Related
So what I want in theory is pretty simple. I want create something where you:
have 2 lines that the user draws
have code that take the first line and take the 2 points from it
take the other lines 2 points
takes the first point from the first line and compares it to the 2 points from (let's call it) line 2
does the same for the 2nd point of line 1
and draws line connecting the points the first point of line 1 to the closest point of line 2, then drawing the 2nd point of line one to the last one (of line 2).
Sounds a bit complicated, but probably because I'm bad a explaining things.
The code I have for this is here
I also just want to add that the function "draw_line()" just takes the inputs of "start_pos", and "end_pos" so it doesn't matter what order the inputs are put in. Also "y2, y1" is the starting point or, 1st, and z is the 2nd.
def find_closer_point(self, origin_point, point1, point2):
# find the distance between both points and then choose the closest one to it
# the variables
x2, x1 = origin_point
y2, y1 = point1
z2, z1 = point2
# the distance formula
calc1 = ((x2 - x1) ^ 2) + ((y2 - y1) ^ 2)
calc2 = ((x2 - x1) ^ 2) + ((z2 - z1) ^ 2)
# to compare the distances (which seems to be the hard part)
try:
if math.sqrt(calc1) > math.sqrt(calc2):
self.draw_line((x2, x1), (z2, z1))
except ValueError:
new_calc1 = math.sqrt(abs(calc1))
new_calc2 = math.sqrt(abs(calc2))
if new_calc1 > new_calc2:
self.draw_line((x2, x1), (z2, z1))
if new_calc1 < new_calc2:
self.draw_line((x2, x1), (y2, y1))
I can't seem to find what's wrong. Also, if I need to provide more code as context or if something else is wrong then just ask.
Use the Dot product to find the point with the pointe which is "nearer" in a certain direction
def dot(p1, p2):
return p1[0]*p2[0] + p1[1]*p2[1]
if dot(origin_point, point1) < dot(origin_point, point2):
self.draw_line(origin_point, point1)
else:
self.draw_line(origin_point, point2)
The Dot product of 2 vectors is equal the cosine of the angle between the 2 vectors multiplied by the magnitude (length) of both vectors.
dot(A, B) == | A | * | B | * cos(angle_A_B)
This follows, that the dot product of 2 unit vectors is equal the cosine of the angle between the 2 vectors, because the length of a unit vector is 1.
uA = normalize(A)
uB = normalize(B)
cos(angle_A_B) == dot(uA, uB)
So I'm not that well versed in linear algebra so I'm struggling with this.
I have a unit vectors v. I want to find two angles(angle 1, rotation around x-axis, and angle 2, rotation around z-axis) such that when I rotate v by them it aligns the vector v with the y-axis. From this question I have a function that can find the angle between arbitrary vectors and returns a rotation. But this function returns 3 angles. Essentially there is an infinite number of 3d rotation that aligns v with the y-axis so I want the two unique angles.
This the code I have now, it requires numpy and scipy:
import numpy as np
import random
from scipy.spatial.transform import Rotation as R
def rotation_from_unit_vectors(a, b):
v = np.cross(a, b)
c = np.dot(a, b)
s = np.linalg.norm(v)
kmat = np.array([[0, -v[2], v[1]], [v[2], 0, -v[0]], [-v[1], v[0], 0]])
rotation_matrix = np.eye(3) + kmat + kmat.dot(kmat) * ((1 - c) / (s ** 2))
return R.from_matrix(rotation_matrix)
y_axis = np.asarray([0.0, 1.0, 0.0])
alpha = random.uniform(0, 10)
beta = random.uniform(0, 10)
gamma = random.uniform(0, 10)
v = np.asarray([alpha, beta, gamma])
v = v / np.linalg.norm(v)
r = rotation_from_unit_vectors(v, y_axis)
print(r.as_euler('xyz', degrees = True))
print(r.apply(v))
Taking advantage of the fixed target alignment, this can be done in a straightforward manner with just trigonometry:
import math
def to_y(x,y,z):
rx=-math.atan2(z,y) # or +math.atan2(z,-y)
y2=y*math.cos(rx)-z*math.sin(rx) # -> (x,y2,0)
return rx,math.atan2(x,y2)
The rotations are defined as counterclockwise when looking at the origin from +x or +z (the right-hand rule); the rotation direction is always that with the smaller magnitude, but it may be possible to find a physically smaller rotation as indicated in the comment. Note that the input need not be normalized, and NaN is never produced (unless it appears in the input).
Hum, non-standard problem, required thinking a little.
Given v1 and v2 you want to rotate_z(rotate_x(v1, alpha), beta) to be on the same direction as v2.
We know that the aligned vector can be obtained by simply scaling scaling v2, this will gives x1,y3,z3 = v3 = v2 * |v1| / |v2|. Since rotation around z-axis, does not affect the z coordinate, we can determine alpha such that the z coordinate of rotate_x(v1, alpha) equals z3. After that we determine the angle beta to align place the X and Y coordinates properly
import numpy as np
def alignment_angles(v1, v2):
x1,y1,z1 = v1 # initial vector
x2,y2,z2 = v2 # reference vector
# magnitude of the two vectors
r1 = np.sqrt(x1**2 + y1**2 + z1**2)
r2 = np.sqrt(x2**2 + y2**2 + z2**2)
# this will be the result when aligning v1 to v2
# it has the magnitude of v1 and the direction of v2
x3,y3,z3 = x2*r1/r2, y2*r1/r2, z2*r1/r2
# the rotation in x must set the z coordinate to the
# final value, since the rotation over the z axis will
# not affect the z coordinate (this have two solutions)
rho1 = np.sqrt(z1**2 + y1**2)
if(abs(z3 / rho1) > 1):
raise ValueError('Cannot align these vectors')
alpha = np.arcsin(z3 / rho1) - np.arctan2(z1, y1);
# apply the rotation to make easier to calcualte the next stage
y1, z1 = (y1 * np.cos(alpha) - z1 * np.sin(alpha),
y1 * np.sin(alpha) + z1 * np.cos(alpha))
np.allclose(rho1, np.sqrt(z1**2 + y1**2))
#assert(np.allclose(z1, z3))
# now it is just a matter of aligning (x1, y1) to (x3, y3)
beta = np.arctan2(y3, x3) - np.arctan2(y1, x1)
x1, y1 = (x1 * np.cos(beta) - y1 * np.sin(beta),
x1 * np.sin(beta) + y1 * np.cos(beta))
# ensure the fotated v1 was correctly aligned
assert(np.allclose([x1, y1, z1], [x3, y3, z3]))
return alpha, beta
Then you just call
alignment_angles((1,2,3), (3,4,5))
or you can also use numpy arrays with 3 rows.
Initially I thought it would be an application of spherical coordinates, that would be the case if the axis for the second rotation was the z-axis rotated accordingly to the first rotation.
Edit
There are some vectors that cannot be aligned with a rotation on x and a rotation on y.
Suppose you want to align the vector v1 = (1, 0, 0) to the vector v2 = (0, 0, 1) the rotation in x will not affect v1, it will always point in the direction x, then when you rotate around the z axis it will always be on the XY plan.
The example you gave was really giving the wrong answer because asin is not injective.
I changed the function to raise a value error when you cannot align the given vectors.
I am not that experienced in python but improving it thanks to this community! I desperately need a function which takes the input and gives the ouput below:
Input:
1- Latitude/longitude coordinates of the center of circle 1 (e.g. (50.851295, 5.667969) )
2- The radius of circle 1 in meters (e.g. 200)
3- Latitude/longitude coordinates of the center of circle 2 (e.g. (50.844101, 5.725889) )
4- The radius of circle 2 in meters (e.g. 300)
Output: Possible output examples can be;
The intersection points are (50.848295, 5.707969) and (50.849295, 5.717969)
The circles are overlapping
The circles are tangential and the intersection point is (50.847295, 5.705969)
The circles do not intersect
I have examined the similar topics in this platform, other platforms, libraries, tried to combine different solutions but couldn't succeed. Any help is much appreciated!
EDIT:
The problem is solved many thanks to Ture Pålsson who commented below and directed me to whuber's brilliant work in this link https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles Based on that work, I wrote the code below and as far as I tested it works. I want to share it here in case someone might find it helpful. Any feedback is appreciated.
'''
FINDING THE INTERSECTION COORDINATES (LAT/LON) OF TWO CIRCLES (GIVEN THE COORDINATES OF THE CENTER AND THE RADII)
Many thanks to Ture Pålsson who directed me to the right source, the code below is based on whuber's brilliant logic and
explanation here https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles
The idea is that;
1. The points in question are the mutual intersections of three spheres: a sphere centered beneath location x1 (on the
earth's surface) of a given radius, a sphere centered beneath location x2 (on the earth's surface) of a given radius, and
the earth itself, which is a sphere centered at O = (0,0,0) of a given radius.
2. The intersection of each of the first two spheres with the earth's surface is a circle, which defines two planes.
The mutual intersections of all three spheres therefore lies on the intersection of those two planes: a line.
Consequently, the problem is reduced to intersecting a line with a sphere.
Note that "Decimal" is used to have higher precision which is important if the distance between two points are a few
meters.
'''
from decimal import Decimal
from math import cos, sin, sqrt
import math
import numpy as np
def intersection(p1, r1_meter, p2, r2_meter):
# p1 = Coordinates of Point 1: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r1_meter = Radius of circle 1 in meters
# p2 = Coordinates of Point 2: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r2_meter = Radius of circle 2 in meters
'''
1. Convert (lat, lon) to (x,y,z) geocentric coordinates.
As usual, because we may choose units of measurement in which the earth has a unit radius
'''
x_p1 = Decimal(cos(math.radians(p1[1]))*cos(math.radians(p1[0]))) # x = cos(lon)*cos(lat)
y_p1 = Decimal(sin(math.radians(p1[1]))*cos(math.radians(p1[0]))) # y = sin(lon)*cos(lat)
z_p1 = Decimal(sin(math.radians(p1[0]))) # z = sin(lat)
x1 = (x_p1, y_p1, z_p1)
x_p2 = Decimal(cos(math.radians(p2[1]))*cos(math.radians(p2[0]))) # x = cos(lon)*cos(lat)
y_p2 = Decimal(sin(math.radians(p2[1]))*cos(math.radians(p2[0]))) # y = sin(lon)*cos(lat)
z_p2 = Decimal(sin(math.radians(p2[0]))) # z = sin(lat)
x2 = (x_p2, y_p2, z_p2)
'''
2. Convert the radii r1 and r2 (which are measured along the sphere) to angles along the sphere.
By definition, one nautical mile (NM) is 1/60 degree of arc (which is pi/180 * 1/60 = 0.0002908888 radians).
'''
r1 = Decimal(math.radians((r1_meter/1852) / 60)) # r1_meter/1852 converts meter to Nautical mile.
r2 = Decimal(math.radians((r2_meter/1852) / 60))
'''
3. The geodesic circle of radius r1 around x1 is the intersection of the earth's surface with an Euclidean sphere
of radius sin(r1) centered at cos(r1)*x1.
4. The plane determined by the intersection of the sphere of radius sin(r1) around cos(r1)*x1 and the earth's surface
is perpendicular to x1 and passes through the point cos(r1)x1, whence its equation is x.x1 = cos(r1)
(the "." represents the usual dot product); likewise for the other plane. There will be a unique point x0 on the
intersection of those two planes that is a linear combination of x1 and x2. Writing x0 = ax1 + b*x2 the two planar
equations are;
cos(r1) = x.x1 = (a*x1 + b*x2).x1 = a + b*(x2.x1)
cos(r2) = x.x2 = (a*x1 + b*x2).x2 = a*(x1.x2) + b
Using the fact that x2.x1 = x1.x2, which I shall write as q, the solution (if it exists) is given by
a = (cos(r1) - cos(r2)*q) / (1 - q^2),
b = (cos(r2) - cos(r1)*q) / (1 - q^2).
'''
q = Decimal(np.dot(x1, x2))
if q**2 != 1 :
a = (Decimal(cos(r1)) - Decimal(cos(r2))*q) / (1 - q**2)
b = (Decimal(cos(r2)) - Decimal(cos(r1))*q) / (1 - q**2)
'''
5. Now all other points on the line of intersection of the two planes differ from x0 by some multiple of a vector
n which is mutually perpendicular to both planes. The cross product n = x1~Cross~x2 does the job provided n is
nonzero: once again, this means that x1 and x2 are neither coincident nor diametrically opposite. (We need to
take care to compute the cross product with high precision, because it involves subtractions with a lot of
cancellation when x1 and x2 are close to each other.)
'''
n = np.cross(x1, x2)
'''
6. Therefore, we seek up to two points of the form x0 + t*n which lie on the earth's surface: that is, their length
equals 1. Equivalently, their squared length is 1:
1 = squared length = (x0 + t*n).(x0 + t*n) = x0.x0 + 2t*x0.n + t^2*n.n = x0.x0 + t^2*n.n
'''
x0_1 = [a*f for f in x1]
x0_2 = [b*f for f in x2]
x0 = [sum(f) for f in zip(x0_1, x0_2)]
'''
The term with x0.n disappears because x0 (being a linear combination of x1 and x2) is perpendicular to n.
The two solutions easily are t = sqrt((1 - x0.x0)/n.n) and its negative. Once again high precision
is called for, because when x1 and x2 are close, x0.x0 is very close to 1, leading to some loss of
floating point precision.
'''
if (np.dot(x0, x0) <= 1) & (np.dot(n,n) != 0): # This is to secure that (1 - np.dot(x0, x0)) / np.dot(n,n) > 0
t = Decimal(sqrt((1 - np.dot(x0, x0)) / np.dot(n,n)))
t1 = t
t2 = -t
i1 = x0 + t1*n
i2 = x0 + t2*n
'''
7. Finally, we may convert these solutions back to (lat, lon) by converting geocentric (x,y,z) to geographic
coordinates. For the longitude, use the generalized arctangent returning values in the range -180 to 180
degrees (in computing applications, this function takes both x and y as arguments rather than just the
ratio y/x; it is sometimes called "ATan2").
'''
i1_lat = math.degrees( math.asin(i1[2]))
i1_lon = math.degrees( math.atan2(i1[1], i1[0] ) )
ip1 = (i1_lat, i1_lon)
i2_lat = math.degrees( math.asin(i2[2]))
i2_lon = math.degrees( math.atan2(i2[1], i2[0] ) )
ip2 = (i2_lat, i2_lon)
return [ip1, ip2]
elif (np.dot(n,n) == 0):
return("The centers of the circles can be neither the same point nor antipodal points.")
else:
return("The circles do not intersect")
else:
return("The centers of the circles can be neither the same point nor antipodal points.")
'''
Example: the output of below is [(36.989311051533505, -88.15142628069133), (38.2383796094578, -92.39048549120287)]
intersection_points = intersection((37.673442, -90.234036), 107.5*1852, (36.109997, -90.953669), 145*1852)
print(intersection_points)
'''
Depending on the precision you need, you may or may not consider the Earth as a sphere. In the second case, calculations become more complex.
The best option for precise measurements when the radius is small (as in your example) is to use a projection (UTM for example) and then apply the common flat euclidean calculations.
Let's first copy the flat circle intersection function from https://stackoverflow.com/a/55817881/2148416:
def circle_intersection(x0, y0, r0, x1, y1, r1):
d = math.sqrt((x1 - x0) ** 2 + (y1 - y0) ** 2)
if d > r0 + r1: # non intersecting
return None
if d < abs(r0 - r1): # one circle within other
return None
if d == 0 and r0 == r1: # coincident circles
return None
a = (r0 ** 2 - r1 ** 2 + d ** 2) / (2 * d)
h = math.sqrt(r0 ** 2 - a ** 2)
x2 = x0 + a * (x1 - x0) / d
y2 = y0 + a * (y1 - y0) / d
x3 = x2 + h * (y1 - y0) / d
y3 = y2 - h * (x1 - x0) / d
x4 = x2 - h * (y1 - y0) / d
y4 = y2 + h * (x1 - x0) / d
return (x3, y3), (x4, y4)
The precise calculation for a small radius (up to a few kilometers) can be done in UTM coordinates with the help of the utm library. It handles all the complications regarding the fact the the Earth is more an ellipsoid than a sphere:
import utm
def geo_circle_intersection(latlon0, radius0, latlon1, radius1):
# Convert lat/lon to UTM
x0, y0, zone, letter = utm.from_latlon(latlon0[0], latlon0[1])
x1, y1, _, _ = utm.from_latlon(latlon1[0], latlon1 [1], force_zone_number=zone)
# Calculate intersections in UTM coordinates
a_utm, b_utm = circle_intersection(x0, y0, r0, x1, y1, r1)
# Convert intersections from UTM back to lat/lon
a = utm.to_latlon(a_utm[0], a_utm[1], zone, letter)
b = utm.to_latlon(b_utm[0], b_utm[1], zone, letter)
return a, b
Using your example (with slightly larger radii):
>>> p0 = 50.851295, 5.667969
>>> r0 = 2000
>>> p1 = 50.844101, 5.725889
>>> r1 = 3000
>>> a, b = geo_circle_intersection(p0, r0, p1, r1)
>>> print(a)
(50.836848562566004, 5.684869539768468)
>>> print(b)
(50.860635308778285, 5.692236858407678)
I have generated a random point named y0=(a,b) in xy-plane , How can I generate another random point (x,y) 10 steps apart from y0?
note: by 10 steps apart from the firt point I don't mean the Euclidean distance. I mean the number of steps on lattice between the two point (a,b) and (x,y) which is given by |x-a|+|y-b|=10
My attempt(sometimes gives wrong result).
import random
y0=(random.randint(0,50),random.randint(0,50))# here I generated the first point.
y=random.randint(0,50)
# I used the formula |x-a|+|y-b|=10.
x=(10 -abs(y-y0[1]))+y0[0] or x=-(10 -abs(y-y0[1]))+y0[0]
x0=(x,y)
Let's say you have a point (x, y)
create another random point anywhere on the plane: (x1, y2) = (random(), random())
take the vector from your point to the new point: (vx, vy) = (x1-x, y1-y)
get the length l of the vector: l = sqrt(vx * vx + vy * vy)
use l to normalise the vector (so it has a length of 1): (vx, vy) = (vx / l, vy / l)
make the vector 10 steps long: (vx, vy) = (vx * 10, vy * 10)
add it to your original point to get to the desired point: (x1, y2) = (x + vx, y + vy)
voilá :)
from random import random
from math import sqrt
# Deviation
dev = 50
# Required distance between points
l = 10
if __name__ == '__main__':
# First random point
x0, y0 = dev*random(), dev*random()
# Second point
x1 = dev*random()
y1 = y0 + sqrt(l**2 - (x1 - x0)**2)
# Output
print "First point (%s, %s)" % (x0, y0)
print "Second point (%s, %s)" % (x1, y1)
print "Distance: %s" % (sqrt((x1 - x0)**2 + (y1 - y0)**2))
Let's say that your new point (x, y) is on a cercle of radius 10 and center (x0, y0). The random component is the angle.
import math as m
# radius of the circle
r = 10
# create random angle and compute coordinates of the new point
theta = 2*m.pi*random.random()
x = x0 + r*m.cos(theta)
y = y0 + r*m.sin(theta)
# test if the point created is in the domain [[0,50], [0, 50]] (see comments of PM2Ring)
while not ( 0<=x<=50 and 0<=y<=50 ) :
# update theta: add pi/2 until the new point is in the domain (see HumanCatfood's comment)
theta += 0.5*m.pi
x = x0 + r*m.cos(theta)
y = y0 + r*m.sin(theta)
So, you got the formula d=d1+d2=|x-x0|+|y-y0| , for d=10
Let's examine what's going on with this formula:
Let's say we generate a random point P at (0,0)
Let's say we generate y=random.randint(0,50) and let's imagine the value is 50.
What does this mean?
d1=|x-p[0]|=50 and your original formula is d=d1+d2=|x-x0|+|y-y0|, so
that means d2=|y-y0|=10-50 and d2=|y-y0|=-40. Is this possible? Absolutely not! An absolute value |y-y0| will always be positive, that's why your formula won't work for certain random points, you need to make sure (d-d1)>0, otherwise your equation won't have solution.
If you wanted to consider Euclidean distance you just need to generate random points in a circle where your original point will be the center, something like this will do:
import random
import math
def random_point(p, r=10):
theta = 2 * math.pi * random.random()
return (p[0] + r * math.cos(theta), p[1] + r * math.sin(theta))
If you draw a few random points you'll see more and more how the circle shape is created, let's try with N=10, N=50, N=1000:
Now, it seems you need the generated circle to be constrained at certain area region. One possible choice (not the most optimal though) would be generating random points till they meet those constraints, something like this would do:
def random_constrained_point(p, r=10, x_limit=50, y_limit=50):
i = 0
MAX_ITERATIONS = 100
while True:
x0, y0 = random_point(p, r)
if (0 <= x0 <= x_limit and 0 <= y0 <= y_limit):
return (x0, y0)
if i == MAX_ITERATIONS:
return p
i += 1
Once you got this, it's interesting to check what shape is created when you increase more and more the circle radius (10,20,50):
As you can see, your generated random constrained points will form a well_defined subarc.
this code generate a random point xy-plane named y0 then generate another point x0 10 steps apart from y0 in taxi distance .
------- begining of the code--------
import random
y0=(random.randint(0,50),random.randint(0,50))
while True:
y=random.randint(0,50)
x=(10 -abs(y-y0[1]))+y0[0]
if (abs(x-y0[0])+abs(y-y0[1]))==10:
x0=(x,y)
break
abs(x)+abs(y)=10 defines a square, so all you need to do is pick a random value along the perimeter of the square (40 units long), and map that random distance back to your x,y coordinate pair.
Something like (untested):
x = random.randint(-10,9)
y = 10 - abs(x)
if (random.randint(0,1) == 0):
x = -x
y = -y
x = x + y0[0]
y = y + y0[1]
x0=(x,y)
Clipping the x range that way ensures that all points are picked uniformly. Otherwise you can end up with (-10,0) and (10,0) having twice the chance of being picked compared to any other coordinate.
I have a polygon consists of lots of points. I want to find the intersection of the polygon and a circle. Providing the circle center of [x0,y0] and radius of r0, I have wrote a rough function to simply solve the quadratic equation of the circle and a line. But what about the efficiency of find the intersection of every line segment of the polygon one by one? Is there more efficient way?
I know sympy already provide the feature to get the intersections of different geometry. But also what about the efficiency of external library like sympy compared to calculate it by my own function, if I want to deal with lots of polygons?
def LineIntersectCircle(p,lsp,lep):
# p is the circle parameter, lsp and lep is the two end of the line
x0,y0,r0 = p
x1,y1 = lsp
x2,y2 = esp
if x1 == x2:
if abs(r0) >= abs(x1 - x0):
p1 = x1, y0 - sqrt(r0**2 - (x1-x0)**2)
p2 = x1, y0 + sqrt(r0**2 - (x1-x0)**2)
inp = [p1,p2]
# select the points lie on the line segment
inp = [p for p in inp if p[1]>=min(y1,y2) and p[1]<=max(y1,y2)]
else:
inp = []
else:
k = (y1 - y2)/(x1 - x2)
b0 = y1 - k*x1
a = k**2 + 1
b = 2*k*(b0 - y0) - 2*x0
c = (b0 - y0)**2 + x0**2 - r0**2
delta = b**2 - 4*a*c
if delta >= 0:
p1x = (-b - sqrt(delta))/(2*a)
p2x = (-b + sqrt(delta))/(2*a)
p1y = k*x1 + b0
p2y = k*x2 + b0
inp = [[p1x,p1y],[p2x,p2y]]
# select the points lie on the line segment
inp = [p for p in inp if p[0]>=min(x1,x2) and p[0]<=max(x1,x2)]
else:
inp = []
return inp
I guess maybe your question is about how to in theory do this in the fastest manner. But if you want to do this quickly, you should really use something which is written in C/C++.
I am quite used to Shapely, so I will provide an example of how to do this with this library. There are many geometry libraries for python. I will list them at the end of this answer.
from shapely.geometry import LineString
from shapely.geometry import Point
p = Point(5,5)
c = p.buffer(3).boundary
l = LineString([(0,0), (10, 10)])
i = c.intersection(l)
print i.geoms[0].coords[0]
(2.8786796564403576, 2.8786796564403576)
print i.geoms[1].coords[0]
(7.121320343559642, 7.121320343559642)
What is a little bit counter intuitive in Shapely is that circles are the boundries of points with buffer areas. This is why I do p.buffer(3).boundry
Also the intersection i is a list of geometric shapes, both of them points in this case, this is why I have to get both of them from i.geoms[]
There is another Stackoverflow question which goes into details about these libraries for those interested.
SymPy
CGAL Python bindings
PyEuclid
PythonOCC
Geometry-simple
EDIT because comments:
Shapely is based on GEOS (trac.osgeo.org/geos) which is built in C++ and considerably faster than anything you write natively in python. SymPy seems to be based on mpmath (mpmath.org) which also seems to be python, but seems to have lots of quite complex math integrated into it. Implementing that yourself may require a lot of work, and will probably not be as fast as GEOS C++ implementations.
Here's a solution that computes the intersection of a circle with either a line or a line segment defined by two (x, y) points:
def circle_line_segment_intersection(circle_center, circle_radius, pt1, pt2, full_line=True, tangent_tol=1e-9):
""" Find the points at which a circle intersects a line-segment. This can happen at 0, 1, or 2 points.
:param circle_center: The (x, y) location of the circle center
:param circle_radius: The radius of the circle
:param pt1: The (x, y) location of the first point of the segment
:param pt2: The (x, y) location of the second point of the segment
:param full_line: True to find intersections along full line - not just in the segment. False will just return intersections within the segment.
:param tangent_tol: Numerical tolerance at which we decide the intersections are close enough to consider it a tangent
:return Sequence[Tuple[float, float]]: A list of length 0, 1, or 2, where each element is a point at which the circle intercepts a line segment.
Note: We follow: http://mathworld.wolfram.com/Circle-LineIntersection.html
"""
(p1x, p1y), (p2x, p2y), (cx, cy) = pt1, pt2, circle_center
(x1, y1), (x2, y2) = (p1x - cx, p1y - cy), (p2x - cx, p2y - cy)
dx, dy = (x2 - x1), (y2 - y1)
dr = (dx ** 2 + dy ** 2)**.5
big_d = x1 * y2 - x2 * y1
discriminant = circle_radius ** 2 * dr ** 2 - big_d ** 2
if discriminant < 0: # No intersection between circle and line
return []
else: # There may be 0, 1, or 2 intersections with the segment
intersections = [
(cx + (big_d * dy + sign * (-1 if dy < 0 else 1) * dx * discriminant**.5) / dr ** 2,
cy + (-big_d * dx + sign * abs(dy) * discriminant**.5) / dr ** 2)
for sign in ((1, -1) if dy < 0 else (-1, 1))] # This makes sure the order along the segment is correct
if not full_line: # If only considering the segment, filter out intersections that do not fall within the segment
fraction_along_segment = [(xi - p1x) / dx if abs(dx) > abs(dy) else (yi - p1y) / dy for xi, yi in intersections]
intersections = [pt for pt, frac in zip(intersections, fraction_along_segment) if 0 <= frac <= 1]
if len(intersections) == 2 and abs(discriminant) <= tangent_tol: # If line is tangent to circle, return just one point (as both intersections have same location)
return [intersections[0]]
else:
return intersections
A low cost spacial partition might be to divide the plane into 9 pieces
Here is a crappy diagram. Imagine, if you will, that the lines are just touching the circle.
| |
__|_|__
__|O|__
| |
| |
8 of the areas we are interested in are surrounding the circle. The square in the centre isn't much use for a cheap test, but you can place a square of r/sqrt(2) inside the circle, so it's corners just touch the circle.
Lets label the areas
A |B| C
__|_|__
D_|O|_E
| |
F |G| H
And the square of r/sqrt(2) in the centre we'll call J
We will call the set of points in the centre square shown in the diagram that aren't in J, Z
Label each vertex of the polygon with it's letter code.
Now we can quickly see
AA => Outside
AB => Outside
AC => Outside
...
AJ => Intersects
BJ => Intersects
...
JJ => Inside
This can turned into a lookup table
So depending on your dataset, you may have saved yourself a load of work. Anything with an endpoint in Z will need to be tested however.
I think that the formula you use to find the coordinates of the two intersections cannot be optimized further. The only improvement (which is numerically important) is to distinguish the two cases: |x_2-x_1| >= |y_2-y_1| and |x_2-x1| < |y_2-y1| so that the quantity k is always between -1 and 1 (in your computation you can get very high numerical errors if |x_2-x_1| is very small). You can swap x-s and y-s to reduce one case to the other.
You could also implement a preliminary check: if both endpoints are internal to the circle there are no intersection. By computing the squared distance from the points to the center of the circle this becomes a simple formula which does not use the square root function. The other check: "whether the line is outside the circle" is already implemented in your code and corresponds to delta < 0. If you have a lot of small segments these two check should give a shortcut answer (no intersection) in most cases.