我试图在Python中找到两个圆之间的交点(使用Matplotlib),但无法获得任何值。
为此,我为每个单独的圆创建了X和Y的列表(当绘制圆时,Matplotlib将第一个参数作为X值,第二个参数作为Y值),然后相应地将列表相交(例如,circle1 x值与circle2 x值)。
import numpy
import math
import matplotlib.pyplot as plt
import random
def origin_circle():
global x_points
global y_points
global r
global n
r=1
n=2**16
x_points=[(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
y_points=[(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
def new_circle(x_offset, y_offset):
global x_points1
global y_points1
x_points1=[x_offset+(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
y_points1=[y_offset+(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
origin_circle()
new_center= random.randint(0, len(x_points))
x_offset = x_points[new_center]
y_offset = y_points[new_center]
new_circle(x_offset, y_offset)
print(set(x_points1).intersection(set(x_points)))
print(set(y_points1).intersection(set(y_points)))
我希望得到返回值,但是返回的集合是空的。
发布于 2019-04-24 02:26:04
如果你正在处理圆,获得交集的正确方法是使用一些代数。有四种可能的情况:没有交集,一个交集(相切),两个交集,和无限的交集(它们是同一个圆).Let我们专注于两个交集的情况。
从https://math.stackexchange.com/a/256123/647423中,您可以获得一个线性方程,该方程沿着通过两个交点的直线将x与y联系起来:
−2x(x1center−x2center)−2y(y1center−y2center) = (r1)^2−(r2)^2−((x1center)^2−(x2center)^2)−((y1center)^2−(y2center)^2).
从这里你可以得到一个关于x的y的公式,然后将y代入你的一个圆公式中以获得x的二次方。如果你不想实现一个二次方程求解器,你可以像这样使用numpy.roots:
root_array = np.roots(quadratic_coeff, linear_coeff, constant_coef)
发布于 2019-11-12 03:34:09
数学/几何问题的语言错误。这是它在更合适的语言(WL)中的样子
Circle @@@ Thread @ {RandomReal[{-1,1},{3,2}], RandomReal[{.5,1},3]} //
Graphics[{
#, Red,
RegionIntersection @@@ #~Subsets~{2}
}]&
发布于 2021-04-29 16:19:04
上面绘制交点的代码没有正确地绘制交点。我调整了代码以绘制它们,如下所示:
import matplotlib.pyplot as plt
import math
def get_intersections(x0, y0, r0, x1, y1, r1):
# circle 1: (x0, y0), radius r0
# circle 2: (x1, y1), radius r1
d=math.sqrt((x1-x0)**2 + (y1-y0)**2)
# non intersecting
if d > r0 + r1 :
return {}
# One circle within other
if d < abs(r0-r1):
return {}
# coincident circles
if d == 0 and r0 == r1:
return {}
else:
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
# intersection circles
x0, y0 = 0, 0
r0 = 5
x1, y1 = 2, 2
r1 = 5
# intersecting with (x1, y1) but not with (x0, y0)
x2, y2 = -1,0
r2 = 2.5
circle1 = plt.Circle((x0, y0), r0, color='b', fill=False)
circle2 = plt.Circle((x1, y1), r1, color='b', fill=False)
circle3 = plt.Circle((x2, y2), r2, color='b', fill=False)
fig = plt.figure(figsize = (10, 10))
plt.grid(True)
ax = fig.add_subplot(111)
ax.set_xlim((-10, 10))
ax.set_ylim((-10, 10))
ax.add_artist(circle1)
ax.add_artist(circle2)
ax.add_artist(circle3)
intersections1 = get_intersections(x0, y0, r0, x1, y1, r1)
print(intersections1)
if len(intersections1) > 0:
print(intersections3)
i_x3, i_y3, i_x4, i_y4 = intersections1
#plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)
intersections2 = get_intersections(x0, y0, r0, x2, y2, r2)
print(intersections2)
if len(intersections2) > 0:
i_x3, i_y3, i_x4, i_y4 = intersections2
plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)
intersections3 = get_intersections(x1, y1, r1, x2, y2, r2)
if len(intersections3) > 0:
print(intersections3)
i_x3, i_y3, i_x4, i_y4 = intersections3
#plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)
plt.gca().set_aspect('equal', adjustable='box')
输出如图所示:
https://stackoverflow.com/questions/55816902
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