0

I need to integrate a certain function that I have specified as discrete values for discrete arguments (I want to count the area under the graph I get). I.e., from the earlier part of the code I have the literal:

args=[a1, a2, a3, a3]
valuses=[v1, v2, v3, v4]

where value v1 corresponds to a1, etc. If it's important, I have args set in advance with a specific discretization width, and I count values with a ready-made function.

I am attaching a figure. fig

And putting this function, which gave me a 'values' array, into integrate.quad() gives me an error:

IntegrationWarning: The maximum number of subdivisions (50) has been achieved. If increasing the limit yields no improvement it is advised to analyze 
the integrand in order to determine the difficulties.  If the position of a 
local difficulty can be determined (singularity, discontinuity) one will 
probably gain from splitting up the interval and calling the integrator on the subranges.  Perhaps a special-purpose integrator should be used.

How can I integrate this? I'm mulling over the scipy documentation, but I can't seem to put it together. Because, after all, args themselves are already discretized by a finite number.

MB-F
  • 22,770
  • 4
  • 61
  • 116
Lilla_mu
  • 1
  • 1
  • 2
    How do you call the integrate function? Can you upload a full testcase that we can reproduce? It shouldn't be hard. We need the input values and the code you use do the integration. – Fra93 Jul 25 '22 at 13:07
  • See https://stackoverflow.com/questions/13320262/calculating-the-area-under-a-curve-given-a-set-of-coordinates-without-knowing-t – Warren Weckesser Jul 25 '22 at 18:01

2 Answers2

1

Since you only have a discrete approximation of the function, integration reduces to summation.

As a simple approximation of an integral, try this:

midpoints = (values[:-1] + values[1:]) / 2
steps = np.diff(args)
area = np.sum(midpoints * steps)

(Assuming args and values are numpy arrays and the function is value = f(arg).)

That approach sums the areas of all trapezoids between adjacent data points (Wikipedia).

MB-F
  • 22,770
  • 4
  • 61
  • 116
1

I am guessing that before passing the integral to quad you did some kind of interpolation on it. In general this is a misguided approach.

Integration and interpolation is very closely related. An integral requires you to compute the area under the curve and thus you must know the value of the function at any given point. Hence, starting from a set of data it is natural to want to interpolate it first. Yet the quad routine does not know that you started with a limited set of data, it just assumes that the function you gave it is "perfect" and it will do its best to compute the area under it! However the interpolated function is just a guess on what the values are between given points and thus integrating an interpolated function is a waste of time.

As MB-F said, in the discrete case you should simply sum up the points while multiplying them by the step size between them. You can do this the naïve way by pretending that the function is just rectangles. Or you can do what MB-F suggested which pretend that all the data points are connected with straight lines. Going one step further is pretending that the line connecting the data points is smooth (often true for physical systems) and use simpson integration implemented by scipy

Simon Tartakovksy
  • 190
  • 7