 is 1x467...
Matt Fig
Matt Fig el 23 de Mzo. de 2011
So then my second example is the solution to your problem, correct?
Y = rand(1,149440);
Y_1 = sum(reshape(Y,320,467));
size(Y_1)
ans =
1 467
osman yurdakul
osman yurdakul el 23 de Mzo. de 2011
yes it is...
i tried and saw it was running :D thank you for your helping...

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