Effacer les filtres
Effacer les filtres

Discretise into Equal Intervals

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simply90
simply90 le 21 Jan 2014
Commenté : simply90 le 21 Jan 2014
I have an array of data, 2 columns and 19 rows of entries, which can be plotted as a loglog curve. However, the x-data I have does not have standard intervals and I need to approximate the curve as a piecewise linear function. Any suggestions as to how to go about this?
  6 commentaires
simply90
simply90 le 21 Jan 2014
Yes. Within x I was considering an interval of 0.07 to fit the data, as it isn't an increasing rate I couldn't make the function interp work as the interval was set, rather than a rate. Thanks so much for your help!
simply90
simply90 le 21 Jan 2014
It works! Thank you so much Bruno and dpb. I was able to start messing with the interp1 function, I was using the wrong one beforehand with your suggestion dpb, and Bruno your answer clarified it from there.
Thanks for your help!

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Bruno Pop-Stefanov
Bruno Pop-Stefanov le 21 Jan 2014
Modifié(e) : Bruno Pop-Stefanov le 21 Jan 2014
You can use several methods to interpolate data points from non-uniformly spaced points. There are listed here for one dimension:
Here is an example using interp1:
vq = interp1(c, v, xq)
where x are the (non-uniformly spaced) input points (the first row in your vector) with values v (the second row in your vector), and xq are the query points (your uniformly spaced points). The function returns the interpolated values at xq in the vector vq.
For uniformly-spaced query points, you can use:
x_begin = x(1,1); % your 0.005
x_end = x(end,1); % your 2.13
numPoints = numel(x(:,1)); % number of points in your original vector; you can change that
xq = linspace(x_begin, x_end, numPoints);
for numPoints linearly spaced between x_begin and x_end. If you prefer to specify the length of each interval instead of the number of points, you could write:
xq = x_begin : int_length : x_end;
with int_length = 0.07.

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