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random numer generation except inside of the circle in python

I want to generate 750 random number in 2D data x (0,1) and y (0,1)
X1 = np.random.random((750,2))
However, I want to make sure I don't have any value in the circular region such as
I can remove the value, but I want to fix the number of the random number to be 750. What would be the best way to generate such list?

import random
points = []
radius = 1
num_points=1500
index = 0
max_coord = 750
while index < num_points:
x=random.random()*max_coord
y=random.random()*max_coord
if x**2 + y**2 > radius**2:
points = points + [[x,y]]
index = index + 1
print(points)
Change max_coor to obtain bigger numbers. This only gives you 1500 points outside the circle of the given radius=1. Each coordinate varies between 0 and 750

you can just apply pythagorean's theorem. Here is my code to do that
import numpy as np
import matplotlib.pyplot as plt
dist = 0.2
X1 = np.random.random((750,2))
X2 = [x for x in X1 if (x[0]-0.5)**2 + (x[1]-0.5)**2 > dist**2] # filter values that are to close to center
# for plotting
x = [x[0] for x in X2]
y = [y[1] for y in X2]
plt.scatter(x,y)
plt.show()

You can do this with a well-known technique called rejection sampling. Generate candidate values and see if they meet your constraints. If so, accept them, otherwise reject and repeat. With 2-d sampling, the probability of acceptance on a given trial is P{Accept} = Area(acceptance region) / Area(generating region). Repetitions due to rejection are independent, so the number of trials has a geometric distribution with parameter p = P{Accept}
and Expected # trials = 1 / P{Accept}. For example, if the rejection region is a circle with radius 1/2 centered in a unit square, P{Accept} = (4 - Pi)/4 andon average it will take 4/(4 - Pi) (approximately 4.66) attempts per value generated. Obviously this won't work if the radius is >= sqrt(2)/2, and gets expensive as you approach that limit.
Note that the rejection circle does not need to be centered in the square, you just reject any candidate that is less than radius away from the circle's center.
Here's the code:
import numpy as np
def generate_pt(exclusion_radius):
while True:
candidate = np.random.sample(2)
# The following assumes the rejection circle is centered at (0.5, 0.5).
# Adjust by subtracting individual x/y coordinates for the center of
# the circle if you don't want it centered. Acceptance/rejection
# distance is determined by Pythagorean theorem.
if np.sum(np.square(candidate - 0.5)) >= exclusion_radius * exclusion_radius:
return candidate
# Generate 750 points outside the centered circle of radius 0.3.
data = np.array([generate_pt(0.3) for _ in range(750)])

Random point inside annulus with a shifted hole

First of all, will appreciate if someone will give me a proper term for "annulus with a shifted hole", see exactly what kind of shape I mean on a picture below.
Back to main question: I want to pick a random point in the orange area, uniform distribution is not required. For a case of a usual annulus I would've picked random point in (r:R) range and a random angle, then transform those to x,y and it's done. But for this unusual shape... is there even a "simple" formula for that, or should I approach it by doing some kind of polygonal approximation of a shape?
I'm interested in a general approach but will appreciate an example in python, javascript or any coding language of your choice.

Here's a simple method that gives a uniform distribution with no resampling.
For simplicity assume that the center of the outer boundary circle (radius r_outer) is at (0, 0) and that the center of the inner circular boundary (radius r_inner) lies at (x_inner, y_inner).
Write D for the outer disk, H1 for the subset of the plane given by the off-center inner hole, and H2 for the central disk of radius r_inner, centered at (0, 0).
Now suppose that we ignore the fact that the inner circle is not central, and instead of sampling from D-H1 we sample from D-H2 (which is easy to do uniformly). Then we've made two mistakes:
there's a region A = H1 - H2 that we might sample from, even though those samples shouldn't be in the result.
there's a region B = H2 - H1 that we never sample from, even though we should
But here's the thing: the regions A and B are congruent: given any point (x, y) in the plane, (x, y) is in H2 if and only if (x_inner - x, y_inner - y) is in H1, and it follows that (x, y) is in A if and only if (x_inner - x, y_inner - y) is in B! The map (x, y) -> (x_inner - x, y_inner - y) represents a rotation by 180 degress around the point (0.5*x_inner, 0.5*y_inner). So there's a simple trick: generate from D - H2, and if we end up with something in H1 - H2, rotate to get the corresponding point of H2 - H1 instead.
Here's the code. Note the use of the square root of a uniform distribution to choose the radius: this is a standard trick. See this article, for example.
import math
import random
def sample(r_outer, r_inner, x_inner, y_inner):
"""
Sample uniformly from (x, y) satisfiying:
x**2 + y**2 <= r_outer**2
(x-x_inner)**2 + (y-y_inner)**2 > r_inner**2
Assumes that the inner circle lies inside the outer circle;
i.e., that hypot(x_inner, y_inner) <= r_outer - r_inner.
"""
# Sample from a normal annulus with radii r_inner and r_outer.
rad = math.sqrt(random.uniform(r_inner**2, r_outer**2))
angle = random.uniform(-math.pi, math.pi)
x, y = rad*math.cos(angle),rad*math.sin(angle)
# If we're inside the forbidden hole, reflect.
if math.hypot(x - x_inner, y - y_inner) < r_inner:
x, y = x_inner - x, y_inner - y
return x, y
And an example plot, generated by the following:
import matplotlib.pyplot as plt
samples = [sample(5, 2, 1.0, 2.0) for _ in range(10000)]
xs, ys = zip(*samples)
plt.scatter(xs, ys, s=0.1)
plt.axis("equal")
plt.show()

Do you really need exact sampling? Because with acceptance/rejection it should work just fine. I assume big orange circle is located at (0,0)
import math
import random
def sample_2_circles(xr, yr, r, R):
"""
R - big radius
r, xr, yr - small radius and its position
"""
x = xr
y = yr
cnd = True
while cnd:
# sample uniformly in whole orange circle
phi = 2.0 * math.pi * random.random()
rad = R * math.sqrt(random.random())
x = rad * math.cos(phi)
y = rad * math.sin(phi)
# check condition - if True we continue in the loop with sampling
cnd = ( (x-xr)**2 + (y-yr)**2 < r*r )
return (x,y)

Since you have shown no equation, algorithm, or code of your own, but just an outline of an algorithm for center-aligned circles, I'll also just give the outline of an algorithm here for the more general case.
The smaller circle is the image of the larger circle under a similarity transformation. I.e. there is a fixed point in the larger circle and a ratio (which is R/r, greater than one) such that you can take any point on the smaller circle, examine the vector from the fixed point to that point, and multiply that vector by the ratio, then the end of that vector when it starts from the fixed point is a point on the larger circle. This transformation is one-to-one.
So you can choose a random point on the smaller circle (choose the angle at random between 0 and two-pi) and choose a ratio at random between 1 and the proportionality ratio R/r between the circles. Then use that the similarity transformation with the same fixed point but using the random ratio to get the image point of the just-chosen point on the smaller circle. This is a random point in your desired region.
This method is fairly simple. In fact, the hardest mathematical part is finding the fixed point of the similarity transformation. But this is pretty easy, given the centers and radii of the two circles. Hint: the transformation takes the center of the smaller circle to the center of the larger circle.
Ask if you need more detail. My algorithm does not yield a uniform distribution: the points will be more tightly packed where the circles are closest together and less tightly packed where the circles are farthest apart.
Here is some untested Python 3.6.2 code that does the above. I'll test it and show a graphic for it when I can.
import math
import random
def rand_pt_between_circles(x_inner,
y_inner,
r_inner,
x_outer,
y_outer,
r_outer):
"""Return a random floating-point 2D point located between the
inner and the outer circles given by their center coordinates and
radii. No error checking is done on the parameters."""
# Find the fixed point of the similarity transformation from the
# inner circle to the outer circle.
x_fixed = x_inner - (x_outer - x_inner) / (r_outer - r_inner) * r_inner
y_fixed = y_inner - (y_outer - y_inner) / (r_outer - r_inner) * r_inner
# Find a a random transformation ratio between 1 and r_outer / r_inner
# and a random point on the inner circle
ratio = 1 + (r_outer - r_inner) * random.random()
theta = 2 * math.pi * random.random()
x_start = x_inner + r_inner * math.cos(theta)
y_start = y_inner + r_inner * math.sin(theta)
# Apply the similarity transformation to the random point.
x_result = x_fixed + (x_start - x_fixed) * ratio
y_result = y_fixed + (y_start - y_fixed) * ratio
return x_result, y_result

The acceptance/rejection method as described by Severin Pappadeux is probably the simplest.
For a direct approach, you can also work in polar coordinates, with the center of the hole as the pole.
The polar equation (Θ, σ) (sorry, no rho) of the external circle will be
(σ cosΘ - xc)² + (σ sinΘ - yc)² = σ² - 2(cosΘ xc + sinΘ yc)σ + xc² + yc² = R²
This is a quadratic equation in σ, that you can easily solve in terms of Θ. Then you can draw an angle in 0, 2π an draw a radius between r and σ.
This won't give you a uniform distribution, because the range of σ is a function of Θ and because of the polar bias. This might be fixed by computing a suitable transfer function, but this is a little technical and probably not tractable analytically.

HOG Feature extraction

I want to extract HOG features of Line images of Arabic Handwriting. The code is as follows. So , I want help regarding how to input the image and how to output the features . Can anyone please help me regarding this.
import numpy as np
from scipy import sqrt, pi, arctan2, cos, sin
from scipy.ndimage import uniform_filter
def hog(image, orientations=9, pixels_per_cell=(8, 8),
cells_per_block=(3, 3), visualise=False, normalise=False):
"""Extract Histogram of Oriented Gradients (HOG) for a given image.
Compute a Histogram of Oriented Gradients (HOG) by
1. (optional) global image normalisation
2. computing the gradient image in x and y
3. computing gradient histograms
4. normalising across blocks
5. flattening into a feature vector
Parameters
----------
image : (M, N) ndarray
Input image (greyscale).
orientations : int
Number of orientation bins.
pixels_per_cell : 2 tuple (int, int)
Size (in pixels) of a cell.
cells_per_block : 2 tuple (int,int)
Number of cells in each block.
visualise : bool, optional
Also return an image of the HOG.
normalise : bool, optional
Apply power law compression to normalise the image before
processing.
Returns
-------
newarr : ndarray
HOG for the image as a 1D (flattened) array.
hog_image : ndarray (if visualise=True)
A visualisation of the HOG image.
References
----------
* http://en.wikipedia.org/wiki/Histogram_of_oriented_gradients
* Dalal, N and Triggs, B, Histograms of Oriented Gradients for
Human Detection, IEEE Computer Society Conference on Computer
Vision and Pattern Recognition 2005 San Diego, CA, USA
"""
image = np.atleast_2d(image)
"""
The first stage applies an optional global image normalisation
equalisation that is designed to reduce the influence of illumination
effects. In practice we use gamma (power law) compression, either
computing the square root or the log of each colour channel.
Image texture strength is typically proportional to the local surface
illumination so this compression helps to reduce the effects of local
shadowing and illumination variations.
"""
if image.ndim > 3:
raise ValueError("Currently only supports grey-level images")
if normalise:
image = sqrt(image)
"""
The second stage computes first order image gradients. These capture
contour, silhouette and some texture information, while providing
further resistance to illumination variations. The locally dominant
colour channel is used, which provides colour invariance to a large
extent. Variant methods may also include second order image derivatives,
which act as primitive bar detectors - a useful feature for capturing,
e.g. bar like structures in bicycles and limbs in humans.
"""
gx = np.zeros(image.shape)
gy = np.zeros(image.shape)
gx[:, :-1] = np.diff(image, n=1, axis=1)
gy[:-1, :] = np.diff(image, n=1, axis=0)
"""
The third stage aims to produce an encoding that is sensitive to
local image content while remaining resistant to small changes in
pose or appearance. The adopted method pools gradient orientation
information locally in the same way as the SIFT [Lowe 2004]
feature. The image window is divided into small spatial regions,
called "cells". For each cell we accumulate a local 1-D histogram
of gradient or edge orientations over all the pixels in the
cell. This combined cell-level 1-D histogram forms the basic
"orientation histogram" representation. Each orientation histogram
divides the gradient angle range into a fixed number of
predetermined bins. The gradient magnitudes of the pixels in the
cell are used to vote into the orientation histogram.
"""
magnitude = sqrt(gx ** 2 + gy ** 2)
orientation = arctan2(gy, (gx + 1e-15)) * (180 / pi) + 90
sy, sx = image.shape
cx, cy = pixels_per_cell
bx, by = cells_per_block
n_cellsx = int(np.floor(sx // cx)) # number of cells in x
n_cellsy = int(np.floor(sy // cy)) # number of cells in y
# compute orientations integral images
orientation_histogram = np.zeros((n_cellsy, n_cellsx, orientations))
for i in range(orientations):
#create new integral image for this orientation
# isolate orientations in this range
temp_ori = np.where(orientation < 180 / orientations * (i + 1),
orientation, 0)
temp_ori = np.where(orientation >= 180 / orientations * i,
temp_ori, 0)
# select magnitudes for those orientations
cond2 = temp_ori > 0
temp_mag = np.where(cond2, magnitude, 0)
orientation_histogram[:,:,i] = uniform_filter(temp_mag, size=(cy, cx))[cy/2::cy, cx/2::cx]
# now for each cell, compute the histogram
#orientation_histogram = np.zeros((n_cellsx, n_cellsy, orientations))
radius = min(cx, cy) // 2 - 1
hog_image = None
if visualise:
hog_image = np.zeros((sy, sx), dtype=float)
if visualise:
from skimage import draw
for x in range(n_cellsx):
for y in range(n_cellsy):
for o in range(orientations):
centre = tuple([y * cy + cy // 2, x * cx + cx // 2])
dx = radius * cos(float(o) / orientations * np.pi)
dy = radius * sin(float(o) / orientations * np.pi)
rr, cc = draw.bresenham(centre[0] - dx, centre[1] - dy,
centre[0] + dx, centre[1] + dy)
hog_image[rr, cc] += orientation_histogram[y, x, o]
"""
The fourth stage computes normalisation, which takes local groups of
cells and contrast normalises their overall responses before passing
to next stage. Normalisation introduces better invariance to illumination,
shadowing, and edge contrast. It is performed by accumulating a measure
of local histogram "energy" over local groups of cells that we call
"blocks". The result is used to normalise each cell in the block.
Typically each individual cell is shared between several blocks, but
its normalisations are block dependent and thus different. The cell
thus appears several times in the final output vector with different
normalisations. This may seem redundant but it improves the performance.
We refer to the normalised block descriptors as Histogram of Oriented
Gradient (HOG) descriptors.
"""
n_blocksx = (n_cellsx - bx) + 1
n_blocksy = (n_cellsy - by) + 1
normalised_blocks = np.zeros((n_blocksy, n_blocksx,
by, bx, orientations))
for x in range(n_blocksx):
for y in range(n_blocksy):
block = orientation_histogram[y:y + by, x:x + bx, :]
eps = 1e-5
normalised_blocks[y, x, :] = block / sqrt(block.sum() ** 2 + eps)
"""
The final step collects the HOG descriptors from all blocks of a dense
overlapping grid of blocks covering the detection window into a combined
feature vector for use in the window classifier.
"""
if visualise:
return normalised_blocks.ravel(), hog_image
else:
return normalised_blocks.ravel()

You can use the OpenCV library to read image files into NumPy arrays.

How can I set a minimum distance constraint for generating points with numpy.random.rand?

I am trying to generate an efficient code for generating a number of random position vectors which I then use to calculate a pair correlation function. I am wondering if there is straightforward way to set a constraint on the minimum distance allowed between any two points placed in my box.
My code currently is as follows:
def pointRun(number, dr):
"""
Compute the 3D pair correlation function
for a random distribution of 'number' particles
placed into a 1.0x1.0x1.0 box.
"""
## Create array of distances over which to calculate.
r = np.arange(0., 1.0+dr, dr)
## Generate list of arrays to define the positions of all points,
## and calculate number density.
a = np.random.rand(number, 3)
numberDensity = len(a)/1.0**3
## Find reference points within desired region to avoid edge effects.
b = [s for s in a if all(s > 0.4) and all(s < 0.6) ]
## Compute pairwise correlation for each reference particle
dist = scipy.spatial.distance.cdist(a, b, 'euclidean')
allDists = dist[(dist < np.sqrt(3))]
## Create histogram to generate radial distribution function, (RDF) or R(r)
Rr, bins = np.histogram(allDists, bins=r, density=False)
## Make empty containers to hold radii and pair density values.
radii = []
rhor = []
## Normalize RDF values by distance and shell volume to get pair density.
for i in range(len(Rr)):
y = (r[i] + r[i+1])/2.
radii.append(y)
x = np.average(Rr[i])/(4./3.*np.pi*(r[i+1]**3 - r[i]**3))
rhor.append(x)
## Generate normalized pair density function, by total number density
gr = np.divide(rhor, numberDensity)
return radii, gr
I have previously tried using a loop that calculated all distances for each point as it was made and then accepted or rejected. This method was very slow if I use a lot of points.

Here is a scalable O(n) solution using numpy. It works by initially specifying an equidistant grid of points and then perturbing the points by some amount keeping the distance between the points at most min_dist.
You'll want to tweak the number of points, box shape and perturbation sensitivity to get the min_dist you want.
Note: If you fix the size of a box and specify a minimum distance between every point, it makes sense that there will be a limit to the number of points you can draw satisfying the minimum distance.
import numpy as np
import matplotlib.pyplot as plt
# specify params
n = 500
shape = np.array([64, 64])
sensitivity = 0.8 # 0 means no movement, 1 means max distance is init_dist
# compute grid shape based on number of points
width_ratio = shape[1] / shape[0]
num_y = np.int32(np.sqrt(n / width_ratio)) + 1
num_x = np.int32(n / num_y) + 1
# create regularly spaced neurons
x = np.linspace(0., shape[1]-1, num_x, dtype=np.float32)
y = np.linspace(0., shape[0]-1, num_y, dtype=np.float32)
coords = np.stack(np.meshgrid(x, y), -1).reshape(-1,2)
# compute spacing
init_dist = np.min((x[1]-x[0], y[1]-y[0]))
min_dist = init_dist * (1 - sensitivity)
assert init_dist >= min_dist
print(min_dist)
# perturb points
max_movement = (init_dist - min_dist)/2
noise = np.random.uniform(
low=-max_movement,
high=max_movement,
size=(len(coords), 2))
coords += noise
# plot
plt.figure(figsize=(10*width_ratio,10))
plt.scatter(coords[:,0], coords[:,1], s=3)
plt.show()

Based on #Samir 's answer, and make it a callable function for your convenience :)
import numpy as np
import matplotlib.pyplot as plt
def generate_points_with_min_distance(n, shape, min_dist):
# compute grid shape based on number of points
width_ratio = shape[1] / shape[0]
num_y = np.int32(np.sqrt(n / width_ratio)) + 1
num_x = np.int32(n / num_y) + 1
# create regularly spaced neurons
x = np.linspace(0., shape[1]-1, num_x, dtype=np.float32)
y = np.linspace(0., shape[0]-1, num_y, dtype=np.float32)
coords = np.stack(np.meshgrid(x, y), -1).reshape(-1,2)
# compute spacing
init_dist = np.min((x[1]-x[0], y[1]-y[0]))
# perturb points
max_movement = (init_dist - min_dist)/2
noise = np.random.uniform(low=-max_movement,
high=max_movement,
size=(len(coords), 2))
coords += noise
return coords
coords = generate_points_with_min_distance(n=8, shape=(2448,2448), min_dist=256)
# plot
plt.figure(figsize=(10,10))
plt.scatter(coords[:,0], coords[:,1], s=3)
plt.show()

As I understood, you're looking for an algorithm to create many random points in a box such that no two points are closer than some minimum distance. If this is your problem, then you can take advantage of statistical physics, and solve it using molecular dynamics software. Moreover, you do need molecular dynamics or Monte Carlo to obtain exact solution of this problem.
You place N atoms in a rectangular box, create a repulsive interaction of a fixed radius between them (such as shifted Lennard-Jones interaction), and run simulation for some time (untill you see that the points spread out uniformly throughout the box). By laws of statistical physics you can show that positions of the points would be maximally random given the constraint that points cannot be close than some distance. This would not be true if you use iterative algorithm, such as placing points one-by-one and rejecting them if they overlap
I would estimate a runtime of several seconds for 10000 points, and several minutes for 100k. I use OpenMM for all my moelcular dynamics simulations.

#example of generating 50 points in a square of 4000x4000 and with minimum distance of 400
import numpy as np
import random as rnd
n_points=50
x,y = np.zeros(n_points),np.zeros(n_points)
x[0],y[0]=np.round(rnd.uniform(0,4000)),np.round(rnd.uniform(0,4000))
min_distances=[]
i=1
while i<n_points :
x_temp,y_temp=np.round(rnd.uniform(0,4000)),np.round(rnd.uniform(0,4000))
distances = []
for j in range(0,i):
distances.append(np.sqrt((x_temp-x[j])**2+(y_temp-y[j])**2))
min_distance = np.min(distances)
if min_distance>400 :
min_distances.append(min_distance)
x[i]=x_temp
y[i]=y_temp
i = i+1
print(x,y)

Calculating element position by computing transformation

This question is related to Transformation between two set of points . Hovewer this is better specified, and some assumptions added.
I have element image and some model.
I've detected contours on both
contoursModel0, hierarchyModel = cv2.findContours(model.copy(), cv2.RETR_LIST,
cv2.CHAIN_APPROX_SIMPLE);
contoursModel = [cv2.approxPolyDP(cnt, 2, True) for cnt in contoursModel0];
contours0, hierarchy = cv2.findContours(canny.copy(), cv2.RETR_LIST,
cv2.CHAIN_APPROX_SIMPLE);
contours = [cv2.approxPolyDP(cnt, 2, True) for cnt in contours0];
Then I've matched each contour to each other
modelMassCenters = [];
imageMassCenters = [];
for cnt in contours:
for cntModel in contoursModel:
result = cv2.matchShapes(cnt, cntModel, cv2.cv.CV_CONTOURS_MATCH_I1, 0);
if(result != 0):
if(result < 0.05):
#Here are matched contours
momentsModel = cv2.moments(cntModel);
momentsImage = cv2.moments(cnt);
massCenterModel = (momentsModel['m10']/momentsModel['m00'],
momentsModel['m01']/momentsModel['m00']);
massCenterImage = (momentsImage['m10']/momentsImage['m00'],
momentsImage['m01']/momentsImage['m00']);
modelMassCenters.append(massCenterModel);
imageMassCenters.append(massCenterImage);
Matched contours are something like features.
Now I want to detect transformation between this two sets of points.
Assumptions: element is rigid body, only rotation, displacement and scale change.
Some features may be miss detected how to eliminate them. I've once used cv2.findHomography and it takes two vectors and calculates homography between them even there are some miss matches.
cv2.getAffineTransformation takes only three points (can't cope missmatches) and here I have multiple features.
Answer in my previous question says how to calculate this transformation but does not take missmatches. Also I think that it is possible to return some quality level from algorithm (by checking how many points are missmatched, after computing some transformation from the rest)
And the last question: should I take all vector points to compute transformation or treat only mass centers of this shapes as feature?
To show it I've added simple image. Features with green are good matches in red bad matches. Here match should be computed from 3 green featrues and red missmatches should affect match quality.
I'm adding fragments of solution I've figured out for now (but I think it could be done much better):
for i in range(0, len(modelMassCenters) - 1):
for j in range(i + 1, len(modelMassCenters) - 1 ):
x1, y1 = modelMassCenters[i];
x2, y2 = modelMassCenters [j];
modelVec = (x2 - x1, y2 - y1);
x1, y1 = imageMassCenters[i];
x2, y2 = imageMassCenters[j];
imageVec = (x2 - x1, y2 - y1);
rotation = angle(modelVec,imageVec);
rotations.append((i, j, rotation));
scale = length(modelVec)/length(imageVec);
scales.append((i, j, scale));
After computing scales and rotation given by each pair of corresponding lines I'm going to find median value and average values of rotation which does not differ more than some delta from median value. The same thing with scale. Then points which are making those values taken to computation will be used to compute displacement.

Your second step (match contours to each other by doing a pairwise shape comparison) sounds very vulnerable to errors if features have a similar shape, e.g., you have several similar-sized circular contours. Yet if you have a rigid body with 5 circular features in one quadrant only, you could get a very robust estimate of the affine transform if you consider the body and its features as a whole. So don't discard information like a feature's range and direction from the center of the whole body when matching features. Those are at least as important in correlating features as size and shape of the individual contour.
I'd try something like (untested pseudocode):
"""
Convert from rectangular (x,y) to polar (r,w)
r = sqrt(x^2 + y^2)
w = arctan(y/x) = [-\pi,\pi]
"""
def polar(x, y): # w in radians
from math import hypot, atan2, pi
return hypot(x, y), atan2(y, x)
model_features = []
model = params(model_body_contour) # return tuple (center_x, center_y, area)
for contour in model_feature_contours:
f = params(countour)
range, angle = polar(f[0]-model[0], f[1]-model[1])
model_features.append((angle, range, f[2]))
image_features = []
image = params(image_body_contour)
for contour in image_feature_contours:
f = params(countour)
range, angle = polar(f[0]-image[0], f[1]-image[1])
image_features.append((angle, range, f[2]))
# sort image_features and model_features by angle, range
#
# correlate image_features against model_features across angle offsets
# rotation = angle offset of max correlation
# scale = average(model areas and ranges) / average(image areas and ranges)
If you have very challenging images, such as a ring of 6 equally-spaced similar-sized features, 5 of which have the same shape and one is different (e.g. 5 circles and a star), you could add extra parameters such as eccentricity and sharpness to the list of feature parameters, and include them in the correlation when searching for the rotation angle.