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I have parsed HL7 file and have generated some values. So that now, I have series of values over time for different identifiers of OBX segment of HL7 file.

Now, as per requirement I want to apply following expressions over this timely generated data:

Slope: dv/dt

Acceleration: Change in slope over time

Integration

Bandpass: Filter function that will limit values in particular range

For slope calculation I have referred http://dynamicnotions.blogspot.in/2009/05/linear-regression-in-c.html. My small code will simply calculate slope of graph generated for values of one object identifier of HL7 file(like HR, PVC). I just want to know how can I apply all remaining expressions. Please tell me if there are any libraries I can refer for this.

Ajinkya
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2 Answers2

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try these sites for integration and differentiation :

http://www.mathdotnet.com/

or

http://www.extremeoptimization.com/

  • Link-only answers are discouraged. If you can summarize the content you're linking to, it would go a long way toward improving this answer. – David Nov 08 '12 at 04:58
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See answer https://stackoverflow.com/a/4362748/380384 with cubic spline fit solution to this problem.

Once the cubic spline is defined from a set of points, then the slope and acceleration can be calculated for any point with the methods .GetYp() and .GetYpp(). Integration can also be made with .Integrate().

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John Alexiou
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  • Will check out and let you know. :) – Ajinkya Nov 05 '12 at 15:31
  • Here slope value I am getting for the data is different than value I am getting with function I have created (referencing URL specified in question) for the same data. Here it requires one particular value as a parameter for calculating slope. – Ajinkya Nov 06 '12 at 10:17
  • How different? Can you update your post with the new results and what you expect. The paramater for .`GetYp()` the independent variable (time) for which you want an answer for. – John Alexiou Nov 06 '12 at 20:24
  • I have given the link that I referred for calculating slope. I have used that function as it is. Slope getting calculated for that data is different than slope getting calculated by the method you have suggested. Is it giving slope value with respect to value of "t"? – Ajinkya Nov 08 '12 at 11:44