special kind of interpolation

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Ingrid
Ingrid 2015-5-27
编辑： Ingrid 2015-5-27
I have some data shown in the example figure below for which I want to perform an interpolation to calculate the average value. However, I need a linear interpolation between the datapoints, except for points that are zero, there are values should be put at zero between the two surrounding points. How can I achieve this?

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Thorsten
Thorsten 2015-5-27
编辑：Thorsten 2015-5-27
This "interpolation" is basically a plot with values added where y is zero:
x = x(:)'; y = y(:)'; % force row vectors
ind = find(diff(y == 0) == -1); % find the index where y changes from zero to non-zero
% add points before and after ind with zero y value
for i = 1:numel(ind)
indi = ind(i);
x = [x(1:indi-1) x(indi-1) x(indi) x(indi+1) x(indi+1:end)];
y = [y(1:indi-1) 0 y(indi) 0 y(indi+1:end)];
end
plot(x, y) % plot the curve
  1 个评论
Ingrid
Ingrid 2015-5-27
this looks pretty good, but it is not the plot that I am interested in, I want to calculate the mean by using the interpolated values (I just use a x-vector with spacing 1 to do the interpolation). I thought it was not possible to use the interp1 function with two x-values equal to each other but with different y-values but I guess it is possible as my tests gives values as expected
I also have to this more than a million times so speed is an important factor so therefore I was looking for a vectorized form but I guess this comes pretty close so thanks

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Walter Roberson
Walter Roberson 2015-5-27
Search the data for 0's. Suppose you find it at index J. Then introduce a new timepoint just after t(J-1), at time t(J-1)*(1+eps), with value 0, and introduce another just before t(J+1), at time t(J+1)*(1-eps), again with value 0. Now that I think of it, you might as well change the time t(J) to be t(J-1)*(1+eps) to move the 0 to right beside the previous point, and then you would only need to insert one new point.
Doing the stitching together in vectorized form could be a bit tricky. Mumble...
tidx = sort([(1:length(t)).', find(datapoints(:)==0)]);
newt = t(tidx);
newdata = datapoints(tidx);
new_at = find(diff(tidx)==0);
newt(new_at) = newt(new_at-1)*(1+eps);
newt(new_at+1) = newt(new_at+2)*(1-eps);
newdata(new_at) = 0;
newdata(new_at+1) = 0;
There, that should be pretty close.
  1 个评论
Ingrid
Ingrid 2015-5-27
编辑：Ingrid 2015-5-27
indeed I was having problems with vectorizing my code as with a for-loop it would be straightforward so thank you for your contribution. However, Thorsten's code was approximately two times faster which in my case where I have to repeat this a million times, is the crucial factor, I have accepted his answer
Elapsed time is 0.000208 seconds. -> walter
Elapsed time is 0.000109 seconds. -> thorsten
oh, and your code works and gives the same final answer, AFTER correction of the typo (just in case someone else want to try it):
tidx = sort([(1:length(t)).'; find(datapoints(:)==0)]);

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