[英]Detect points upstream and downstream of an intersection between two curves
I have two curves, defined by我有两条曲线,定义为
X1=[9, 10.5, 11, 12, 12, 11, 10, 8, 7, 7]
Y1=[-5, -3.5, -2.5, -0.7, 1, 3, 4, 5, 5, 5]
X2=[5, 7, 9, 9.5, 10, 11, 12]
Y2=[-2, 4, 1, 0, -0.5, -0.7, -3]
They intersect each other它们相互交叉
and by a function which is written in the system code I am using, I can have the coordinates of the intersection.并且通过我正在使用的系统代码中编写的函数,我可以获得交点的坐标。
loop1=Loop([9, 10.5, 11, 12, 12, 11, 10, 8, 7, 7],[-5, -3.5, -2.5, -0.7, 1, 3, 4, 5, 5, 5])
loop2=Loop([5, 7, 9, 9.5, 10, 11, 12], [-2, 4, 1, 0, -0.5, -0.7, -3])
x_int, y_int = get_intersect(loop1,loop2)
Intersection = [[],[]]
Intersection.append(x_int)
Intersection.append(y_int)
for both curves, I need to find the points which are upstream and downstream the intersection identified by (x_int, y_int).对于两条曲线,我需要找到由 (x_int, y_int) 标识的交点上游和下游的点。
What I tried is something like:我试过的是这样的:
for x_val, y_val, x, y in zip(Intersection[0], Intersection[1], loop1[0], loop1[1]):
if abs(x_val - x) < 0.5 and abs(y_val - y) < 0.5:
print(x_val, x, y_val, y)
The problem is that the result is extremely affected by the delta that I decide (0.5 in this case) and this gives me wrong results especially if I work with more decimal numbers (which is actually my case).问题是结果受到我决定的 delta 的极大影响(在这种情况下为 0.5),这给了我错误的结果,特别是如果我使用更多的十进制数(这实际上是我的情况)。
How can I make the loop more robust and actually find all and only the points which are upstream and downstream the intersection?我怎样才能使循环更加健壮并实际找到所有且仅找到交叉点上游和下游的点?
Many thanks for your help非常感谢您的帮助
TL;TR: loop over poly line segments and test if the intersection is betwwen the segment end points . TL;TR:循环多段线段并测试交点是否在段端点之间。
A more robust (than "delta" in OP) approach is to find a segment of the polyline, which contains the intersection (or given point in general).更强大(比 OP 中的“delta”)方法是找到折线的一段,其中包含交点(或一般给定的点)。 This segment should IMO be part of the get_intersect
function, but if you do not have access to it, you have to search the segment yourself.该段应 IMO 成为get_intersect
函数的一部分,但如果您无权访问它,则必须自己搜索该段。
Because of roundoff errors, the given point does not exactly lie on the segment, so you still have some tol
parameter, but the results should be "almost-insensitive" to its (very low) value.由于舍入误差,给定的点并不完全位于线段上,因此您仍然有一些tol
参数,但结果应该对其(非常低的)值“几乎不敏感”。
The approach uses simple geometry, namely dot product and cross product and their geometric meaning:该方法使用简单的几何,即点积和叉积及其几何含义:
a
and b
divided by |a|
向量a
和b
点积除以|a|
is projection (length) of b
onto the direction of a
.是的投影(长度) b
到的方向a
。 Once more dividing by |a|
再次除以|a|
normalizes the value to the range [0;1]
将值归一化到范围[0;1]
a
and b
is the area of the parallelogram having a and b as sides . a
和b
叉积是a和b为边的平行四边形的面积。 Dividing it by square of length make it some dimensionless factor of distance.将其除以长度的平方使其成为距离的无量纲因子。 If a point lies exactly on the segment, the cross product is zero.如果点正好位于线段上,则叉积为零。 But a small tolerance is needed for floating point numbers.但是浮点数需要一个小的容差。X1=[9, 10.5, 11, 12, 12, 11, 10, 8, 7, 7]
Y1=[-5, -3.5, -2.5, -0.7, 1, 3, 4, 5, 5, 5]
X2=[5, 7, 9, 9.5, 10, 11, 12]
Y2=[-2, 4, 1, 0, -0.5, -0.7, -3]
x_int, y_int = 11.439024390243903, -1.7097560975609765
def splitLine(X,Y,x,y,tol=1e-12):
"""Function
X,Y ... coordinates of line points
x,y ... point on a polyline
tol ... tolerance of the normalized distance from the segment
returns ... (X_upstream,Y_upstream),(X_downstream,Y_downstream)
"""
found = False
for i in range(len(X)-1): # loop over segments
# segment end points
x1,x2 = X[i], X[i+1]
y1,y2 = Y[i], Y[i+1]
# segment "vector"
dx = x2 - x1
dy = y2 - y1
# segment length square
d2 = dx*dx + dy*dy
# (int,1st end point) vector
ix = x - x1
iy = y - y1
# normalized dot product
dot = (dx*ix + dy*iy) / d2
if dot < 0 or dot > 1: # point projection is outside segment
continue
# normalized cross product
cross = (dx*iy - dy*ix) / d2
if abs(cross) > tol: # point is perpendicularly too far away
continue
# here, we have found the segment containing the point!
found = True
break
if not found:
raise RuntimeError("intersection not found on segments") # or return None, according to needs
i += 1 # the "splitting point" has one higher index than the segment
return (X[:i],Y[:i]),(X[i:],Y[i:])
# plot
import matplotlib.pyplot as plt
plt.plot(X1,Y1,'y',linewidth=8)
plt.plot(X2,Y2,'y',linewidth=8)
plt.plot([x_int],[y_int],"r*")
(X1u,Y1u),(X1d,Y1d) = splitLine(X1,Y1,x_int,y_int)
(X2u,Y2u),(X2d,Y2d) = splitLine(X2,Y2,x_int,y_int)
plt.plot(X1u,Y1u,'g',linewidth=3)
plt.plot(X1d,Y1d,'b',linewidth=3)
plt.plot(X2u,Y2u,'g',linewidth=3)
plt.plot(X2d,Y2d,'b',linewidth=3)
plt.show()
Result:结果:
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