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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.

Mapping Degrees to (x,y) pixels on a 360 Image

I am looking to plot out points on a 2d spherical image from a GoPro Max given an Altitude° and an Azimuth°.
Here's what you can assume:
The center of the image is always facing south
the image is always level looking straight at the equator.
Here's what I know:
The image size is 5760x2880 which means the equator is at 1440px vertically.
I have found an image online and labeled it with the Azimuth° going left to right and the Altitude° going up and down. You can find that image here
I hope that this will give you a better idea of what I'm trying to do here.
Put simply, I'm imagining a method something like:
ConvertCoords(Azimuth°, Altitude°){
...
return (x,y)
}
Is this possible? I will probably implement this using Python but I'm open to suggestions. Any sort of information to point me in the right direction is greatly appreciated!
Edit: I should mention that from my research I believe the GoPro Max uses an equirectangular projection. I was able to overlay that image I attached on one of the 360 photos and plot out some data manually which seemed to come out correct.

With an equirectangular projection, the mapping is direct and linear.
def convertCoords( azimuth, altitude ):
# Assumptions:
# - Both values are in radians
# - South is an azimuth of 0, negative to the left, range -pi to +pi
# - Altitude range is -pi/2 to +pi/2, negative down
x = 2880 + (azimuth * 2880 / math.pi)
y = 1440 - (altitude * 2880 / math.pi)
return x,y
If you'd rather use degrees, change "math.pi" to "180".

How to get rid of extraneous points outside of a closed curve using Python

I am trying to "shrink" the size of the outer boundary (in red) that is a set of pixels along the edge of an image. I already shrunk the boundary using normal vectors, and I have a set of points that are a certain distance inside the outer boundary. However, there are a lot of points that I could get rid of to make the inner boundary (in blue) smoother. So, how do I get rid of the extraneous points?
I tried converting everything to polar coordinates, and then taking the point that is closest to to the center of the image for each interval of pi/60 radians from 0 to 2pi. However, I realized that this wouldn't work for something like boomerang-ish shape, where there would be two points that I would want to keep in the pi/60 interval.
polarShrunk is the set of shrunk points converted to polar coordinates
cell_img is the object with a distanceBetween method that finds the distance between two points and a findMoments method which finds the "center of mass" of the image when it binarized to black and white. pol2cart converts polar coordinates to cartesian coordinates
i = 0
smoothArray = []
for x in range(120):
i += 1
theta1 = (i - 1)*np.pi/60
theta2 = i*np.pi/60
simAnglePoints = []
for y in polarShrunk:
if (y[0] < theta2) and (y[1] > theta1):
simAnglePoints.append(y)
closestPoint = np.array([10000, 10000])
for z in np.array(simAnglePoints):
if cell_img.distanceBetween(np.array(cell_img.pol2cart(z)), cell_img.findMoments) < cell_img.distanceBetween(closestPoint, cell_img.findMoments):
closestPoint = z
smoothArray.append(closestPoint)
I want only the points that follow the black line that I drew over the picture, and to get rid of all of the other points from the polarShrunk array.
Any help would be appreciated! I really only need a way to approach this problem, and I would like to figure out how to implement that approach in Python on my own.

Theory behind Wolfenstein-style 3D rendering

I'm currently working on a project about 3D rendering, and I'm trying to make simplistic program that can display a simple 3D room (static shading, no player movement, only rotation) with pygame
So far I've worked through the theory:
Start with a list of coordinates for the X and Z of each "Node"
Nodes are kept in an order which forms a closed loop, so that a pair of nodes will form either side of a wall
The height of the wall is determined when it is rendered, being relative to distance from the camera
Walls are rendered using painter's algorithm, so closer objects are drawn on top of further ones
For shading "fake contrast", which brightens/darkens walls based on the gradient between it's two nodes
While it seems simple enough, the process behind translating the 3D coordinates into 2D points on the screen is proving the difficult for me to understand.
Googling this topic has so far only yeilded these equations:
screenX = (worldX/worldZ)
screenY = (worldY/worldZ)
Which seem flawed to me, as you would get a divide by zero error if any Z coordinate is 0.
So if anyone could help explain this, I'd be really greatful.

Well the
screenX = (worldX/worldZ)
screenY = (worldY/worldZ)
is not the whole stuff that is just the perspective division by z and it is not meant for DOOM or Wolfenstein techniques.
Well in Doom there is only single angle of viewing (you can turn left/right but cannot look up/down only duck or jump which is not the same). So we need to know our player position and direction px,py,pz,pangle. The z is needed only if you want to implement also z axis movement/looking...
If you are looking in a straight line (Red) all the object that cross that line in the 3D are projected to single x coordinate in the player screen...
So if we are looking at some direction (red) any object/point crossing/touching this red line will be place at the center of screen (in x axis). What is left from it will be rendered on the left and similarly whats on right will be rendered on the right too...
With perspective we need to define how large viewing angle we got...
This limits our view so any point touches the green line will be projected on the edge of view (in x axis). From this we can compute screen x coordinate sx of any point (x,y,z) directly:
// angle of point relative to player direction
sx = point_ang - pangle;
if (sx<-M_PI) sx+=2.0*M_PI;
if (sx>+M_PI) sx-=2.0*M_PI;
// scale to pixels
sx = screen_size_x/2 + sx*screen_size_x/FOVx
where screen_size_x is resolution of our view area and point ang is angle of point x,y,z relative to origin px,py,pz. You can compute it like this:
point_ang = atan2(y-py,x-px)
but if you truly do a DOOM ray-casting then you already got this angle.
Now we need to compute the screen y coordinate sy which is dependent on the distance from player and wall size. We can exploit triangle similarity.
so:
sy = screen_size_y/2 (+/-) wall_height*focal_length/distance
Where focal length is the distance at which wall with 100% height will cover exactly the whole screen in y axis. As you can see we dividing by distance which might be zero. Such state must be avoided so you need to make sure your rays will be evaluated at the next cell if standing directly on cell boundary. Also we need to select the focal length so square wall will be projected as square.
Here a piece of code from mine Doom engine (putted all together):
double divide(double x,double y)
{
if ((y>=-1e-30)&&(y<=+1e-30)) return 0.0;
return x/y;
}
bool Doom3D::cell2screen(int &sx,int &sy,double x,double y,double z)
{
double a,l;
// x,y relative to player
x-=plrx;
y-=plry;
// convert z from [cell] to units
z*=_Doom3D_cell_size;
// angle -> sx
a=atan2(y,x)-plra;
if (a<-pi) a+=pi2;
if (a>+pi) a-=pi2;
sx=double(sxs2)*(1.0+(2.0*a/view_ang));
// perpendicular distance -> sy
l=sqrt((x*x)+(y*y))*cos(a);
sy=sys2+divide((double(plrz+_Doom3D_cell_size)-z-z)*wall,l);
// in front of player?
return (fabs(a)<=0.5*pi);
}
where:
_Doom3D_cell_size=100; // [units] cell cube size
view_ang=60.0*deg; // FOVx
focus=0.25; // [cells] view focal length (uncorrected)
wall=double(sxs)*(1.25+(0.288*a)+(2.04*a*a))*focus/double(_Doom3D_cell_size); // [px] projected wall size ratio size = height*wall/distance
sxs,sys = screen resolution
sxs2,sys2 = screen half resolution
pi=M_PI, pi2=2.0*M_PI
Do not forget to use perpendicular distances (multiplied by cos(a) as I did) otherwise serious fish-eye effect will occur. For more info see:
Ray Casting with different height size

Check if points are inside ellipse faster than contains_point method

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.