I am trying to write a code that, for a given list of circles (list1), it is able to find the positions for new circles (list2). list1 and list2 have the same length, because for each circle in list1 there must be a circle from list2.
Each pair of circles (let's say circle1 from list1 and circle2 from list2), must be as close together as possible,
circles from list2 must not overlap with circles from list1, while circles of the single lists can overlap each other.
list1 is fixed, so now I have to find the right position for circles from list2.
I wrote this simple function to recognize if 2 circles overlap:
def overlap(x1, y1, x2, y2, r1, r2):
distSq = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)
radSumSq = (r1 + r2) * (r1 + r2)
if (distSq >= radSumSq):
return False # no overlap
else:
return True #overlap
and this is the list1:
with:
x=[14.11450195 14.14184093 14.15435028 14.16206741 14.16951752 14.17171097
14.18569565 14.19700241 14.23129082 14.24083233 14.24290752 14.24968338
14.2518959 14.26536751 14.27209759 14.27612877 14.2904377 14.29187012
14.29409599 14.29618549 14.30615044 14.31624985 14.3206892 14.3228569
14.36143875 14.36351967 14.36470699 14.36697292 14.37235737 14.41422081
14.42583466 14.43226814 14.43319225 14.4437027 14.4557848 14.46592999
14.47036076 14.47452068 14.47815609 14.52229309 14.53059006 14.53404236
14.5411644 ]
y=[-0.35319126 -0.44222349 -0.44763246 -0.35669261 -0.24366629 -0.3998799
-0.38940558 -0.57744932 -0.45223859 -0.21021004 -0.44250247 -0.45866323
-0.47203487 -0.51684451 -0.44884869 -0.2018993 -0.40296811 -0.23641759
-0.18019417 -0.33391538 -0.53565156 -0.45215255 -0.40939832 -0.26936951
-0.30894437 -0.55504167 -0.47177047 -0.45573688 -0.43100587 -0.5805912
-0.21770373 -0.199422 -0.17372169 -0.38522363 -0.56950212 -0.56947368
-0.48770753 -0.24940367 -0.31492445 -0.54263926 -0.53460872 -0.4053807
-0.43733299]
radius = 0.014
Copy and pasteable...
x = [14.11450195,14.14184093,14.15435028,14.16206741,14.16951752,
14.17171097,14.18569565,14.19700241,14.23129082,14.24083233,
14.24290752,14.24968338,14.2518959,14.26536751,14.27209759,
14.27612877,14.2904377,14.29187012,14.29409599,14.29618549,
14.30615044,14.31624985,14.3206892,14.3228569,14.36143875,
14.36351967,14.36470699,14.36697292,14.37235737,14.41422081,
14.42583466,14.43226814,14.43319225,14.4437027,14.4557848,
14.46592999,14.47036076,14.47452068,14.47815609,14.52229309,
14.53059006,14.53404236,14.5411644]
y = [-0.35319126,-0.44222349,-0.44763246,-0.35669261,-0.24366629,
-0.3998799,-0.38940558,-0.57744932,-0.45223859,-0.21021004,
-0.44250247,-0.45866323,-0.47203487,-0.51684451,-0.44884869,
-0.2018993,-0.40296811,-0.23641759,-0.18019417,-0.33391538,
-0.53565156,-0.45215255,-0.40939832,-0.26936951,-0.30894437,
-0.55504167,-0.47177047,-0.45573688,-0.43100587,-0.5805912,
-0.21770373,-0.199422,-0.17372169,-0.38522363,-0.56950212,
-0.56947368,-0.48770753,-0.24940367,-0.31492445,-0.54263926,
-0.53460872,-0.4053807,-0.43733299]
Now I am not sure about what I have to do, my first idea is to draw circles of list2 taking x and y from list one and do something like x+c and y+c, where c is a fixed value. Then I can call my overlapping function and, if there is overlap I can increase the c value.
In this way I have 2 for loops. Now, my questions are:
There is a way to avoid for loops?
Is there a smart solution to find a neighbor (circle from list2) for each circle from list1 (without overlaps with other circles from list2)?
Using numpy arrays, you can avoid for loops.
Setup from your example.
import numpy as np
#Using your x and y
c1 = np.array([x,y]).T
# random set of other centers within the same range as c1
c2 = np.random.random((10,2))
np.multiply(c2, c1.max(0)-c1.min(0),out = c2)
np.add(c2, c1.min(0), out=c2)
radius = 0.014
r = radius
min_d = (2*r)*(2*r)
plot_circles(c1,c2) # see function at end
An array of distances from each center in c1 to each center in c2
def dist(c1,c2):
dx = c1[:,0,None] - c2[:,0]
dy = c1[:,1,None] - c2[:,1]
return dx*dx + dy*dy
d = dist(c1,c2)
Or you could use scipy.spatial
from scipy.spatial import distance
d = distance.cdist(c1,c2,'sqeuclidean')
Create a 2d Boolean array for circles that intersect.
intersect = d <= min_d
Find the indices of overlapping circles from the two sets.
a,b = np.where(intersect)
plot_circles(c1[a],c2[b])
Using intersect or a and b to index c1,c2, and d you should be able to get groups of intersecting circles then figure out how to move the c2 centers - but I'll leave that for another question/answer. If a list2 circle intersects one list1 circle - find the line between the two and move along that line. If a list2 circle intersects more than one list1 circle - find the line between the two closestlist1circles and move thelitst2` circle along a line perpendicular to that. You didn't mention any constraints on moving the circles so maybe random movement then find the intersects again but that might be problematic. In the following image, it may be trivial to figure out how to move most of the red circles but the group circled in blue might require a different strategy.
Here are some examples for getting groups:
>>> for f,g,h in zip(c1[a],c2[b],d[a,b]):
print(f,g,h)
>>> c1[intersect.any(1)],c2[intersect.any(0)]
>>> for (f,g) in zip(c2,intersect.T):
if g.any():
print(f.tolist(),c1[g].tolist())
import matplotlib as mpl
from matplotlib import pyplot as plt
def plot_circles(c1,c2):
bounds = np.array([c1.min(0),c2.min(0),c1.max(0),c2.max(0)])
xmin, ymin = bounds.min(0)
xmax, ymax = bounds.max(0)
circles1 = [mpl.patches.Circle(xy,radius=r,fill=False,edgecolor='g') for xy in c1]
circles2 = [mpl.patches.Circle(xy,radius=r,fill=False,edgecolor='r') for xy in c2]
fig = plt.figure()
ax = fig.add_subplot(111)
for c in circles2:
ax.add_artist(c)
for c in circles1:
ax.add_artist(c)
ax.set_xlim(xmin-r,xmax+r)
ax.set_ylim(ymin-r,ymax+r)
plt.show()
plt.close()
This problem can very well be seen as an optimization problem. To be more precise, a nonlinear optimization problem with constraints.
Since optimization strategies are not always so easy to understand, I will define the problem as simply as possible and also choose an approach that is as general as possible (but less efficient) and does not involve a lot of mathematics. As a spoiler: We are going to formulate the problem and the minimization process in less than 10 lines of code using the scipy library.
However, I will still provide hints on where you can get your hands even dirtier.
Formulating the problem
As a guide for a formulation of an NLP-class problem (Nonlinear Programming), you can go directly to the two requirements in the original post.
Each pair of circles must be as close together as possible -> Hint for a cost-function
Circles must not overlap with other (moved) circles -> Hint for a constraint
Cost function
Let's start with the formulation of the cost function to be minimized.
Since the circles should be moved as little as possible (resulting in the closest possible neighborhood), a quadratic penalty term for the distances between the circles of the two lists can be chosen for the cost function:
import scipy.spatial.distance as sd
def cost_function(new_positions, old_positions):
new_positions = np.reshape(new_positions, (-1, 2))
return np.trace(sd.cdist(new_positions, old_positions, metric='sqeuclidean'))
Why quadratic? Partly because of differentiability and for stochastic reasons (think of the circles as normally distributed measurement errors -> least squares is then a maximum likelihood estimator). By exploiting the structure of the cost function, the efficiency of the optimization can be increased (elimination of sqrt). By the way, this problem is related to nonlinear regression, where (nonlinear) least squares are also used.
Now that we have a cost function at hand, we also have a good way to evaluate our optimization. To be able to compare solutions of different optimization strategies, we simply pass the newly calculated positions to the cost function.
Let's give it a try: For example, let us use the calculated positions from the Voronoi approach (by Paul Brodersen).
print(cost_function(new_positions, old_positions))
# prints 0.007999244511697411
That's a pretty good value if you ask me. Considering that the cost function spits out zero when there is no displacement at all, this cost is pretty close. We can now try to outperform this value by using classical optimization!
Non-linear constraint
We know that circles must not overlap with other circles in the new set. If we translate this into a constraint, we find that the lower bound for the distance is 2 times the radius and the upper bound is simply infinity.
import scipy.spatial.distance as sd
from scipy.optimize import NonlinearConstraint
def cons_f(x):
x = np.reshape(x, (-1, 2))
return sd.pdist(x)
nonlinear_constraint = NonlinearConstraint(cons_f, 2*radius, np.inf, jac='2-point')
Here we make life easy by approximating the Jacobi matrix via finite differences (see parameter jac='2-point'). At this point it should be said that we can increase the efficiency here, by formulating the derivatives of the first and second order ourselves instead of using approximations. But this is left to the interested reader. (It is not that hard, because we use quite simple mathematical expressions for distance calculation here.)
One additional note: You can also set a boundary constraint for the positions themselves not to exceed a specified region. This can then be used as another parameter. (See scipy.optimize.Bounds)
Minimizing the cost function under constraints
Now we have both ingredients, the cost function and the constraint, in place. Now let's minimize the whole thing!
from scipy.optimize import minimize
res = minimize(lambda x: cost_function(x, positions), positions.flatten(), method='SLSQP',
jac="2-point", constraints=[nonlinear_constraint])
As you can see, we approximate the first derivatives here as well. You can also go deeper here and set up the derivatives yourself (analytically).
Also note that we must always pass the parameters (an nx2 vector specifying the positions of the new layout for n circles) as a flat vector. For this reason, reshaping can be found several times in the code.
Evaluation, summary and visualization
Let's see how the optimization result performs in our cost function:
new_positions = np.reshape(res.x, (-1,2))
print(cost_function(new_positions, old_positions))
# prints 0.0010314079483565686
Starting from the Voronoi approach, we actually reduced the cost by another 87%! Thanks to the power of modern optimization strategies, we can solve a lot of problems in no time.
Of course, it would be interesting to see how the shifted circles look now:
Circles after Optimization
Performance: 77.1 ms ± 1.17 ms
The entire code:
from scipy.optimize import minimize
import scipy.spatial.distance as sd
from scipy.optimize import NonlinearConstraint
# Given by original post
positions = np.array([x, y]).T
def cost_function(new_positions, old_positions):
new_positions = np.reshape(new_positions, (-1, 2))
return np.trace(sd.cdist(new_positions, old_positions, metric='sqeuclidean'))
def cons_f(x):
x = np.reshape(x, (-1, 2))
return sd.pdist(x)
nonlinear_constraint = NonlinearConstraint(cons_f, 2*radius, np.inf, jac='2-point')
res = minimize(lambda x: cost_function(x, positions), positions.flatten(), method='SLSQP',
jac="2-point", constraints=[nonlinear_constraint])
One solution could be to follow the gradient of the unwanted spacing between each circle, though maybe there is a better way. This approach has a few parameters to tune and takes some time to run.
import matplotlib.pyplot as plt
from scipy.optimize import minimize as mini
import numpy as np
from scipy.optimize import approx_fprime
x = np.array([14.11450195,14.14184093,14.15435028,14.16206741,14.16951752,
14.17171097,14.18569565,14.19700241,14.23129082,14.24083233,
14.24290752,14.24968338,14.2518959,14.26536751,14.27209759,
14.27612877,14.2904377,14.29187012,14.29409599,14.29618549,
14.30615044,14.31624985,14.3206892,14.3228569,14.36143875,
14.36351967,14.36470699,14.36697292,14.37235737,14.41422081,
14.42583466,14.43226814,14.43319225,14.4437027,14.4557848,
14.46592999,14.47036076,14.47452068,14.47815609,14.52229309,
14.53059006,14.53404236,14.5411644])
y = np.array([-0.35319126,-0.44222349,-0.44763246,-0.35669261,-0.24366629,
-0.3998799,-0.38940558,-0.57744932,-0.45223859,-0.21021004,
-0.44250247,-0.45866323,-0.47203487,-0.51684451,-0.44884869,
-0.2018993,-0.40296811,-0.23641759,-0.18019417,-0.33391538,
-0.53565156,-0.45215255,-0.40939832,-0.26936951,-0.30894437,
-0.55504167,-0.47177047,-0.45573688,-0.43100587,-0.5805912,
-0.21770373,-0.199422,-0.17372169,-0.38522363,-0.56950212,
-0.56947368,-0.48770753,-0.24940367,-0.31492445,-0.54263926,
-0.53460872,-0.4053807,-0.43733299])
radius = 0.014
x0, y0 = (x, y)
def plot_circles(x, y, name='initial'):
fig, ax = plt.subplots()
for ii in range(x.size):
ax.add_patch(plt.Circle((x[ii], y[ii]), radius, color='b', fill=False))
ax.set_xlim(x.min() - radius, x.max() + radius)
ax.set_ylim(y.min() - radius, y.max() + radius)
fig.savefig(name)
plt.clf()
def spacing(s):
x, y = np.split(s, 2)
dX, dY = [np.subtract(*np.meshgrid(xy, xy, indexing='ij')).T
for xy in [x, y]]
dXY2 = dX**2 + dY**2
return np.minimum(dXY2[np.triu_indices(x.size, 1)] - (2 * radius) ** 2, 0).sum()
plot_circles(x, y)
def spacingJ(s):
return approx_fprime(s, spacing, 1e-8)
s = np.append(x, y)
for ii in range(50):
j = spacingJ(s)
if j.sum() == 0: break
s += .01 * j
x_new, y_new = np.split(s, 2)
plot_circles(x_new, y_new, 'new%i' % ii)
plot_circles(x_new, y_new, 'new%i' % ii)
https://giphy.com/gifs/x0lWDLZBz5O3gWTbLa
This answer implements a variation of the Lloyds algorithm. The basic idea is to compute the Voronoi diagram for your points / circles. This assigns each point a cell, which is a region that includes the point and which has a center that is maximally far away from all other points.
In the original algorithm, we would move each point towards the center of its Voronoi cell. Over time, this results in an even spread of points, as illustrated here.
In this variant, we only move points that overlap another point.
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial import Voronoi
from scipy.spatial.distance import cdist
def remove_overlaps(positions, radii, tolerance=1e-6):
"""Use a variation of Lloyds algorithm to move circles apart from each other until none overlap.
Parameters
----------
positions : array
The (x, y) coordinates of the circle origins.
radii : array
The radii for each circle.
tolerance : float
If all circles overlap less than this threshold, the computation stops.
Higher values leads to faster convergence.
Returns
-------
new_positions : array
The (x, y) coordinates of the circle origins.
See also
--------
https://en.wikipedia.org/wiki/Lloyd%27s_algorithm
"""
positions = np.array(positions)
radii = np.array(radii)
minimum_distances = radii[np.newaxis, :] + radii[:, np.newaxis]
minimum_distances[np.diag_indices_from(minimum_distances)] = 0 # ignore distances to self
# Initialize the first loop.
distances = cdist(positions, positions)
displacements = np.max(np.clip(minimum_distances - distances, 0, None), axis=-1)
while np.any(displacements > tolerance):
centroids = _get_voronoi_centroids(positions)
# Compute the direction from each point towards its corresponding Voronoi centroid.
deltas = centroids - positions
magnitudes = np.linalg.norm(deltas, axis=-1)
directions = deltas / magnitudes[:, np.newaxis]
# Mask NaNs that arise if the magnitude is zero, i.e. the point is already center of the Voronoi cell.
directions[np.isnan(directions)] = 0
# Step into the direction of the centroid.
# Clipping prevents overshooting of the centroid when stepping into the direction of the centroid.
# We step by half the displacement as the other overlapping point will be moved in approximately the opposite direction.
positions = positions + np.clip(0.5 * displacements, None, magnitudes)[:, np.newaxis] * directions
# Initialize next loop.
distances = cdist(positions, positions)
displacements = np.max(np.clip(minimum_distances - distances, 0, None), axis=-1)
return positions
def _get_voronoi_centroids(positions):
"""Construct a Voronoi diagram from the given positions and determine the center of each cell."""
voronoi = Voronoi(positions)
centroids = np.zeros_like(positions)
for ii, idx in enumerate(voronoi.point_region):
region = [jj for jj in voronoi.regions[idx] if jj != -1] # i.e. ignore points at infinity; TODO: compute correctly clipped regions
centroids[ii] = np.mean(voronoi.vertices[region], axis=0)
return centroids
if __name__ == '__main__':
x = np.array([14.11450195,14.14184093,14.15435028,14.16206741,14.16951752,
14.17171097,14.18569565,14.19700241,14.23129082,14.24083233,
14.24290752,14.24968338,14.2518959,14.26536751,14.27209759,
14.27612877,14.2904377,14.29187012,14.29409599,14.29618549,
14.30615044,14.31624985,14.3206892,14.3228569,14.36143875,
14.36351967,14.36470699,14.36697292,14.37235737,14.41422081,
14.42583466,14.43226814,14.43319225,14.4437027,14.4557848,
14.46592999,14.47036076,14.47452068,14.47815609,14.52229309,
14.53059006,14.53404236,14.5411644])
y = np.array([-0.35319126,-0.44222349,-0.44763246,-0.35669261,-0.24366629,
-0.3998799,-0.38940558,-0.57744932,-0.45223859,-0.21021004,
-0.44250247,-0.45866323,-0.47203487,-0.51684451,-0.44884869,
-0.2018993,-0.40296811,-0.23641759,-0.18019417,-0.33391538,
-0.53565156,-0.45215255,-0.40939832,-0.26936951,-0.30894437,
-0.55504167,-0.47177047,-0.45573688,-0.43100587,-0.5805912,
-0.21770373,-0.199422,-0.17372169,-0.38522363,-0.56950212,
-0.56947368,-0.48770753,-0.24940367,-0.31492445,-0.54263926,
-0.53460872,-0.4053807,-0.43733299])
radius = 0.014
positions = np.c_[x, y]
radii = np.full(len(positions), radius)
fig, axes = plt.subplots(1, 2, sharex=True, sharey=True, figsize=(14, 7))
for position, radius in zip(positions, radii):
axes[0].add_patch(plt.Circle(position, radius, fill=False))
axes[0].set_xlim(x.min() - radius, x.max() + radius)
axes[0].set_ylim(y.min() - radius, y.max() + radius)
axes[0].set_aspect('equal')
new_positions = remove_overlaps(positions, radii)
for position, radius in zip(new_positions, radii):
axes[1].add_patch(plt.Circle(position, radius, fill=False))
for ax in axes.ravel():
ax.set_aspect('equal')
plt.show()
I use matplotlib 1.15.1 and I try to generate scattergrams like this:
The ellipses have fixes size and are drawn with center coordinates, width, height and angle (provided from outside): I have no idea what their equotions are.
g_ell_center = (0.8882, 0.8882)
g_ell_width = 0.36401857095483
g_ell_height = 0.16928136341606
g_ellipse = patches.Ellipse(g_ell_center, g_ell_width, g_ell_height, angle=angle, fill=False, edgecolor='green', linewidth=2)
This ellipses should mark normal and semi-normal data on my plot.
Then, I have an array of ~500 points which must be colored according to ellipse they belong to. So I tried to check each point with contains_point method:
colors_array = []
colors_scheme = ['green', 'yellow', 'black']
for point in points_array:
if g_ellipse.contains_point(point, radius=0):
colors_array.append(0)
elif y_ellipse.contains_point(point, radius=0):
colors_array.append(1)
else:
colors_array.append(2)
Finally, points are drawn:
plt.scatter(x_array, y_array, s=10, c=[colors_scheme[x] for x in colors_array], edgecolor="k", linewidths=0.3)
But contains_point is extremely slow! It worked for 5 minutes for 300-points scattergram, and I have to generate thousands of them in parallel. Maybe there's faster approach?
P.S. Whole project is bound to matplotlib, I can't use other libraries.
This approach should test if a point is within an ellipse, given the ellipse's centre, width, height and angle. You find the point's x and y coordinates relative to the ellipse centre, then transform those using the angle to be the coordinates along the major and minor axes. Finally, you find the normalised distance of the point from the cell centre, where a distance of 1 would be on the ellipse, less than 1 is inside, and more than 1 is outside.
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import numpy as np
fig,ax = plt.subplots(1)
ax.set_aspect('equal')
# Some test points
x = np.random.rand(500)*0.5+0.7
y = np.random.rand(500)*0.5+0.7
# The ellipse
g_ell_center = (0.8882, 0.8882)
g_ell_width = 0.36401857095483
g_ell_height = 0.16928136341606
angle = 30.
g_ellipse = patches.Ellipse(g_ell_center, g_ell_width, g_ell_height, angle=angle, fill=False, edgecolor='green', linewidth=2)
ax.add_patch(g_ellipse)
cos_angle = np.cos(np.radians(180.-angle))
sin_angle = np.sin(np.radians(180.-angle))
xc = x - g_ell_center[0]
yc = y - g_ell_center[1]
xct = xc * cos_angle - yc * sin_angle
yct = xc * sin_angle + yc * cos_angle
rad_cc = (xct**2/(g_ell_width/2.)**2) + (yct**2/(g_ell_height/2.)**2)
# Set the colors. Black if outside the ellipse, green if inside
colors_array = np.array(['black'] * len(rad_cc))
colors_array[np.where(rad_cc <= 1.)[0]] = 'green'
ax.scatter(x,y,c=colors_array,linewidths=0.3)
plt.show()
Note, this whole script takes 0.6 seconds to run and process 500 points. That includes creating and saving the figure, etc.
The process of setting the colors_array using the np.where method above takes 0.00007s for 500 points.
Note, in an older implementation shown below, setting the colors_array in a loop took 0.00016 s:
colors_array = []
for r in rad_cc:
if r <= 1.:
# point in ellipse
colors_array.append('green')
else:
# point not in ellipse
colors_array.append('black')
Your current implementation should only be calling contains_point 25,000 to 50,000 times, which isn't a lot. So, I'm guessing that the implementation of contains_point is targeted toward precision rather than speed.
Since you have a distribution of points where only a small percentage will be in any given ellipse, and therefore most will rarely be anywhere near any given ellipse, you can easily use rectangular coordinates as a short-cut to figure out whether the point is close enough to the ellipse to be worth calling contains_point.
Compute the left and right x coordinates and top and bottom y coordinates of the ellipse, possibly with a bit of padding to account for rendering differences, then check if the point is within those, such as the following pseudo-code:
if point.x >= ellipse_left and point.x <= ellipse_right and _
point.y >= ellipse_top and point.y <= ellipse_bottom:
if ellipse.contains_point(point, radius=0):
... use the contained point here
This approach eliminates expensive calculations for most of the points, allowing simple comparisons instead to rule out the obvious mismatches, while preserving the accuracy of the computations where the point is close enough that it might be in the ellipse. If e.g. only 1% of your points are anywhere near a given ellipse, this approach will eliminate 99% of your calls to contains_point and instead replace them with much faster comparisons.
I'm trying to come up with an algorithm that will determine turning points in a trajectory of x/y coordinates. The following figures illustrates what I mean: green indicates the starting point and red the final point of the trajectory (the entire trajectory consists of ~ 1500 points):
In the following figure, I added by hand the possible (global) turning points that an algorithm could return:
Obviously, the true turning point is always debatable and will depend on the angle that one specifies that has to lie between points. Furthermore a turning point can be defined on a global scale (what I tried to do with the black circles), but could also be defined on a high-resolution local scale. I'm interested in the global (overall) direction changes, but I'd love to see a discussion on the different approaches that one would use to tease apart global vs local solutions.
What I've tried so far:
calculate distance between subsequent points
calculate angle between subsequent points
look how distance / angle changes between subsequent points
Unfortunately this doesn't give me any robust results. I probably have too calculate the curvature along multiple points, but that's just an idea.
I'd really appreciate any algorithms / ideas that might help me here. The code can be in any programming language, matlab or python are preferred.
EDIT here's the raw data (in case somebody want's to play with it):
mat file
text file (x coordinate first, y coordinate in second line)
You could use the Ramer-Douglas-Peucker (RDP) algorithm to simplify the path. Then you could compute the change in directions along each segment of the simplified path. The points corresponding to the greatest change in direction could be called the turning points:
A Python implementation of the RDP algorithm can be found on github.
import matplotlib.pyplot as plt
import numpy as np
import os
import rdp
def angle(dir):
"""
Returns the angles between vectors.
Parameters:
dir is a 2D-array of shape (N,M) representing N vectors in M-dimensional space.
The return value is a 1D-array of values of shape (N-1,), with each value
between 0 and pi.
0 implies the vectors point in the same direction
pi/2 implies the vectors are orthogonal
pi implies the vectors point in opposite directions
"""
dir2 = dir[1:]
dir1 = dir[:-1]
return np.arccos((dir1*dir2).sum(axis=1)/(
np.sqrt((dir1**2).sum(axis=1)*(dir2**2).sum(axis=1))))
tolerance = 70
min_angle = np.pi*0.22
filename = os.path.expanduser('~/tmp/bla.data')
points = np.genfromtxt(filename).T
print(len(points))
x, y = points.T
# Use the Ramer-Douglas-Peucker algorithm to simplify the path
# http://en.wikipedia.org/wiki/Ramer-Douglas-Peucker_algorithm
# Python implementation: https://github.com/sebleier/RDP/
simplified = np.array(rdp.rdp(points.tolist(), tolerance))
print(len(simplified))
sx, sy = simplified.T
# compute the direction vectors on the simplified curve
directions = np.diff(simplified, axis=0)
theta = angle(directions)
# Select the index of the points with the greatest theta
# Large theta is associated with greatest change in direction.
idx = np.where(theta>min_angle)[0]+1
fig = plt.figure()
ax =fig.add_subplot(111)
ax.plot(x, y, 'b-', label='original path')
ax.plot(sx, sy, 'g--', label='simplified path')
ax.plot(sx[idx], sy[idx], 'ro', markersize = 10, label='turning points')
ax.invert_yaxis()
plt.legend(loc='best')
plt.show()
Two parameters were used above:
The RDP algorithm takes one parameter, the tolerance, which
represents the maximum distance the simplified path
can stray from the original path. The larger the tolerance, the cruder the simplified path.
The other parameter is the min_angle which defines what is considered a turning point. (I'm taking a turning point to be any point on the original path, whose angle between the entering and exiting vectors on the simplified path is greater than min_angle).
I will be giving numpy/scipy code below, as I have almost no Matlab experience.
If your curve is smooth enough, you could identify your turning points as those of highest curvature. Taking the point index number as the curve parameter, and a central differences scheme, you can compute the curvature with the following code
import numpy as np
import matplotlib.pyplot as plt
import scipy.ndimage
def first_derivative(x) :
return x[2:] - x[0:-2]
def second_derivative(x) :
return x[2:] - 2 * x[1:-1] + x[:-2]
def curvature(x, y) :
x_1 = first_derivative(x)
x_2 = second_derivative(x)
y_1 = first_derivative(y)
y_2 = second_derivative(y)
return np.abs(x_1 * y_2 - y_1 * x_2) / np.sqrt((x_1**2 + y_1**2)**3)
You will probably want to smooth your curve out first, then calculate the curvature, then identify the highest curvature points. The following function does just that:
def plot_turning_points(x, y, turning_points=10, smoothing_radius=3,
cluster_radius=10) :
if smoothing_radius :
weights = np.ones(2 * smoothing_radius + 1)
new_x = scipy.ndimage.convolve1d(x, weights, mode='constant', cval=0.0)
new_x = new_x[smoothing_radius:-smoothing_radius] / np.sum(weights)
new_y = scipy.ndimage.convolve1d(y, weights, mode='constant', cval=0.0)
new_y = new_y[smoothing_radius:-smoothing_radius] / np.sum(weights)
else :
new_x, new_y = x, y
k = curvature(new_x, new_y)
turn_point_idx = np.argsort(k)[::-1]
t_points = []
while len(t_points) < turning_points and len(turn_point_idx) > 0:
t_points += [turn_point_idx[0]]
idx = np.abs(turn_point_idx - turn_point_idx[0]) > cluster_radius
turn_point_idx = turn_point_idx[idx]
t_points = np.array(t_points)
t_points += smoothing_radius + 1
plt.plot(x,y, 'k-')
plt.plot(new_x, new_y, 'r-')
plt.plot(x[t_points], y[t_points], 'o')
plt.show()
Some explaining is in order:
turning_points is the number of points you want to identify
smoothing_radius is the radius of a smoothing convolution to be applied to your data before computing the curvature
cluster_radius is the distance from a point of high curvature selected as a turning point where no other point should be considered as a candidate.
You may have to play around with the parameters a little, but I got something like this:
>>> x, y = np.genfromtxt('bla.data')
>>> plot_turning_points(x, y, turning_points=20, smoothing_radius=15,
... cluster_radius=75)
Probably not good enough for a fully automated detection, but it's pretty close to what you wanted.
A very interesting question. Here is my solution, that allows for variable resolution. Although, fine-tuning it may not be simple, as it's mostly intended to narrow down
Every k points, calculate the convex hull and store it as a set. Go through the at most k points and remove any points that are not in the convex hull, in such a way that the points don't lose their original order.
The purpose here is that the convex hull will act as a filter, removing all of "unimportant points" leaving only the extreme points. Of course, if the k-value is too high, you'll end up with something too close to the actual convex hull, instead of what you actually want.
This should start with a small k, at least 4, then increase it until you get what you seek. You should also probably only include the middle point for every 3 points where the angle is below a certain amount, d. This would ensure that all of the turns are at least d degrees (not implemented in code below). However, this should probably be done incrementally to avoid loss of information, same as increasing the k-value. Another possible improvement would be to actually re-run with points that were removed, and and only remove points that were not in both convex hulls, though this requires a higher minimum k-value of at least 8.
The following code seems to work fairly well, but could still use improvements for efficiency and noise removal. It's also rather inelegant in determining when it should stop, thus the code really only works (as it stands) from around k=4 to k=14.
def convex_filter(points,k):
new_points = []
for pts in (points[i:i + k] for i in xrange(0, len(points), k)):
hull = set(convex_hull(pts))
for point in pts:
if point in hull:
new_points.append(point)
return new_points
# How the points are obtained is a minor point, but they need to be in the right order.
x_coords = [float(x) for x in x.split()]
y_coords = [float(y) for y in y.split()]
points = zip(x_coords,y_coords)
k = 10
prev_length = 0
new_points = points
# Filter using the convex hull until no more points are removed
while len(new_points) != prev_length:
prev_length = len(new_points)
new_points = convex_filter(new_points,k)
Here is a screen shot of the above code with k=14. The 61 red dots are the ones that remain after the filter.
The approach you took sounds promising but your data is heavily oversampled. You could filter the x and y coordinates first, for example with a wide Gaussian and then downsample.
In MATLAB, you could use x = conv(x, normpdf(-10 : 10, 0, 5)) and then x = x(1 : 5 : end). You will have to tweak those numbers depending on the intrinsic persistence of the objects you are tracking and the average distance between points.
Then, you will be able to detect changes in direction very reliably, using the same approach you tried before, based on the scalar product, I imagine.
Another idea is to examine the left and the right surroundings at every point. This may be done by creating a linear regression of N points before and after each point. If the intersecting angle between the points is below some threshold, then you have an corner.
This may be done efficiently by keeping a queue of the points currently in the linear regression and replacing old points with new points, similar to a running average.
You finally have to merge adjacent corners to a single corner. E.g. choosing the point with the strongest corner property.