Making a non-overlapping bubble chart
Asked Answered
A

1

9

I am currently trying to make a bubble chart in Matplotlib where the bubbles do not overlap, so packing the circles/bubbles in the chart, approximately like

enter image description here

What I thought might work:

  • Plot the first data-point using x = 1, y = 1
  • Randomly plotting the other data-points by calculating the radius of the bubble given the scalar value as to avoid overlapping
Admissive answered 9/9, 2017 at 14:12 Comment(0)
M
12

The following would be a brute-force approach.
You can first place all circles on a grid, with a gridspacing as large as twice the maximum radius of any of the circles.
enter image description here

Then you let the circles do a random walk and check in each step if the "potential energy" of the bunch of cicles has become smaller and if the positions obtained are valid (i.e. no overlaps).

if (e < self.E and self.isvalid(i)):

As a "potential" we can simply use a square radial function.

self.p = lambda x,y: np.sum((x**2+y**2)**2)

The code:

import numpy as np
import matplotlib.pyplot as plt

# create 10 circles with different radii
r = np.random.randint(5,15, size=10)

class C():
    def __init__(self,r):
        self.N = len(r)
        self.x = np.ones((self.N,3))
        self.x[:,2] = r
        maxstep = 2*self.x[:,2].max()
        length = np.ceil(np.sqrt(self.N))
        grid = np.arange(0,length*maxstep,maxstep)
        gx,gy = np.meshgrid(grid,grid)
        self.x[:,0] = gx.flatten()[:self.N]
        self.x[:,1] = gy.flatten()[:self.N]
        self.x[:,:2] = self.x[:,:2] - np.mean(self.x[:,:2], axis=0)

        self.step = self.x[:,2].min()
        self.p = lambda x,y: np.sum((x**2+y**2)**2)
        self.E = self.energy()
        self.iter = 1.

    def minimize(self):
        while self.iter < 1000*self.N:
            for i in range(self.N):
                rand = np.random.randn(2)*self.step/self.iter
                self.x[i,:2] += rand
                e = self.energy()
                if (e < self.E and self.isvalid(i)):
                    self.E = e
                    self.iter = 1.
                else:
                    self.x[i,:2] -= rand
                    self.iter += 1.

    def energy(self):
        return self.p(self.x[:,0], self.x[:,1])

    def distance(self,x1,x2):
        return np.sqrt((x1[0]-x2[0])**2+(x1[1]-x2[1])**2)-x1[2]-x2[2]

    def isvalid(self, i):
        for j in range(self.N):
            if i!=j: 
                if self.distance(self.x[i,:], self.x[j,:]) < 0:
                    return False
        return True

    def plot(self, ax):
        for i in range(self.N):
            circ = plt.Circle(self.x[i,:2],self.x[i,2] )
            ax.add_patch(circ)

c = C(r)

fig, ax = plt.subplots(subplot_kw=dict(aspect="equal"))
ax.axis("off")

c.minimize()

c.plot(ax)
ax.relim()
ax.autoscale_view()
plt.show()

enter image description here

Because of the random walk nature of this, finding the solution will take a little time (~10 seconds in this case); you may of course play with the parameters (mainly the number of steps 1000*self.N until a solution is fixed) and see what suits your needs.

Mattheus answered 9/9, 2017 at 18:12 Comment(0)

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