Having problems with loops

Hi, I have a file with a 29661x2 array. From that file the first column consists of numbers from 1 to 29661. The second column, consists of timestamps. What I want is to every 60 seconds calculate the rate using the formula of (number in column 1) divided by (time in column 2). for example the beginning of my data looks like this:

if true
  % code
end
1  1.0388
2  1.222
3  2.691833
4  2.862933
5  7.049433
6  10.501633
7  11.089067
8  11.122933
9  12.9572
10  13.499233
11  13.9066
12  16.985033
13  21.246567
14  21.5275
15  27.307267
16  28.023233
17  32.218567
18  35.5397
19  35.773767
20  45.421833
21  50.658233
22  50.8893
23  51.586667
24  52.077567
25  52.506367
26  52.899867
27  53.974267
28  54.6747
29  54.772033
30  56.044233
31  56.292667
32  56.361633
33  57.394533
34  59.001533
35  59.837167
36  61.208533

So what I want is to the code to go through the second column, look for the closest number to 60 (in this portion would be 59.837167) and calculate the rate (35/59.837167 for example). Then check the closest timestamp to 120 and do the same and continue calculating it every 60 seconds (more or less). Because my timestamps are not integers I can't just make a loop calculating the rate every 60 seconds exactly. Instead I thought of a loop that would check if the value of the timestamp is between 59 and 60 and then calculate the rate. Continuing with then checking next if it is between 119 and 120, and so on. I made this code, but for some reason I can;t figure out it is stopping after after calculating the rate between 359 and 360 seconds. Any idea what Im missing?

if true
clc, clear all, close all  
d = importdata('eightfirsthourv5.txt');
a = 60;
b = 59;
i=1;
array = zeros(30000,2);
for i = 1:1:29661
  if d(i,2)<a && d(i,2)>b
      a = a+60;
      b = b+60;        
      X = d(i,1)/d(i,2);
      array(i,1) = i;
      array(i,2) = X;
  end
end

My variables after running it end up a= 360 array 3000x2 double b = 359 d = 29661x2double i = 29661 X = 1.4031

Answer 1

[Star Strider]

I would use interp1 instead, and do something like this:

Data = [1  1.0388
2  1.222
3  2.691833
4  2.862933
5  7.049433
6  10.501633
7  11.089067
8  11.122933
9  12.9572
10  13.499233
11  13.9066
12  16.985033
13  21.246567
14  21.5275
15  27.307267
16  28.023233
17  32.218567
18  35.5397
19  35.773767
20  45.421833
21  50.658233
22  50.8893
23  51.586667
24  52.077567
25  52.506367
26  52.899867
27  53.974267
28  54.6747
29  54.772033
30  56.044233
31  56.292667
32  56.361633
33  57.394533
34  59.001533
35  59.837167
36  61.208533];

ti = [Data(1,2):60:Data(end,2)]';                                   % Time Interpolation Vector
pos = interp1(Data(:,2), Data(:,1), ti);
Velocity = pos ./ ti

Velocity =
   962.6492e-003
   587.7611e-003

You will have to experiment with this to get the result you want with your code, but it eliminates the problem of having to search through your data with a for loop.

Answer 2

[Roger Stafford]

You can use a second-order approximation for your derivative as follows. If x are the numbers in the first column and t the time values in the second column, then for three values, (x1,t1), (x2,t2), and (x3,t3), a formula for the approximate second order derivative at t2 is:

   dx/dt = (x2-x1)/(t2-t1)*(t3-t2)/(t3-t1) + (x3-x2)/(t3-t2)*(t2-t1)/(t3-t1)

Note that this is a weighted average between the two quantities (x2-x1)/(t2-t1) and (x3-x2)/(t3-t2). This formula has the property that if x is a quadratic function of t, the approximation is actually exact.

If your original array is called A, here is how the code would go:

   n = size(A,1);
   x = A(:,1);
   t = A(:,2);
   xe = [x(3);x;x(n-2)]; % Special treatment of the two endpoints
   te = [t(3),t,t(n-2)];
   d21 = (xe(2:n+1)-xe(1:n))./(te(2:n+1)-te(1:n));
   d32 = (xe(3:n+2)-xe(2:n+1))./(te(3:n+2)-te(2:n+1));
   dxdt = (d21.*(te(3:n+2)-te(2:n+1))+d32.*(te(2:n+1)-te(1:n)))./(te(3:n+2)-te(1:n));

The vector dxdt is then a column vector giving the approximate derivative at each corresponding point in A.

I realize I haven't followed your direction to obtain the derivative every 60 seconds, but hopefully the above might give you some ideas. For example, using 'interp1' on these derivative values with cubic interpolation.