Draw a smooth polygon around the data points in the scatter plot, in matplotlib

Question

Draw a smooth polygon around the data points in the scatter plot, in matplotlib

I have a bunch of cross-plots with two datasets and was looking for a matploltib way to highlight their plotted areas with smooth polygon outlines.

For the moment, I'm just using Adobe Illustrator and changing the saved plot, but this is not perfect. Example: enter image description here

I would be grateful for any pointers / links to examples.

Greetings

+4

python matplotlib polygon scatter-plot polygons

user1665220 Jul 9 '13 at 16:07

source share

1 answer

Pablo · Accepted Answer · 2013-07-09T20:43:50+0000

Here you have an example. I wrote some basic ideas, but obviously you could do it better.

Brief Explanations:

1) You need to compute the convex hull ( http://en.wikipedia.org/wiki/Convex_hull )

2) With the case, you can scale it to save all your data inside.

3) You must interpolate the resulting curve.

The first part was done at http://wiki.scipy.org/Cookbook/Finding_Convex_Hull . The second is trivial. The third is very general, and you can execute any method, there are many different ways to do the same. I took the @Jaime approach ( Smooth spline representation of an arbitrary path, f (length) → x, y ), which, in my opinion, is a very good method.

Hope this helps you ...

#Taken from http://wiki.scipy.org/Cookbook/Finding_Convex_Hull import numpy as n, pylab as p, time def _angle_to_point(point, centre): '''calculate angle in 2-D between points and x axis''' delta = point - centre res = n.arctan(delta[1] / delta[0]) if delta[0] < 0: res += n.pi return res def _draw_triangle(p1, p2, p3, **kwargs): tmp = n.vstack((p1,p2,p3)) x,y = [x[0] for x in zip(tmp.transpose())] p.fill(x,y, **kwargs) def area_of_triangle(p1, p2, p3): '''calculate area of any triangle given co-ordinates of the corners''' return n.linalg.norm(n.cross((p2 - p1), (p3 - p1)))/2. def convex_hull(points, graphic=False, smidgen=0.0075): ''' Calculate subset of points that make a convex hull around points Recursively eliminates points that lie inside two neighbouring points until only convex hull is remaining. :Parameters: points : ndarray (2 xm) array of points for which to find hull graphic : bool use pylab to show progress? smidgen : float offset for graphic number labels - useful values depend on your data range :Returns: hull_points : ndarray (2 xn) convex hull surrounding points ''' if graphic: p.clf() p.plot(points[0], points[1], 'ro') n_pts = points.shape[1] assert(n_pts > 5) centre = points.mean(1) if graphic: p.plot((centre[0],),(centre[1],),'bo') angles = n.apply_along_axis(_angle_to_point, 0, points, centre) pts_ord = points[:,angles.argsort()] if graphic: for i in xrange(n_pts): p.text(pts_ord[0,i] + smidgen, pts_ord[1,i] + smidgen, \ '%d' % i) pts = [x[0] for x in zip(pts_ord.transpose())] prev_pts = len(pts) + 1 k = 0 while prev_pts > n_pts: prev_pts = n_pts n_pts = len(pts) if graphic: p.gca().patches = [] i = -2 while i < (n_pts - 2): Aij = area_of_triangle(centre, pts[i], pts[(i + 1) % n_pts]) Ajk = area_of_triangle(centre, pts[(i + 1) % n_pts], \ pts[(i + 2) % n_pts]) Aik = area_of_triangle(centre, pts[i], pts[(i + 2) % n_pts]) if graphic: _draw_triangle(centre, pts[i], pts[(i + 1) % n_pts], \ facecolor='blue', alpha = 0.2) _draw_triangle(centre, pts[(i + 1) % n_pts], \ pts[(i + 2) % n_pts], \ facecolor='green', alpha = 0.2) _draw_triangle(centre, pts[i], pts[(i + 2) % n_pts], \ facecolor='red', alpha = 0.2) if Aij + Ajk < Aik: if graphic: p.plot((pts[i + 1][0],),(pts[i + 1][1],),'go') del pts[i+1] i += 1 n_pts = len(pts) k += 1 return n.asarray(pts) if __name__ == "__main__": import scipy.interpolate as interpolate # fig = p.figure(figsize=(10,10)) theta = 2*n.pi*n.random.rand(1000) r = n.random.rand(1000)**0.5 x,y = r*p.cos(theta),r*p.sin(theta) points = n.ndarray((2,len(x))) points[0,:],points[1,:] = x,y scale = 1.03 hull_pts = scale*convex_hull(points) p.plot(x,y,'ko') x,y = [],[] convex = scale*hull_pts for point in convex: x.append(point[0]) y.append(point[1]) x.append(convex[0][0]) y.append(convex[0][1]) x,y = n.array(x),n.array(y) #Taken from /questions/442555/smooth-spline-representation-of-an-arbitrary-contour-flength-xy nt = n.linspace(0, 1, 100) t = n.zeros(x.shape) t[1:] = n.sqrt((x[1:] - x[:-1])**2 + (y[1:] - y[:-1])**2) t = n.cumsum(t) t /= t[-1] x2 = interpolate.spline(t, x, nt) y2 = interpolate.spline(t, y, nt) p.plot(x2, y2,'r--',linewidth=2) p.show()