creating a vector from data in a large array

Hello all,

Thanks for this resource, it has been helpful reading but I am now at the point I need to ask a question......

I have a very large array that I need to use periodic values from to create a vector (still a large vector)

my problem, example, I need to take the difference between two sets of numbers and divide them (finding the gradient) and then repeat this every 65 lines till the end of the array, creating a vector.

eg. ((3,2)-(2,2))/((3,4)-(2,4))         this is the first entry of the vector 
    ((68,2)-(67,2))/((68,4)-(67,4))     this is the second entry of the vector 
    ((133,2)-(132,2))/((133,4)-(132,4)) this is the third entry of the vector 
    ((198,2)-(197,2))/((198,4)-(197,4)) this is the forth entry of the vector

and so on and so on, every 65 lines until the end of the 65000 x 8 array creating a 1000 X 1 vector

The array is called AvD_pi_tr_64_7900

I hope someone can help as I feel this is not as hard as I am making it but just don't know the terminology to search for within the sight.

Thank you in advance,

all the best,

Leon

Answer 1

[Image Analyst]

How about a simple for loop:

AvD_pi_tr_64_7900 = rand(65000, 8); % Whatever...sample data.
counter = 1;
for row = 3 : 65 : size(AvD_pi_tr_64_7900, 1)
  output(counter) = (AvD_pi_tr_64_7900(row, 2) - AvD_pi_tr_64_7900(row - 1, 2)) / ...
    (AvD_pi_tr_64_7900(row, 4) - AvD_pi_tr_64_7900(row - 1, 4));
  counter = counter + 1;
end

By the way, 65000 by 8 is not that large.

Answer 2

[dpb]

Start with the "brute force, straightahead" solution...

dN=65;                     % the magic number between observations--don't bury inside code
iy1=2; iy2=4;              % the magic numbers of target columns for difference
[nr,~]=size(x);            % number rows in array
ix=3:dN:nr                 % index vector to rows needed
v=(x(ix+1,iy1)-x(ix,iy1))./(x(ix+1,iy2)-x(ix,iy2));  % compute the derivative vector

ERRATUM

I just realized I forgot to do the differences previously, sorry...see above

Answer 3

[Leon Newton]

Thanks so much for both of the responses.

I can now do what I needed in this bit of my task.

All the best and thanks again.... :-)