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Python - Find x and y values of a 2D gaussian given a value for the function

I have a 2D gaussian function f(x,y). I know the values x₀ and y₀ at which the peak g₀ of the function occurs. But then I want to find xₑ and yₑ values at which f(xₑ, yₑ) = g₀ / e¹. I know there are multiple solutions to this, but at least one is sufficient.
So far I have
def f(x, y, g0,x0,y0,sigma_x,sigma_y,offset):
return offset + g0* np.exp(-(((x-x0)**(2)/(2*sigma_x**(2))) + ((y-y0)**(2)/(2*sigma_y**(2)))))
All variables taken as parameters are known as they were extracted from a curve fit.
I understand that taking the derivative in x and setting f() = 0 and similarly in y, gives a solvable linear system for (x,y), but this seems like overkill to manually implement, there must be some library or tool out there that can do what I am trying to achieve?

There are an infinite number of possibilities (or possibly 1 trivial or none in special cases regarding the value of g0). A solution can be computed analytically in constant time using a direct method. No need for approximations or iterative methods to find roots of a given function. It is just pure maths.
Gaussian kernel have interesting symmetries. One of them is the invariance to the rotation when the peak is translated to (0,0). Another on is that the 1D section of a 2D gaussian surface is a gaussian curve.
Lets ignore offset for a moment: it does not really change the problem (it is just a Z-axis translation) and add additional useless term for the resolution.
The geometric solution to is problem is an ellipse so the solution (xe, ye) follows the conic expression : (xe-x0)² / a² + (ye-y0)² / b² = 1. If sigma_x = sigma_y, then the solution is simpler : this is a circle with the expression (xe-x0)² + (ye-y0)² = r. Note that a, b and r are dependant of the searched value and the kernel parameters (eg. sigma_x). Changing sigma_x and sigma_y is like stretching the space, and so the solution similarly. Changing x0 and y0 is like translating the space and so the solution too.
In fact, we could solve the problem for the simpler case where x0=0, y0=0, sigma_x=1 and sigma_y=1. Then we can apply a translation, followed by a linear transformation using a transformation matrix. A basic multiplication 4x4 matrix can do that. Solving the simpler case is much easier since there are are less parameter to consider. Actually, g0 and offset can also be partially discarded of f since it is on both side of the expression and one just need to solve the linear equation offset + g0 * h(xe,ye) = g0 / e so h(x,y) = 1 / e - offset / g0 where h(xe, ye) = exp(-(xe² + ye²)/2). Assuming we forget the translation and linear transformation for a moment, the problem can be solve quite easily:
h(xe, ye) = 1 / e - offset / g0
exp(-(xe² + ye²)/2) = 1 / e - offset / g0
-(xe² + ye²)/2 = ln(1 / e - offset / g0)
xe² + ye² = -2 * ln(1 / e - offset / g0)
That's it! We got our circle expression where the radius r is -2*ln(1 / e - offset / g0)! Note that ln in the expression is basically the natural logarithm.
Now we could try to find the 4x4 matrix coefficients, or actually try to directly solve the full expression which is finally not so difficult.
offset + g0 * exp(-((x-x0)²/(2*sigma_x²) + (y-y0)²/(2*sigma_y²))) = g0 / e
exp(-((x-x0)²/(2*sigma_x²) + (y-y0)²/(2*sigma_y²))) = 1 / e - offset / g0
-((x-x0)²/(2*sigma_x²) + (y-y0)²/(2*sigma_y²)) = ln(1 / e - offset / g0)
((x-x0)²/sigma_x² + (y-y0)²/sigma_y²)/2 = -ln(1 / e - offset / g0)
(x-x0)²/sigma_x² + (y-y0)²/sigma_y² = -2 * ln(1 / e - offset / g0)
That's it! We got you conic expression where r = -2 * ln(1 / e - offset / g0) is a constant, a = sigma_x and b = sigma_y are the unknown parameter in the above expression. It can be normalized using a = sigma_x/sqrt(r) and b = sigma_y/sqrt(r) so the right hand side is 1 fitting exactly with the above expression but this is just some math details.
You can find one point of the ellipse easily since you know the centre of the ellipse (x0, y0) and there is at least 1 point at the intersection of the line y=y0 and the above conic expression. Lets find it:
(x-x0)²/sigma_x² + (y0-y0)²/sigma_y² = -2 * ln(1 / e - offset / g0)
(x-x0)²/sigma_x² = -2 * ln(1 / e - offset / g0)
(x-x0)² = -2 * ln(1 / e - offset / g0) * sigma_x²
x = sqrt(-2 * ln(1 / e - offset / g0) * sigma_x²) + x0
Note there are two solutions (-sqrt(...) + x0) but you only need one of them. I hope I did not make any mistake in the computation (at least the details should be enough to find it easily) and the solution is not a complex number in your case. The benefit of this solution is that it is very very fast to compute.
The final solution is:
(xe, ye) = (sqrt(-2*ln(1/e-offset/g0)*sigma_x²)+x0, y0)

How to fit a piecewise (alternating linear and constant segments) function to a parabolic function?

I do have a function, for example , but this can be something else as well, like a quadratic or logarithmic function. I am only interested in the domain of . The parameters of the function (a and k in this case) are known as well.
My goal is to fit a continuous piece-wise function to this, which contains alternating segments of linear functions (i.e. sloped straight segments, each with intercept of 0) and constants (i.e. horizontal segments joining the sloped segments together). The first and last segments are both sloped. And the number of segments should be pre-selected between around 9-29 (that is 5-15 linear steps + 4-14 constant plateaus).
Formally
The input function:
The fitted piecewise function:
I am looking for the optimal resulting parameters (c,r,b) (in terms of least squares) if the segment numbers (n) are specified beforehand.
The resulting constants (c) and the breakpoints (r) should be whole natural numbers, and the slopes (b) round two decimal point values.
I have tried to do the fitting numerically using the pwlf package using a segmented constant models, and further processed the resulting constant model with some graphical intuition to "slice" the constant steps with the slopes. It works to some extent, but I am sure this is suboptimal from both fitting perspective and computational efficiency. It takes multiple minutes to generate a fitting with 8 slopes on the range of 1-50000. I am sure there must be a better way to do this.
My idea would be to instead using only numerical methods/ML, the fact that we have the algebraic form of the input function could be exploited in some way to at least to use algebraic transforms (integrals) to get to a simpler optimization problem.
import numpy as np
import matplotlib.pyplot as plt
import pwlf
# The input function
def input_func(x,k,a):
return np.power(x,1/a)*k
x = np.arange(1,5e4)
y = input_func(x, 1.8, 1.3)
plt.plot(x,y);
def pw_fit(func, x_r, no_seg, *fparams):
# working on the specified range
x = np.arange(1,x_r)
y_input = func(x, *fparams)
my_pwlf = pwlf.PiecewiseLinFit(x, y_input, degree=0)
res = my_pwlf.fit(no_seg)
yHat = my_pwlf.predict(x)
# Function values at the breakpoints
y_isec = func(res, *fparams)
# Slope values at the breakpoints
slopes = np.round(y_isec / res, decimals=2)
slopes = slopes[1:]
# For the first slope value, I use the intersection of the first constant plateau and the input function
slopes = np.insert(slopes,0,np.round(y_input[np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0]] / np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0], decimals=2))
plateaus = np.unique(np.round(yHat))
# If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
to_del = np.argwhere(np.diff(slopes) == 0).flatten()
slopes = np.delete(slopes,to_del + 1)
plateaus = np.delete(plateaus,to_del)
breakpoints = [np.ceil(plateaus[0]/slopes[0])]
for idx, j in enumerate(slopes[1:-1]):
breakpoints.append(np.floor(plateaus[idx]/j))
breakpoints.append(np.ceil(plateaus[idx+1]/j))
breakpoints.append(np.floor(plateaus[-1]/slopes[-1]))
return slopes, plateaus, breakpoints
slo, plat, breaks = pw_fit(input_func, 50000, 8, 1.8, 1.3)
# The piecewise function itself
def pw_calc(x, slopes, plateaus, breaks):
x = x.astype('float')
cond_list = [x < breaks[0]]
for idx, j in enumerate(breaks[:-1]):
cond_list.append((j <= x) & (x < breaks[idx+1]))
cond_list.append(breaks[-1] <= x)
func_list = [lambda x: x * slopes[0]]
for idx, j in enumerate(slopes[1:]):
func_list.append(plateaus[idx])
func_list.append(lambda x, j=j: x * j)
return np.piecewise(x, cond_list, func_list)
y_output = pw_calc(x, slo, plat, breaks)
plt.plot(x,y,y_output);

(Not important, but I think the fitted piecewise function is not continuous as it is. Intervals should be x<=r1; r1<x<=r2; ....)
As Anatolyg has pointed out, it looks to me that in the optimal solution (for the function posted at least, and probably for any where the derivative is different from zero), the horizantal segments will collapse to a point or the minimum segment length (in this case 1).
EDIT---------------------------------------------
The behavior above could only be valid if the slopes could have an intercept. If the intercepts are zero, as posted in the question, one consideration must be taken into account: Is the initial parabolic function defined in zero or nearby? Imagine the function y=0.001 *sqrt(x-1000), then the segments defined as b*x will have a slope close to zero and will be so similar to the constant segments that the best fit will be just the line that without intercept that fits better all the function.
Provided that the function is defined in zero or nearby, you can start by approximating the curve just by linear segments (with intercepts):
divide the function domain in N intervals(equal intervals or whose size is a function of the average curvature (or second derivative) of the function along the domain).
linear fit/regression in each intervals
for each interval, if a point (or bunch of points) in the extreme of any interval is better fitted by the line of the neighbor interval than the line of its interval, this point is assigned to the neighbor interval.
Repeat from 2) until no extreme points are moved.
Linear regressions might be optimized not to calculate all the covariance matrixes from scratch on each iteration, but just adding the contributions of the moved points to the previous covariance matrixes.
Then each linear segment (LSi) is replaced by a combination of a small constant segment at the beginning (Cbi), a linear segment without intercept (Si), and another constant segment at the end (Cei). This segments are easy to calculate as Si will contain the middle point of LSi, and Cbi and Cei will have respectively the begin and end values of the segment LSi. Then the intervals of each segment has to be calculated as an intersection between lines.
With this, the constant end segment will be collinear with the constant begin segment from the next interval so they will merge, resulting in a series of constant and linear segments interleaved.
But this would be a floating point start solution. Next, you will have to apply all the roundings which will mess up quite a lot all the segments as the conditions integer intervals and linear segments without slope can be very confronting. In fact, b,c,r are not totally independent. If ci and ri+1 are known, then bi+1 is already fixed
If nothing is broken so far, the final task will be to minimize the error/cost function (I assume that it will be the integral of the error between the parabolic function and the segments). My guess is that gradients here will be quite a pain, as if you change for example one ci, all the rest of the bj and cj will have to adapt as well due to the integer intervals restriction. However, if you can generalize the derivatives between parameters ( how much do I have to adapt bi+1 if ci changes a unit), you can propagate the change of one parameter to all other parameters and have kind of a gradient. Then for each interval, you can estimate what would be the ideal parameter and averaging all intervals calculate the best gradient step. Let me illustrate this:
Assuming first that r parameters are fixed, if I change c1 by one unit, b2 changes by 0.1, c2 changes by -0.2 and b3 changes by 0.2. This would be the gradient.
Then I estimate, comparing with the parabolic curve, that c1 should increase 0.5 (to reduce the cost by 10 points), b2 should increase 0.2 (to reduce the cost by 5 points), c2 should increase 0.2 (to reduce the cost by 6 points) and b3 should increase 0.1 (to reduce the cost by 9 points).
Finally, the gradient step would be (0.5/1·10 + 0.2/0.1·5 - 0.2/(-0.2)·6 + 0.1/0.2·9)/(10 + 5 + 6 + 9)~= 0.45. Thus, c1 would increase 0.45 units, b2 would increase 0.45·0.1, and so on.
When you add the r parameters to the pot, as integer intervals do not have an proper derivative, calculation is not straightforward. However, you can consider r parameters as floating points, calculate and apply the gradient step and then apply the roundings.

We can integrate the squared error function for linear and constant pieces and let SciPy optimize it. Python 3:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize
xl = 1
xh = 50000
a = 1.3
p = 1 / a
n = 8
def split_b_and_c(bc):
return bc[::2], bc[1::2]
def solve_for_r(b, c):
r = np.empty(2 * n)
r[0] = xl
r[1:-1:2] = c / b[:-1]
r[2::2] = c / b[1:]
r[-1] = xh
return r
def linear_residual_integral(b, x):
return (
(x ** (2 * p + 1)) / (2 * p + 1)
- 2 * b * x ** (p + 2) / (p + 2)
+ b ** 2 * x ** 3 / 3
)
def constant_residual_integral(c, x):
return x ** (2 * p + 1) / (2 * p + 1) - 2 * c * x ** (p + 1) / (p + 1) + c ** 2 * x
def squared_error(bc):
b, c = split_b_and_c(bc)
r = solve_for_r(b, c)
linear = np.sum(
linear_residual_integral(b, r[1::2]) - linear_residual_integral(b, r[::2])
)
constant = np.sum(
constant_residual_integral(c, r[2::2])
- constant_residual_integral(c, r[1:-1:2])
)
return linear + constant
def evaluate(x, b, c, r):
i = 0
while x > r[i + 1]:
i += 1
return b[i // 2] * x if i % 2 == 0 else c[i // 2]
def main():
bc0 = (xl + (xh - xl) * np.arange(1, 4 * n - 2, 2) / (4 * n - 2)) ** (
p - 1 + np.arange(2 * n - 1) % 2
)
bc = scipy.optimize.minimize(
squared_error, bc0, bounds=[(1e-06, None) for i in range(2 * n - 1)]
).x
b, c = split_b_and_c(bc)
r = solve_for_r(b, c)
X = np.linspace(xl, xh, 1000)
Y = [evaluate(x, b, c, r) for x in X]
plt.plot(X, X ** p)
plt.plot(X, Y)
plt.show()
if __name__ == "__main__":
main()

I have tried to come up with a new solution myself, based on the idea of #Amo Robb, where I have partitioned the domain, and curve fitted a dual - constant and linear - piece together (with the help of np.maximum). I have used the 1 / f(x)' as the function to designate the breakpoints, but I know this is arbitrary and does not provide a global optimum. Maybe there is some optimal function for these breakpoints. But this solution is OK for me, as it might be appropriate to have a better fit at the first segments, at the expense of the error for the later segments. (The task itself is actually a cost based retail margin calculation {supply price -> added margin}, as the retail POS software can only work with such piecewise margin function).
The answer from #David Eisenstat is correct optimal solution if the parameters are allowed to be floats. Unfortunately the POS software can not use floats. It is OK to round up c-s and r-s afterwards. But the b-s should be rounded to two decimals, as those are inputted as percents, and this constraint would ruin the optimal solution with long floats. I will try to further improve my solution with both Amo's and David's valuable input. Thank You for that!
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
# The input function f(x)
def input_func(x,k,a):
return np.power(x,1/a) * k
# 1 / f(x)'
def one_per_der(x,k,a):
return a / (k * np.power(x, 1/a-1))
# 1 / f(x)' inverted
def one_per_der_inv(x,k,a):
return np.power(a / (x*k), a / (1-a))
def segment_fit(start,end,y,first_val):
b, _ = curve_fit(lambda x,b: np.maximum(first_val, b*x), np.arange(start,end), y[start-1:end-1])
b = float(np.round(b, decimals=2))
bp = np.round(first_val / b)
last_val = np.round(b * end)
return b, bp, last_val
def pw_fit(end_range, no_seg, **fparams):
y_bps = np.linspace(one_per_der(1, **fparams), one_per_der(end_range,**fparams) , no_seg+1)[1:]
x_bps = np.round(one_per_der_inv(y_bps, **fparams))
y = input_func(x, **fparams)
slopes = [np.round(float(curve_fit(lambda x,b: x * b, np.arange(1,x_bps[0]), y[:int(x_bps[0])-1])[0]), decimals = 2)]
plats = [np.round(x_bps[0] * slopes[0])]
bps = []
for i, xbp in enumerate(x_bps[1:]):
b, bp, last_val = segment_fit(int(x_bps[i]+1), int(xbp), y, plats[i])
slopes.append(b); bps.append(bp); plats.append(last_val)
breaks = sorted(list(x_bps) + bps)[:-1]
# If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
to_del = np.argwhere(np.diff(slopes) == 0).flatten()
breaks_to_del = np.concatenate((to_del * 2, to_del * 2 + 1))
slopes = np.delete(slopes,to_del + 1)
plats = np.delete(plats[:-1],to_del)
breaks = np.delete(breaks,breaks_to_del)
return slopes, plats, breaks
def pw_calc(x, slopes, plateaus, breaks):
x = x.astype('float')
cond_list = [x < breaks[0]]
for idx, j in enumerate(breaks[:-1]):
cond_list.append((j <= x) & (x < breaks[idx+1]))
cond_list.append(breaks[-1] <= x)
func_list = [lambda x: x * slopes[0]]
for idx, j in enumerate(slopes[1:]):
func_list.append(plateaus[idx])
func_list.append(lambda x, j=j: x * j)
return np.piecewise(x, cond_list, func_list)
fparams = {'k':1.8, 'a':1.2}
end_range = 5e4
no_steps = 10
x = np.arange(1, end_range)
y = input_func(x, **fparams)
slopes, plats, breaks = pw_fit(end_range, no_steps, **fparams)
y_output = pw_calc(x, slopes, plats, breaks)
plt.plot(x,y_output,y);

is there a way to examine if an angle is a right angle programmatically?

i plot an angle in python
here is the code
x = [0,0.5,1]
y = [0,0.5,0]
plt.scatter(x,y)
plt.plot(x,y)
plt.show()
is there a way to examine if the angle is a right angle programmatically?

The easiest way is to test if the dot product of the vectors is 0.
In your case, you simply compute:
v1 = ( (x[1]-x[0]), (y[1]-y[0]) ) <- (0.5, 0.5)
v2 = ( (x[2]-x[1]), (y[2]-y[1]) ) <- (0.5, -0.5)
dot_product = v1[0]*v2[0] + v1[1]*v2[1] <- 0.5² - 0.5² = 0

The other answers do not really care about possible inaccuracies and truncation errors, nor efficiency.
Rather than an exact comparison to 90° (or 0° in case of dot product), it is wiser to check for a small angle difference to 90° (resp. 0°).
Also wise to avoid divisions, square roots and trigonometric functions. The cross-product method is among the most attractive.
Compute the cross-product of the sides of the angle and their squared lengths, and compare
with a precomputed tolerance:
(ABx . BCy - ABy . BCx)² ≥ α.(ABx² + ABy²).(BCx² + BCy²)
with α = cos²δ where δ is the angle tolerance.

You can try to calculate the angle, but an easier way could be to check whether the Pythagorean Theorem applies. For that you'll need to calculate the size of the three edges and then check whether A^2 + B^2 ~= C^2

Yes, there is.
x = [0,0.5,1]
y = [0,0.5,0]
points = [np.array(point) for point in zip(x,y)]
a, b, c = points
ba = a - b
bc = c - b
cosine_angle = np.dot(ba, bc) / (np.linalg.norm(ba) * np.linalg.norm(bc))
angle_rad = np.arccos(cosine_angle)
angle_deg = np.rad2deg(angle_rad)
print(angle_deg) # 90.0

You can compute the angle between the two vectors as following: first, get the two vectors v1 and v2 and then use np.arccos() which returns the angle in radians. Convert it to degrees to check if it is 90 degrees. The formulae for computing angle between two vectors can be found on this Wiki link
import numpy as np
x = np.array([0,0.5,1])
y = np.array([0,0.5,0])
vecs = np.vstack((x, y))
v1 = vecs[:, 1] - vecs[:, 0]
v2 = vecs[:, 2] - vecs[:, 1]
angle_rad = np.arccos(np.dot(v1, v2) / (np.linalg.norm(v1) * np.linalg.norm(v2)))
angle_deg = np.rad2deg(angle_rad)
# 90.0

Calculating 'Diagonal Distance' in 3 dimensions for A* path-finding heuristic

I'm running A* Path finding over a 3D grid of data. Available movements are the 26 surrounding nodes (i.e. you can move diagonally) I've been using euclidean distance as the heuristic and this works well but I'd also like to try 'diagonal distance' to see how that works (and if there are any speed gains)
I've found some logic online to do this in 2 dimensions...
function heuristic(node) =
dx = abs(node.x - goal.x)
dy = abs(node.y - goal.y)
return D * (dx + dy) + (D2 - 2 * D) * min(dx, dy)
..,where D is the north/east/south/west distance (for example 1m) and D2 is the diagonal distance (for example sqrt(2))
I'm not exactly sure how to convert this to 3 dimensions - so any help would be greatly appreciated
As an extra question (as this is how much grid is in reality...) suppose nodes on the x and y axis are 5m apart, but 2m apart on the z axis...how would the formula work then?
Thanks for any help!

It can be extended to 3D relatively easily. It does require finding the "middle" of 3 values, there is a trick for that given that we have the minimum and maximum.
dx = absdiff(node.x, goal.x)
dy = absdiff(node.y, goal.y)
dz = absdiff(node.z, goal.z)
dmin = min(dx, dy, dz)
dmax = max(dx, dy, dz)
dmid = dx + dy + dz - dmin - dmax
This works for Python style integers and even for Java style int, for floats it can cause some rounding though.
Combine them like this:
return (D3 - D2) * dmin + (D2 - D1) * dmid + D1 * dmax

am i right?
Depth camera show us the depth of object.
The depth data is a diagonal distance from the center of camera.
It is not a z-axis distance which is moving control.
Therefore, it has to be calculating using depth, pixel X and Pixel Y.
I know that we don't know the pysical unit of

Signed angle between directed line segments

I have an algorithm in which I need to work out the signed angle (-180 to 180) between edges in a graph. I've done some research and found plenty of specific answers but can't figure out how to relate them to my situation (e.g. this question which uses atan2, however the OP wanted only positive angles)
I've tried implementing a few different ways (using atan2 or arccos) but I'm struggling to relate the examples to my specific problem. I've tried treating the edges as vectors but got strange results.
Given a graph with points (A, B, C, D, E), and the average of those points (avg)... how do I find the signed angle between one of those points (e.g. A) and the other points (e.g. B, C, D, E), taking the angle from the current origin (A) to the 'avg' point as equal to 0 degrees. Example below...
...in this example, the anti-clockwise angle from (A, avg) to (A, B) would be positive something (between 0 and 180), and the angle from (A, avg) to (A, E) would be negative something (between 0 and -180).
Ideally I want a formula which I could also apply to defining any of the points as the origin, for example taking point C as the origin.. the 'zero angle' would be (C, avg) and the angle between (C, avg) and (C, A) would be negative (0 to -180) and the angle between (C, avg) and (C, E) would be positive (0 to 180).
I haven't studied math beyond high-school so I find it hard to decipher equations with symbols I don't understand.
UPDATE: Thought I'd clean this up to make it more obvious what the conclusion was.
I made two small changes to the accepted answer, resulting in the below snippet:
def angle(vertex, start, dest):
AhAB = math.atan2((dest.y - vertex.y), (dest.x - vertex.x))
AhAO = math.atan2((start.y - vertex.y), (start.x - vertex.x))
AB = AhAB - AhAO
# in between 0-math.pi = do nothing, more than math.pi = +(-2 * math.pi), less than zero = do nothing
AB = math.degrees(AB + (-2 * math.pi if AB > math.pi else (2 * math.pi if AB < 0 - math.pi else 0)))
return AB
...the final one-liner may be a bit much to grok after a few months of not working on this, so I turned it into it's own function, taking the result of AB = AhAB - AhAO as it's argument...
def calc(ab):
if ab > math.pi:
return ab + (-2 * math.pi)
else:
if ab < 0 - math.pi:
return ab + (2 * math.pi)
else:
return ab + 0
I thought this was a little clearer to read, though more lines.
The final function in full:
def angle(vertex, start, dest):
"""Calculates the signed angle between two edges with the same origin.
Origin is the 'vertex' argument, 'start' is the bounding point of the edge to calculate the angle from.
Positively signed result means anti-clockwise rotation about the vertex."""
def calc_radians(ab):
if ab > math.pi:
return ab + (-2 * math.pi)
else:
if ab < 0 - math.pi:
return ab + (2 * math.pi)
else:
return ab + 0
AhAB = math.atan2((dest.y - vertex.y), (dest.x - vertex.x))
AhAO = math.atan2((start.y - vertex.y), (start.x - vertex.x))
res = calc_radians(AhAB - AhAO)
return math.degrees(res)
Note: The function assumes the three arguments will all be instances of a typical Point class with x and y attributes.
Also, the example graph above has only positive values, but I am fairly sure that this works with graphs that involve negative values too.

I read your problem statement as follows: given 2 points A and B, and a center O, find the angle A to B as the angle, positive if anticlockwise, between the vectors A→O and A→B.
If my premises are correct, then you can
find the angle between A→B and a horizontal, rightward line passing in A,
find the angle between A→O and a horizontal, rightward line passing in A,
find the angle A to B as the difference of said angles,
normalize the result range so that it's between -π and +π.
What I've said can be visualized as follows
or in code (assuming a Point class with attributes x and y)
AhAB = math.atan2((B.y-A.y), (B.x-A.x)) # -π < AhAB ≤ +π
AhAO = math.atan2((O.y-A.y), (O.x-A.x)) # -π < AhA) ≤ +π
AB = AhAB - AhAO # -2π < AB ≤ +2π
AB = AB + ( 2*math.pi if AB < math.pi else (-2*math.pi if AB> math.pi else 0))
Addendum
Here it is a small code example, the position of the points is just similar to what you can see in the picture
In [18]: from math import atan2, pi
In [21]: class Point():
...: def __init__(self, x, y):
...: self.x, self.y = x, y
...: def __repr__(self):
...: return '(%s, %s)'%(self.x, self.y)
In [22]: A = Point(0.0, 0.0)
In [23]: B = Point(-2.0, 2.0)
In [24]: O = Point(0.0, -3.0)
In [25]: AhAB = atan2((B.y-A.y), (B.x-A.x)) ; print(3/4, AhAB/pi)
0.75 0.75
In [26]: AhAO = atan2((O.y-A.y), (O.x-A.x)) ; print(-1/2, AhAO/pi)
-0.5 -0.5
In [27]: AB = AhAB - AhAO ; print(5/4, AB/pi)
1.25 1.25
In [28]: AB = AB + ( 2*pi if AB < pi else (-2*pi if AB> pi else 0)) ; print(AB/pi)
-0.75
In [29]:
The last line normalize your result AB to be in the correct range -π < AB ≤ π, adding or subtracting 2π that doesn't change the meaning of the measured angle.

The definition of positive and negative angles is heavily depending on the reference system or reference point. Despite of its 'correct' definition, the basic calculation can be pretty much done based on the slope between two points and the resulting angle of incline which can be calculated by applying the inverse tan to the slope.
Applying the inverse tan in programming can be a bit annoying since many programming languages provide two different functions for this:
arctan or atan which is implemented in Python's math.atan() or numpy.atan()
arctan2 or atan2 which is delivered by math.atan2() or numpy.atan2()
Both of these functions, regardless of the implementation in the math module or numpy package, return the calculated angle in radians which is basically based on the number Pi instead of degrees which makes some further conversion necessary. This can either be done manually or by applying a function like numpy.rad2deg(). To get a basic idea of the data points and to get some eye-balled estimation for the calculated results, I suggest plotting the data point by using matplotlib.
Glueing all the before-mentioned considerations into code can look like this:
import pandas as pd
import matplotlib
import numpy as np
%matplotlib inline
import matplotlib.pyplot as plt
# Define some sample data points
coords = {
'A': (1.5, 3.0),
'B': (3.0, 5.0),
'C': (5.5, 4.5),
'D': (5.8, 2.2),
'E': (2.8, 1.2)
}
# Extract data values from `coords` dict
values = np.array(list(coords.values()))
# Calculate the averaged point of all data points
avg = np.mean(values, axis=0)
# Plot sample data for better overview
for k, v in coords.items():
plt.plot(*v, marker='o', linestyle='')
plt.text(*v, k)
plt.plot(*avg, marker='o', linestyle='')
plt.text(*avg, 'avg')
plt.show()
# For further information about slope and angle of incline
# see Wikipedia (https://en.wikipedia.org/wiki/Slope).
# Calculating the angle from `avg` to each point. Please adopt
# to your own needs if needed for other pairs of points.
# Calculate the distance in x- and y-direction from each point to point `avg`
distances_x_y = (values - avg)
# Depending on your definition of the 'reference point' consider using
# distances_x_y = (avg - values)
# For further explanation on `atan` and `atan2` see
# https://stackoverflow.com/q/35749246/3991125 and
# https://en.wikipedia.org/wiki/Atan2 .
# Using a for loop instead of numpy's array/vectors is not very elegant,
# but easy to understand and therefore has potential for improvements.
# Calculate angle from point `avg` to each other point based on distances
angle_radians = np.array([np.arctan2(element[1], element[0]) for element in distances_x_y])
# since `.arctan2()` or `.arctan()` return the angle in radians,
# we need to convert to degrees
angle_degrees = np.rad2deg(angle_radians)
# print results
print(angle_degrees)

If you consider the coordinates x0=xavg-xA, y0=yavg-yA and x=xPoint-xA,y=yPoint-yA, the formula f(x,y) gives the signed angle that is positive as counter clockwise.
f(x,y)=pi()/2*((1+sign(x0))* (1-sign(y0^2))-(1+sign(x))* (1-sign(y^2)))
+pi()/4*((2+sign(x0))*sign(y0)-(2+sign(x))*sign(y))
+sign(x0*y0)*atan((abs(x0)-abs(y0))/(abs(x0)+abs(y0)))
-sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))