How do I make matlab read certain lines and do calculations only when a condition is true?

Please see attached file. variables t and v represent time and velocity respectively. Z being the condition. I want to calculate average acceleration over the points where the condition is '1', and not when the condition is '0'.

I also want the calculation to be specific to the '1' sections. In other words, calculate the average acceleration for the first section where z is 1, and make that a vector value. Then calculate a separate average acceleration for the second section where z is '1', and so on for the rest of the data.

clear;
clc;

%----------------------------------------------------
uiimport %importing said csv file
while (1) %keeps code from running while selecting a file

cont=input('Press 1 then enter to continue:','s');
if cont=='1'
break
end
end

wrdvec = size(t);

a = repmat('1',[wrdvec,1]);

b = int2str(z)

abc = a==b

av = v(2:end,:) - v(1:end-1,:)
at = t(2:end,:) - t(1:end-1,:)
av = av./at;
avv = zeros(20,1);

for f = find(abc)
    avv(f) = av
end

[Abraham Boayue]

Hi Dylan, your problem seems pretty simple, but I'm not sure if I understand what you want. If Z is the condition, then it has values between 5 and -4 including 0 and 1 for ex1.csv. If Z = 1, you will have a single acceleration for t = 5s and v= 6m/s which will also be the case for ex2.csv.In both cases, the acceleration is a scaler value function. Is this what you want? Cause that's how I understand your question. You can get an acceleration vector for all positive values of Z including or not including 0.

[Dylan Mecca]

Hi Abraham. Sorry for my delayed response! I have updated the question. It looked like I had the wrong files for example up. I've also included my code so far. Right now, I'm getting an error where f isnt the same dimension as av. How can I solve this? Once that gets fixed, do you think I'm on the right track?

[Bob Thompson]

av is an array, while f I'm assuming is a single value. Unless avv is a cell array then you cannot put an entire array, av, into a single value, f. That's why you're getting the error.

[Dylan Mecca]

f shouldn't be a single value. It comes in as an 18x1 double. avv is a column of 20 zeros, so the matrix agreement isn't there. Do you know how I can fix this?

[Bob Thompson]

Does avv need to be 20x1? You could just initiate the size of avv to be based on f. Additionally, there still needs to be agreement between av and f. Even if you have two arrays if you have one that's 18x1 and one that is 16x2 then Matlab is not going to want to join them.

[Dylan Mecca]

I understand that there is a size disagreement. This is why I come for help on this website. avv does not need to be a 20x1. It is just written in for the sake of being written in. Ideally, it would be zeros(size(z)). I understand with calculating acceleration by using the previous line, there will always be a one line discrepancy.

Answer 1

[Abraham Boayue]

clear  variables

% Test the loop with these values of t, v and z, it should work just fine
% for any data.
t = 1:10;
v = [ 3 4 5 7 3 2 8 2 1 6];
M = length(t);
z = [1 1 0 1 1 0 1 0 1 1];
A = [];

for k = 1:M
    if(z(k)== 1)
        a = v(k)/t(k);
        A = [A; k a];
   fprintf('A section (eg. 1 and 2) occurs after a number skip  :counter of 1s before a 0 = %12.8f, a = %12.8f\n', k, a);
    end

      continue

end
%disp(A);

[Abraham Boayue]

Hey Dylan, here I am with another simple code again. This time, I am positive that it does the job. See the output of the print statement. You did mention that you wanted a vector of the average acceleration for each section of ones, however, it is the value of z that decides whether you will get a vector or scalar. Eg. Z = [1 1 0 1 1 0 1] will give two vectors and a scalar.

Answer 2

[Abraham Boayue]

t = 0.1:0.1:2.1;
v= 1:21;
z = [ones(1,8) 0 ones(1,6) 0 0 ones(1,4)];
N=length (t)
a = zeros(1,N);
A1 = []; 
A2 = [];
A3 = [];

for i =1:N
    if(z==1 && v <=8) % condition 
                      % for the 
                     % 1st sect.
    a(i) = v(i)/t(i);
    A1 = [A1 a];
    elseif ( z==1 ||v <=15)
     a(i) = v(i)/t(i);
     A2 = [V2 a];
    elseif (z==1)
    a(i)= v(i)/a(i);
    A3 = [A3 a];
 end
end

dis (' a for the first section ')
disp(A1)
disp('a for the second section )
disp(A2)
disp(' and a for the last section)
disp(A3)

[Abraham Boayue]

Hey Dylan, Why don't you take a look at this simple code I sent; it does the job, although I haven't tested it as I do not have access to my laptop to test it. The data that you are dealing with can be entered into matlab by hand and thereby making life easier. I can make it more advanced than that if you want but that may take a few days from now. I wrote this one on my phone. Hope that helps.

[Dylan Mecca]

Thank you for helping, Abraham. I am looking for something that would be able to automatically pick out the random interval of 1's and 0's in the z vector. Do you have any ideas?

[Abraham Boayue]

Yes, I do have an idea. I just to have access to pc at this point, I will look into that when I do.

[Dylan Mecca]

Excellent, thank you!