Python中的区域交集 [英] Area intersection in Python

问题描述

我有一个将条件C作为输入的代码，并将我的问题的解决方案计算为(x，y)空间上的允许区域"A.该区域由数个管"组成，由两条永不交叉的线定义.

I have a code that takes a condition C as an input, and computes the solution to my problem as an 'allowed area' A on the (x,y) space. This area is made of several 'tubes', which are defined by 2 lines that can never cross.

我要寻找的最终结果必须满足k个条件{C1，..，Ck}，因此是k个区域{A1，..，Ak}之间的交点S.

The final result I'm looking for must satisfy k conditions {C1, .., Ck}, and is therefore an intersection S between k areas {A1, .. , Ak}.

这里是一个有2个条件的示例(A1:绿色，3个试管.A2:紫色，1个试管)；溶液S为红色.

Here is an example with 2 conditions (A1: green, 3 tubes. A2: purple, 1 tube); the solution S is in red.

当我处理每个大约10个试管的4个区域时，如何找到S? (最后的情节太糟糕了！)

How can I find S when I'm dealing with 4 areas of around 10 tubes each? (The final plot is awful!)

我需要能够绘制它，并找到平均坐标和S点的方差(每个坐标的方差). [如果有一种知道点P是否属于S的有效方法，我将仅使用蒙特卡洛方法.]

I would need to be able to plot it, and to find the mean coordinate and the variance of the points in S (variance of each coordinate). [If there is an efficient way of knowing whether a point P belongs to S or not, I’ll just use a Monte Carlo method].

理想情况下，我还希望能够实现从S中删除的禁止使用的管道" [这可能比将S与禁止区域的外部相交要复杂得多，因为两个来自同一禁止使用的管道区域可以交叉(即使定义管的线永不交叉).

Ideally, I’d also like to be able to implement "forbidden tubes" that I would remove from S [it might be a bit more complicated than intersecting S with the outside of my forbidden area, since two tubes from the same area can cross (even if the lines defining a tube never cross)].

注意:

• 该代码还存储直线的弧长.

• The code also stores the arc length of the lines.

线被存储为点数组(每行大约1000点).定义管的两条线不一定具有相同的点数，但是Python可以根据其弧长在1秒内对所有点进行插值.

The lines are stored as arrays of points (around 1000 points per line). The two lines defining a tube do not necessarily have the same number of points, but Python can interpolate ALL of them as a function of their arc length in 1 second.

这些线是参数函数(即我们不能写y = f(x)，因为允许这些线是垂直的).

The lines are parametric functions (i.e. we cannot write y = f(x), since the lines are allowed to be vertical).

使用油漆编辑了该图，以使结果在右侧...效率不高！

The plot was edited with paint to get the result on the right... Not very efficient!

• 我不知道如何为多个交叉点使用plt.fill_between(我可以在这里针对2个条件进行此操作，但是当需要太多行进行眼图判断时，我需要代码自动执行此操作).

• I don't know how I can use plt.fill_between for a multiple intersection (I can do it here for 2 conditions, but I need the code to do it automatically when there are too many lines for eye judgement).

现在我只生成行.我没有写任何东西来寻找最终解决方案，因为我绝对不知道哪种结构最适合此解决方案. [但是，该代码的先前版本能够找到2条不同管的线之间的交点，并且我打算将它们作为多边形传递给形状，但这隐含了其他一些问题.]

For now I just generate the lines. I didn’t write anything for finding the final solution since I absolutely don’t know which structure is the most adapted for this. [However, a previous version of the code was able to find the intersection points between the lines of 2 different tubes, and I was planning to pass them as polygons to shapely, but this implied several other problems..]

我不认为我可以用sets做到这一点:以所需的精度扫描整个(x，y)区域代表大约6e8点... [由于变量，这些行只有1e3点步长(适应曲率)，但整个问题都很大]

I don't think I can do it with sets: scanning the whole (x,y) area at required precision represents around 6e8 points... [The lines have only 1e3 points thanks to a variable step size (adapts to the curvature), but the whole problem is quite large]

推荐答案

Shapely解决了问题！

Problem solved with Shapely!

我将每个管定义为Polygon，区域A是作为其管的并集构建的MultiPolygon对象.

I defined each tube as a Polygon, and an area A is a MultiPolygon object built as the union of its tubes.

然后，intersection方法计算出我正在寻找的解决方案(所有区域之间的重叠).

The intersection method then computes the solution I was looking for (the overlap between all areas).

整个过程几乎是瞬时的.我不知道大型物体的形状是否如此好(每根管大约2000个点，每个区域10个管，4个区域).

The whole thing is almost instantaneous. I didn't know shapely was so good with large objects [around 2000 points per tube, 10 tubes per area, 4 areas].

谢谢您的帮助！ :)

一个有效的例子.

import matplotlib.pyplot as plt
import shapely
from shapely.geometry import Polygon
from descartes import PolygonPatch
import numpy as np

def create_tube(a,height):
    x_tube_up = np.linspace(-4,4,300)
    y_tube_up = a*x_tube_up**2 + height
    x_tube_down = np.flipud(x_tube_up)          #flip for correct definition of polygon
    y_tube_down = np.flipud(y_tube_up - 2)

    points_x = list(x_tube_up) + list(x_tube_down)
    points_y = list(y_tube_up) + list(y_tube_down)

    return Polygon([(points_x[i], points_y[i]) for i in range(600)])

def plot_coords(ax, ob):
    x, y = ob.xy
    ax.plot(x, y, '+', color='grey')


area_1 = Polygon()          #First area, a MultiPolygon object
for h in [-5, 0, 5]:
    area_1 = area_1.union(create_tube(2, h))

area_2 = Polygon()
for h in [8, 13, 18]:
    area_2 = area_2.union(create_tube(-1, h))

solution = area_1.intersection(area_2)      #What I was looking for

##########  PLOT  ##########

fig = plt.figure()
ax = fig.add_subplot(111)

for tube in area_1:
    plot_coords(ax, tube.exterior)
    patch = PolygonPatch(tube, facecolor='g', edgecolor='g', alpha=0.25)
    ax.add_patch(patch)

for tube in area_2:
    plot_coords(ax, tube.exterior)
    patch = PolygonPatch(tube, facecolor='m', edgecolor='m', alpha=0.25)
    ax.add_patch(patch)

for sol in solution:
    plot_coords(ax, sol.exterior)
    patch = PolygonPatch(sol, facecolor='r', edgecolor='r')
    ax.add_patch(patch)

plt.show()

情节:

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