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Calculate the area enclosed by a 2D array of unordered points in python

I am trying to calculate the area of a shape enclosed by a large set of unordered points in python. I have a 2D array of points which I can plot as a scatterplot like this.
There are several ways to calculate the area enclosed by points, but these all assume ordered points, such as here and here. This method calculates the area unordered points, but it doesn't appear to work for complex shapes, as seen here. How would I calculate this area from unordered points in python?
Sample data looks like this:
[[225.93459 -27.25677 ]
[226.98128 -32.001945]
[223.3623 -34.119724]
[225.84741 -34.416553]]
From pen and paper one can see that this shape contains an area of ~12 (unitless) but putting these coordinates into one of the algorithms linked to previously returns an area of ~0.78.

Let's first mention that in the question How would I calculate this area from unordered points in python? used phrase 'unordered points' in the context of calculation of an area usually means that given are points of a contour enclosing an area which area is to calculate.
But in the question provided data sample are not points of a contour but just a cloud of points, which if visualized using a scatterplot results in a visually perceivable area.
The above is the reason why in the question provided links to algorithms calculating areas from 'unordered points' don't apply at all to what the question is about.
In other words, the actual title of the question I will answer below will be:
Calculate the visually perceivable area a cloud of (x,y) points is forming when visualized as a scatterplot
One of the possible options is mentioned in a comment to the question:
Honestly, you might consider taking THAT graph as a bitmap, and counting the number of non-white pixels in it. That is probably as close as you can get. – Tim Roberts
Given the image perfectly covering (without any margin) all the non-white pixels you can calculate the area the image rectangle is covering in units used in the underlying (x,y) data by calculating the area TA of the rectangle visible in the image from the underlying list of points P with (x,y) point coordinates ( P = [(x1,y1), (x2,y2), ...] ) as follows:
X = [x for x,y in P]
Y = [y for x,y in P]
TA = (max(X)-min(X))*(max(Y)-min(Y))
Assuming N_white is the number of all white pixels in the image with N pixels the actual area A covered by non-white pixels expressed in units used in the list of points P will be:
A = TA*(N-N_white)/N
Another approach using a list of points P with (x,y) point coordinates only ( without creation of an image ) consists of following steps:
decide which area Ap a point is covering and calculate half of the size h2 of a rectangle with this area around that point ( h2 = 0.5*sqrt(Ap) )
create a list R with rectangles around all points in the list P: R = [(x-h2, y+h2, x+h2, y-h2) for x,y in P]
use the code provided through a link listed in the stackoverflow question
Area of Union Of Rectangles using Segment Trees to calculate the total area covered by the rectangles in the list R.
The above approach has the advantage over the graphical one obtained from the scatterplot that with the choice of the area covered by a point you directly influence the used precision/resolution/granularity for the area calculation.

Given a 2D array of points the area covered by the points can be calculated with help of the return value of the same hist2d() function provided in the matplotlib module (as matplotlib.pyplot.hist2d()) which is used to show the scatterplot.
The 'trick' is to set the cmin parameter value of the function to 1 ( cmin=1 ) and then calculate the number of numpy.nan values in the by the function returned array setting them in relation to entire amount of array values.
In other words all what is necessary to calculate the area when creating the scatterplot is already there for easy use in a simple area calculation formulas if you know that the histogram creating function provide as return value all what is therefore necessary.
Below code of a ready to use function for the area calculation along with demonstration of function usage:
def area_of_points(points, grid_size = [1000, 1000]):
"""
Returns the area covered by N 2D-points provided in a 'points' array
points = [ (x1,y1), (x2,y2), ... , (xN, yN) ]
'grid_size' gives the number of grid cells in x and y direction the
'points' bounding box is divided into for calculation of the area.
Larger 'grid_size' values mean smaller grid cells, higher precision
of the area calculation and longer runtime.
area_of_points() requires installed matplotlib module. """
import matplotlib.pyplot as plt
import numpy as np
pts_x = [x for x,y in points]
pts_y = [y for x,y in points]
pts_bb_area = (max(pts_x)-min(pts_x))*(max(pts_y)-min(pts_y))
h2D,_,_,_ = plt.hist2d( pts_x, pts_y, bins = grid_size, cmin=1)
numberOfWhiteBins = np.count_nonzero(np.isnan(h2D))
numberOfAll2Dbins = h2D.shape[0]*h2D.shape[1]
areaFactor = 1.0 - numberOfWhiteBins/numberOfAll2Dbins
pts_pts_area = areaFactor * pts_bb_area
print(f'Areas: b-box = {pts_bb_area:8.4f}, points = {pts_pts_area:8.4f}')
plt.show()
return pts_pts_area
#:def area_of_points(points, grid_size = [1000, 1000])
import numpy as np
np.random.seed(12345)
x = np.random.normal(size=100000)
y = x + np.random.normal(size=100000)
pts = [[xi,yi] for xi,yi in zip(x,y)]
print(area_of_points(pts))
# ^-- prints: Areas: b-box = 114.5797, points = 7.8001
# ^-- prints: 7.800126875291629
The above code creates following scatterplot:
Notice that the printed output Areas: b-box = 114.5797, points = 7.8001 and the by the function returned area value 7.800126875291629 give the area in units in which the x,y coordinates in the array of points are specified.
Instead of usage of a function when utilizing the know how you can play around with the parameter of the scatterplot calculating the area of what can be seen in the scatterplot.
Below code which changes the displayed scatterplot using the same underlying point data:
import numpy as np
np.random.seed(12345)
x = np.random.normal(size=100000)
y = x + np.random.normal(size=100000)
pts = [[xi,yi] for xi,yi in zip(x,y)]
pts_values_example = \
[[0.53005, 2.79209],
[0.73751, 0.18978],
... ,
[-0.6633, -2.0404],
[1.51470, 0.86644]]
# ---
pts_x = [x for x,y in pts]
pts_y = [y for x,y in pts]
pts_bb_area = (max(pts_x)-min(pts_x))*(max(pts_y)-min(pts_y))
# ---
import matplotlib.pyplot as plt
bins = [320, 300] # resolution of the grid (for the scatter plot)
# ^-- resolution of precision for the calculation of area
pltRetVal = plt.hist2d( pts_x, pts_y, bins = bins, cmin=1, cmax=15 )
plt.colorbar() # display the colorbar (for a 2d density histogram)
plt.show()
# ---
h2D, xedges1D, yedges1D, h2DhistogramObject = pltRetVal
numberOfWhiteBins = np.count_nonzero(np.isnan(h2D))
numberOfAll2Dbins = (len(xedges1D)-1)*(len(yedges1D)-1)
areaFactor = 1.0 - numberOfWhiteBins/numberOfAll2Dbins
area = areaFactor * pts_bb_area
print(f'Areas: b-box = {pts_bb_area:8.4f}, points = {area:8.4f}')
# prints "Areas: b-box = 114.5797, points = 20.7174"
creating following scatterplot:
Notice that the calculated area is now larger due to smaller values used for grid resolution resulting in more of the area colored.

Efficient algorithm to find number density of points in 3D space

I have the position data for particles in 3D space. The particles are in random positions in the 3D box and I am trying to find the position of the maximum number density. Is there a simple algorithm to do this efficiently (I have a few million particles)? I have tried to use a similar idea to the centre of mass of the system (code is below). This gives me the centre of mass..is there a similar approach to find the position of the maximum number density?
I was thinking of making some 3d cube and separating it out into smaller cubes to the the number of particles within each cube....but that will take very long for many particles.
import numpy as np
X_data = np.random.random(100000) # x coordinates
Y_data = np.random.random(100000) # y-coordinates
Z_data = np.random.random(100000) # z-coordinates
#Assume all points are weighted equally
com_x = np.mean(X_data)
com_y = np.mean(Y_data)
com_z = np.mean(Z_data)
#Now have the centre of mass position

Regions of very dense points for sphere generation despite using uniform distributions for theta and phi

I'm setting up a data set, and the goal is to make a sphere with a normal distribution along the radial direction, and uniform theta and phi distributions. Despite using uniform distributions for theta and phi, I keep getting regions of very dense points at the poles (symmetrically near +/- radius on the z axis) that have a small but noticeable size. I am also writing to a file and using ROOT to plot my results.
I've tried reducing the range of theta and phi to graph half a sphere, but the problem still persists in both cases. I have also set the radius to a constant to make sure that wasn't interfering with the formation of the poles. The poles still formed.
#Size is the number of points to generate, and n is the dimension of the problem (n=2 for circle, n=3 for sphere, etc.)
hs_points = np.zeros((size, n))
for i in range(size):
hs_point = hs_points[i]
for j in range(n):
if j == 0:
# normal distribution on radius
coord = np.random.normal(mu, sigma)
elif j < n-1:
coord=round(random.uniform(0,np.pi),15)
else:
coord = np.random.rand()*(2*np.pi)
hs_point[j] = coord
hs_points[i] = hs_point
return hs_points
c_points = np.zeros((size, n))
# translate each hyperspherical point into a cartesian point
for i in range(size):
hs_point = hs_points[i]
xCoord=hs_point[0]*np.sin(hs_point[1])*np.cos(hs_point[2])
yCoord=hs_point[0]*np.sin(hs_point[1])*np.sin(hs_point[2])
zCoord=hs_point[0]*np.cos(hs_point[1])
c_points[i,0]=xCoord
c_points[i,1]=yCoord
c_points[i,2]=zCoord
I expect the program to output a "fuzzy" sphere with a normal distribution along the radial direction, but with uniform behavior everywhere else (i.e. no particularly dense regions). Actual behavior is the generation of dense poles with uniform behavior everywhere else.

Need help weighting (scaling) each of the bins in a histogram by a different factor

I'm trying to make a histogram of the radial distribution of a circular scatterring of particles, and I'm trying to scale the histogram so that the radial distribution is in particles per unit area.
Disclaimer: If you don't care about the math behind what I'm talking about, just skip over this section:
I'm splitting the radial distribution in to annuluses of equal width, going out from the center. So, in the center, I will have a circle of some radius, a. The area of this inner most portion will be $\pi a^{2}$.
Now if we want to know the area of the annulus going from radial distance a to 2a, we do $$ \int_{a}^{2a} 2 \pi r \ dr = 3 \pi a^{2} $$
Continuing in a similar fashion (going from 2a to 3a, 3a to 4a, etc.) we see that the areas increase as follows: $$ Areas = \pi a^{2}, 3 \pi a^{2}, 5 \pi a^{2}, 7 \pi a^{2}, ... $$
So, when I weight the histogram for the radial distribution of my scatter, going out from the center, each bin will have to be weighted so that the count of first bin is left alone, the count of the second bin is divided by 3, the count of the third bin is divided by 5, etc, etc.
So: Here's my try at the code:
import numpy as np
import matplotlib.pyplot as plt
# making random sample of 100000 points between -2.5 and 2.5
y_vec = 5*np.random.random(100000) - 2.5
z_vec = 5*np.random.random(100000) - 2.5
# blank canvasses for the y, z, and radial arrays
y_vec2 = []
z_vec2 = []
R_vec = []
# number of bins I want in the ending histogram
bns = 40
# cutting out the random samplings that aren't in a circular distribution
# and making the radial array
for i in range(0, 100000):
if np.sqrt((y_vec[i]*y_vec[i] + z_vec[i]*z_vec[i])) <= 2.5:
y_vec2.append(y_vec[i])
z_vec2.append(z_vec[i])
R_vec.append(np.sqrt(y_vec[i]*y_vec[i] + z_vec[i]*z_vec[i]))
# setting up the figures and plots
fig, ax = plt.subplots()
fig2, hst = plt.subplots()
# creating a weighting array for the histogram
wghts = []
i = 0
c = 1
# making the weighting array so that each of the bins will be weighted correctly
# (splitting the radial array up evenly in to groups of the size the bins will be
# and weighting them appropriately). I assumed the because the documentation says
# the "weights" array has to be the same size as the "x" initial input, that the
# weights act on each point individually...
while i < bns:
wghts.extend((1/c)*np.ones(len(R_vec)/bns))
c = c + 2
i = i + 1
# Making the plots
ax.scatter(y_vec2, z_vec2)
hst.hist(R_vec, bins = bns, weights = wghts)
# plotting
plt.show()
The scatter plot looks great:
But, the radial plot suggest that I got the weighting wrong. It should be constant across all annuli, but it is increasing, as though it were not weighted at all:
The erratic look of the Radial Distribution suggests to me that the weighting function in the "hist" operator weights each member of R_vec individually instead of weighting the bins.
How would I weight the bins by the factors I need to scale them by? Any help?

You are correct when you surmise that the weights weight the individual values and not the bins. This is documented:
Each value in x only contributes its associated weight towards the bin count (instead of 1).
Therefore the basic problem is that, in calculating the weights, you aren't taking account of the order of the points. You created points at random, but then you create the weights in sequence from greatest to least. This means you're not assigning the right weights to the right points.
The way you should create the weights is by directly computing each point's weight from its radius. The way you seem to want to do this is by discretizing the radius into a binned radius, then weighting inversely by that. Instead of what you're doing for the weights, try this:
R_vec = np.array(R_vec)
wghts = 1 / (2*(R_vec//(2.5/bns))+1)
This gives me the right result:
You can also get essentially the same result without doing the binning in the weighting --- that is, just directly weight each point by the reciporcal of its radius:
R_vec = np.array(R_vec)
wghts = 1 / R_vec
The advantage of doing this is that you can then plot a histogram a different number of bins without recomputing the weights. It also makes somewhat more conceptual sense to weight each point by how far out it is in a continuous sense, not by whether it falls on one side or the other of a discrete bin boundary.

When you want to plot something "per unit area", use area as your independent variable.
This way, you can still use a histogram if you like, but you don't have to worry about non-uniform binning or weighting.
I replaced your line:
hst.hist(R_vec, bins = bns, weights = wghts)
with:
hst.hist(np.pi*np.square(R_vec),bins=bns)

calculate turning points / pivot points in trajectory (path)

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.