Storing data in for loop

Here are the script and the mat. file

What I wanna do is if I run the script file,and type 1 and 600 as a input I got

I got three ans but only last one ans = 0.2515 is stored as a variable I need to store 'ans' as

ans =

      - 0.4992
       0.3636
       0.2515

how can I do this?

[KSSV]

Error while running the script:

Undefined function or variable 'magvector'.

YOur code need to be refined a lot. No initialization, you need to initialize the arrays before loops.

[Yunseo Choi]

function Rmag = magvector(r)

Rmag=sqrt(r(1)^2+r(2)^2+r(3)^2);

[Yunseo Choi]

Sorry I cannot attach the file anymore

[Stephen23]

Or even better, get rid of the loops entirely. Most of them are waste of time, as vectorized code is simpler. The first two loops can be replaced by these two lines:

diffe = diff(A);
subbond = (diffe(1:end-1,:)+diffe(2:end,:))/2;

Possibly the rest of the code could be simplified as well, but until we have magvector we can't do much more.

Answer 1

[Jan]

Either:

result = nan(598, size(sbz, 2));   % Pre-allocate
count  = 0;
for n = 1:598
  if c(1)<=position2(n,1)&&c(2)>=position2(n,1)
    count            = count + 1;
    result(count, :) = sbz(n,:)
  end    
end
result = result(1:count, :);  % Crop unneeded memory

or more efficient with logical indexing:

index  = (c(1) <= position2(:,1) & c(2) >= position2(:,1));
result = sbz(index, :);

Note: Do not use "ans" directly, because it is very volatile. Assign the values to a variable explicitly instead.

Answer 2

[Stephen23]

Get rid of the most of the loops, they don't help you. Learn how to write vectorized code. Here is your code, simplified:

magvector = @(r) sqrt(r(1).^2+r(2).^2+r(3).^2);
%
S = load('position2.mat');
%
diffe = diff(S.position2);                     % 1st loop
subbond = (diffe(1:end-1,:)+diffe(2:end,:))/2; % 2nd loop
% 
for j = size(diffe,1)-1:-1:1
    sbz(j,:) = (3*((dot(subbond(j,:), [1,0,0])/magvector(subbond(j,:)))^2)-1)/2;
end
%
c = linspace(min(S.position2(:,1)),max(S.position2(:,1))+0.0001,1*ceil(max(S.position2(:,1))-min(S.position2(:,1))));
%
idx = c(1)<=S.position2(:,1) & c(2)>=S.position2(:,1); % 4th loop
out = sbz(idx,:)

Giving:

out =
  -0.49921
   0.36363
   0.25148

[Yunseo Choi]

thank you so much!