Thread Subject:

Reading from beginning and end of an array and repeat

Hello everyone
I want to select an array(s)with length n out of
a bigger array (z) in a way that the
s =[162 320 169 280 172]'
or(get one from first row and one from end and repeat until it reaches n)
How to write a code to get array s?
below is what I tried.

z=[162 169 172 184 199 205 220 240 280 320]'
n=5;
[p,q]=size(z);
s=zeros(n,1);
[d,e]=size(s);


for i=1:p;
    for j=p:p-(i-1);
        for d=1:n;
            for f=d+1:n;
                s(d)=z(i);
                s(f)=z(j);
            end
        end
    end
end
s

enviro <farhadnejadkoorki@yahoo.co.uk> wrote in message <33098052.1235314424604.JavaMail.jakarta@nitrogen.mathforum.org>...
> Hello everyone
> I want to select an array(s)with length n out of
> a bigger array (z) in a way that the
> s =[162 320 169 280 172]'
> or(get one from first row and one from end and repeat until it reaches n)
> How to write a code to get array s?
> below is what I tried.
>
> z=[162 169 172 184 199 205 220 240 280 320]'
> n=5;
> [p,q]=size(z);
> s=zeros(n,1);
> [d,e]=size(s);
>
>
> for i=1:p;
> for j=p:p-(i-1);
> for d=1:n;
> for f=d+1:n;
> s(d)=z(i);
> s(f)=z(j);
> end
> end
> end
> end
> s

???

I do not understand what you want. Can you please rephrase your problem? And given an example of the input and expected output?

Jos

"Jos " <#10584@fileexchange.com> wrote in message <gnruiq$cr2$1@fred.mathworks.com>...
> enviro <farhadnejadkoorki@yahoo.co.uk> wrote in message <33098052.1235314424604.JavaMail.jakarta@nitrogen.mathforum.org>...
> > Hello everyone
> > I want to select an array(s)with length n out of
> > a bigger array (z) in a way that the
> > s =[162 320 169 280 172]'
> > or(get one from first row and one from end and repeat until it reaches n)
> > How to write a code to get array s?
> > below is what I tried.
> >
> > z=[162 169 172 184 199 205 220 240 280 320]'
> > n=5;
> > [p,q]=size(z);
> > s=zeros(n,1);
> > [d,e]=size(s);
> >
> >
> > for i=1:p;
> > for j=p:p-(i-1);
> > for d=1:n;
> > for f=d+1:n;
> > s(d)=z(i);
> > s(f)=z(j);
> > end
> > end
> > end
> > end
> > s
>
> ???
>
> I do not understand what you want. Can you please rephrase your problem? And given an example of the input and expected output?
>
> Jos
-------------------------------------------------------------
input=[162 169 172 184 199 205 220 240 280 320]'
output=[162 320 169 280 172]'
I want to select 5 number from input as above. selection must happens (input(1),
input(10),input(2),input(9),input(3)...).
Please let me know if it is not clear. thanks in advance

input=[162 169 172 184 199 205 220 240 280 320]
outputlong = reshape([input;input(end:-1:1)],1,[])

% Now take however many you want.

output = outputlong(1:5)






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enviro <farhadnejadkoorki@yahoo.co.uk> wrote in message <33098052.1235314424604.JavaMail.jakarta@nitrogen.mathforum.org>...
> Hello everyone
> I want to select an array(s)with length n out of
> a bigger array (z) in a way that the
> s =[162 320 169 280 172]'
> or(get one from first row and one from end and repeat until it reaches n)
> How to write a code to get array s?
> below is what I tried.
>
> z=[162 169 172 184 199 205 220 240 280 320]'
> n=5;
> [p,q]=size(z);
> s=zeros(n,1);
> [d,e]=size(s);
>
>
> for i=1:p;
> for j=p:p-(i-1);
> for d=1:n;
> for f=d+1:n;
> s(d)=z(i);
> s(f)=z(j);
> end
> end
> end
> end
> s

  Do this:

 n = 5;
 m = size(z,2);
 s = z((2*m+3+(2*m-2*(1:n)+1).*(-1).^(1:n))/4);

Roger Stafford

Thanks Roger,

Could you please let me know where I can find further readings on this kind of engines?

enviro

enviro <farhadnejadkoorki@yahoo.co.uk> wrote in message <24635655.1235394672434.JavaMail.jakarta@nitrogen.mathforum.org>...
> Thanks Roger,
>
> Could you please let me know where I can find further readings on this kind of engines?
>
> enviro

  I don't quite know how to answer your question. I just looked at the sequence

 1 m 2 m-1 3 m-2 4 m-3 ....

of indices you needed to accomplish your task, and looked for some pattern one could generate easily. I know from experience that (-1)^(1:n) has useful properties for producing alternating features in a sequence. I finally set down the equations

 a*(1:4) + b + (c*(1:4)+d)*(-1)^(1:4) = [1 m 2 m-1]

which gives four equations and four unknowns and lo and behold that solution also held true for all the other indices in (1:n). (I had an intuitive feeling it would work before I tried it.)

  I have no idea where one can read about such things. I am depending on things I have learned over a period of many years.

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <gns1ae$dk5$1@fred.mathworks.com>...

> n = 5;
> m = size(z,2);
> s = z((2*m+3+(2*m-2*(1:n)+1).*(-1).^(1:n))/4);

Another way of creating the indices IDX into z:

z = 10:20 ;
n = 5 ;

a = rem(1:n,2)
b = ceil((1:n)/2)
idx = a .* b + ~a .* (numel(z)-b+1)
z(idx)

Jos

Is it possible to make a loop to do this? I should go that way as it is needed to check pair values for my original script. thanks