For loop skipping with if statement

I have a 'for loop' designed to go through graph data (col1 = time, col2 = y), with an 'if statement' so if it finds a value equal to 0 in col2 it will calculate an average between the values in col2 before and after the 0 at 'a' (the time in col1):

for a = 2:length(pk)-1
    if pk(a,2) == 0
        m = (pk(a+1,2) - pk(a-1,2))/(pk(a+1,1) - pk(a-1,1));    % calculate gradient of values before an after 'a'
        pk(a,2) = pk(a+1,2) - m*(pk(a+1,1)-pk(a,1));               % calculate value at 'a' [there are no 0's in col1]
    else
    end
end

However in col2 of the data I have found consecutive runs of 0's which means I cannot apply the code above. How can I alter the code so as to skip over the runs of 0 to create average values for them also?

[Torsten]

if it finds a value equal to 0 in col2 it will calculate an average between the values in col2 before and after the 0 at 'a'

And why do you do that ?

[Sam Hurrell]

I'm comparing 2 different graphs, but they sometimes have different time points (hence why col2 sometimes equals 0). All I need is an average as a stand in.

[Sam Hurrell]

That's not really the question here though, your average still wouldn't work with runs of 0's

Answer 1

[Jan]

This is a linear interpolation. This is implemented without a loop already:

pk = [1, 2; 3, 0; 4, 0; 5, 3; 7, 2; 10, 0; 13, 5];
plot(pk(:, 1), pk(:,2), 'bo');
hold('on');

miss        = (pk(:, 2) == 0);
pk(miss, 2) = interp1(pk(~miss, 1), pk(~miss, 2), pk(miss, 1));

plot(pk(:, 1), pk(:, 2), 'r-');
plot(pk(miss, 1), pk(miss, 2), 'ro');

[Sam Hurrell]

I'm not trying to ignore the values equal to zero, I'm trying to calculate values for them, but I run into trouble when there are runs of 0

[Jan]

Yes, and this is what I have posted already. The zeros are not ignored, but the average using the surrounding non-zero values is used. If there is one zero, the result is the same as with your code, but the interp1 approach allows a sequence of zeros also.

I've added red circles in the diagram to show the replaced zero values.

Answer 2

[Torsten]

The usual average would be

pk(a,2) = ( pk(a-1,2)*(pk(a,1)-pk(a-1,1)) + pk(a+1,2)*(pk(a+1,1)-pk(a,1)) )/(pk(a+1,1)-pk(a-1,1)) 

But I think this will result in the same problem you already have.

To compare two graphs, use interp1 to find values for the union of the time points of graph 1 and graph 2.