For loop skipping with if statement

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Sam Hurrell
Sam Hurrell 2022-10-11
评论： Jan 2022-10-12
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?
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Sam Hurrell
Sam Hurrell 2022-10-11
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
Sam Hurrell 2022-10-11
编辑：Sam Hurrell 2022-10-11
That's not really the question here though, your average still wouldn't work with runs of 0's

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回答（2 个）

Jan
Jan 2022-10-11
编辑：Jan 2022-10-12
Ran in:
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');
  2 个评论
Sam Hurrell
Sam Hurrell 2022-10-11
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
Jan 2022-10-12
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.

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Torsten
Torsten 2022-10-11
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.

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