在python中通过不规则网格集成2D数据

Question

So I have 2D function which is sampled irregularly over a domain, and I want to calculate the volume underneath the surface. The data is organised in terms of [x,y,z] , taking a simple example:

def f(x,y):
    return np.cos(10*x*y) * np.exp(-x**2 - y**2)

datrange1 = np.linspace(-5,5,1000)
datrange2 = np.linspace(-0.5,0.5,1000)

ar = []
for x in datrange1:
    for y in datrange2:
        ar += [[x,y, f(x,y)]]


for x in xrange2:
    for y in yrange2:
        ar += [[x,y, f(x,y)]] 

val_arr1 = np.array(ar)

data = np.unique(val_arr1)


xlist, ylist, zlist = data.T 

where np.unique sorts the data in the first column then the second. The data is arranged in this way as I need to sample more heavily around the origin as there is a sharp feature that must be resolved.

Now I wondered about constructing a 2D interpolating function using scipy.interpolate.interp2d , then integrating over this using dblquad . As it turns out, this is not only inelegant and slow, but also kicks out the error:

RuntimeWarning: No more knots can be added because the number of B-spline
coefficients already exceeds the number of data points m. 

Is there a better way to integrate data arranged in this fashion or overcoming this error?

Answer 1

If you can sample the data with high enough resolution around the feature of interest, then more sparsely everywhere else, the problem definition then becomes how to define the area under each sample. This is easy with regular rectangular samples, and could likely be done stepwise in increments of resolution around the origin. The approach I went after is to generate the 2D Voronoi cells for each sample in order to determine their area. I pulled most of the code from this answer, as it had almost all the components needed already.

import numpy as np
from scipy.spatial import Voronoi

#taken from: # https://stackoverflow.com/questions/28665491/getting-a-bounded-polygon-coordinates-from-voronoi-cells
#computes voronoi regions bounded by a bounding box
def square_voronoi(xy, bbox): #bbox: (min_x, max_x, min_y, max_y)
    # Select points inside the bounding box
    points_center = xy[np.where((bbox[0] <= xy[:,0]) * (xy[:,0] <= bbox[1]) * (bbox[2] <= xy[:,1]) * (bbox[2] <= bbox[3]))]
    # Mirror points
    points_left = np.copy(points_center)
    points_left[:, 0] = bbox[0] - (points_left[:, 0] - bbox[0])
    points_right = np.copy(points_center)
    points_right[:, 0] = bbox[1] + (bbox[1] - points_right[:, 0])
    points_down = np.copy(points_center)
    points_down[:, 1] = bbox[2] - (points_down[:, 1] - bbox[2])
    points_up = np.copy(points_center)
    points_up[:, 1] = bbox[3] + (bbox[3] - points_up[:, 1])
    points = np.concatenate((points_center, points_left, points_right, points_down, points_up,), axis=0)
    # Compute Voronoi
    vor = Voronoi(points)
    # Filter regions (center points should* be guaranteed to have a valid region)
    # center points should come first and not change in size
    regions = [vor.regions[vor.point_region[i]] for i in range(len(points_center))]
    vor.filtered_points = points_center
    vor.filtered_regions = regions
    return vor

#also stolen from: https://stackoverflow.com/questions/28665491/getting-a-bounded-polygon-coordinates-from-voronoi-cells
def area_region(vertices):
    # Polygon's signed area
    A = 0
    for i in range(0, len(vertices) - 1):
        s = (vertices[i, 0] * vertices[i + 1, 1] - vertices[i + 1, 0] * vertices[i, 1])
        A = A + s
    return np.abs(0.5 * A)

def f(x,y):
    return np.cos(10*x*y) * np.exp(-x**2 - y**2)

#sampling could easily be shaped to sample origin more heavily
sample_x = np.random.rand(1000) * 10 - 5 #same range as example linspace
sample_y = np.random.rand(1000) - .5
sample_xy = np.array([sample_x, sample_y]).T

vor = square_voronoi(sample_xy, (-5,5,-.5,.5)) #using bbox from samples
points = vor.filtered_points
sample_areas = np.array([area_region(vor.vertices[verts+[verts[0]],:]) for verts in vor.filtered_regions])
sample_z = np.array([f(p[0], p[1]) for p in points])

volume = np.sum(sample_z * sample_areas)

I haven't exactly tested this, but the principle should work, and the math checks out.