连接墙壁端点

如何解决连接墙壁端点

我有以下代表墙的点的列表：

walls = [[  76.7403,1283.2495,894.5198,1310.0122],[ 864.3867,415.2406,891.3730,1287.3810],[  80.0975,402.9981,899.7578,428.0596],[  75.9861,404.3390,101.6148,1292.0922],[ 469.9275,1118.6304,481.1644,1300.6523],[ 500.6000,897.3496,876.6120,908.2079],[ 102.0548,753.6523,402.9403,764.0107],[ 247.2823,968.0134,320.4769,977.5980],[ 503.2201,898.0987,515.2742,1136.6272],[ 400.9270,689.1513,411.8462,907.2622],[ 353.0995,1123.1868,362.8233,1293.9554],[ 248.6094,967.5870,261.0458,1124.8018],[ 246.0828,896.9332,411.2629,906.4570],[ 880.6799,60.5249,892.5886,411.2219],[ 249.5214,1116.5377,259.7009,1290.2698],[ 309.1409,899.7199,320.3546,978.5977],[  85.6537,1118.2689,259.5562,1128.9491],[ 249.0276,763.1385,261.5212,905.9616],[  86.2850,1117.8192,501.5602,1128.5548],[ 240.1799,1117.4913,385.9280,1128.2761],[ 288.5830,1117.4141,515.1036,1128.3738],[ 512.2842,1015.8013,592.7465,1022.5025],[ 249.2041,748.7802,260.6894,1140.3009],[ 401.1811,687.2712,411.9943,779.8344],[ 380.8932,1040.3300,386.9687,1126.2324]]  

我正在使用以下方法从这些墙形成一组多边形：

polys = []
for w in walls:
  x0,y0,x1,y1 = w
  poly = [
            (x0,y0),(x1,y1),(x0,y1)
        ]
  poly = list(itertools.chain.from_iterable(poly))
  polys.append(poly)

这些多边形最终看起来如下：

我想做的是连接墙壁端点，以使它们不会重叠，如图所示。这样做的最佳算法是什么？

以下是生成图像的代码：

<svg width="1200" height="1300">
<polygon points='76.7403,1310.0122,76.7403,1310.0122'/>
<polygon points='864.3867,891.373,1287.381,864.3867,1287.381'/>
<polygon points='80.0975,428.0596,80.0975,428.0596'/>
<polygon points='75.9861,404.339,1292.0922,75.9861,1292.0922'/>
<polygon points='469.9275,1300.6523,469.9275,1300.6523'/>
<polygon points='500.6,876.612,908.2079,500.6,908.2079'/>
<polygon points='102.0548,764.0107,102.0548,764.0107'/>
<polygon points='247.2823,977.598,247.2823,977.598'/>
<polygon points='503.2201,1136.6272,503.2201,1136.6272'/>
<polygon points='400.927,907.2622,400.927,907.2622'/>
<polygon points='353.0995,1293.9554,353.0995,1293.9554'/>
<polygon points='248.6094,967.587,1124.8018,248.6094,1124.8018'/>
<polygon points='246.0828,906.457,246.0828,906.457'/>
<polygon points='880.6799,411.2219,880.6799,411.2219'/>
<polygon points='249.5214,1290.2698,249.5214,1290.2698'/>
<polygon points='309.1409,978.5977,309.1409,978.5977'/>
<polygon points='85.6537,1128.9491,85.6537,1128.9491'/>
<polygon points='249.0276,905.9616,249.0276,905.9616'/>
<polygon points='86.285,1128.5548,86.285,1128.5548'/>
<polygon points='240.1799,385.928,1128.2761,240.1799,1128.2761'/>
<polygon points='288.583,1128.3738,288.583,1128.3738'/>
<polygon points='512.2842,1022.5025,512.2842,1022.5025'/>
<polygon points='249.2041,1140.3009,249.2041,1140.3009'/>
<polygon points='401.1811,779.8344,401.1811,779.8344'/>
<polygon points='380.8932,1040.33,1126.2324,380.8932,1126.2324'/>
</svg>

解决方法

目前还不清楚您的最终目标是什么。 您要剪切哪几行？所有松散的末端还是只是有些太长的末端？那线的不同厚度呢？

我想到的最简单的方法是将walls转换为具有特定宽度的线，而不是离散化这些线的端点，这样，生成的图像看起来会更清晰一些：

import matplotlib.pyplot as plt
import numpy as np


def walls2lines(walls):
    lines = []
    widths = []
    for w in walls:
        x0,y0,x1,y1 = w
        if (x1-x0) > (y1 - y0):
            widths.append(y1-y0)
            lines.append(((x0,y0 + widths[-1] / 2),(x1,y0 + widths[-1] / 2)))
        else:
            widths.append(x1-x0)
            lines.append(((x0 + widths[-1] / 2,y0),(x0 + widths[-1] / 2,y1)))

    return np.array(lines),np.array(widths)


def discretize_np(x,step):
    difference = x % step  # distance to the next discrete value
    difference[difference > step / 2] -= step  # round correctly
    return x - difference


lines,widths = walls2lines(walls=walls)
lines = discretize_np(lines,step=50)
widths *= 10 / widths.max()

fig = plt.figure()
ax = fig.subplots()

for l,w in zip(lines,widths):
    ax.plot(*l.T,lw=w,alpha=0.5)

另一种可能性是真正查看不同矩形的重叠部分并尝试移动尊重的边界。但是，您遇到了许多不同的情况，因此很难找到一个好的通用解决方案，这实际上取决于您要寻找的内容。

我在eps上添加了walls2overlap，以也包括不太相交的矩形。我评论了一条线，以不同的方式处理自由端。 很多微调和反复试验...

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from matplotlib.collections import PatchCollection


def walls2overlap(walls,eps=0):

    idx_overlap = np.ones((len(walls),len(walls)),dtype=bool)
    for i,w_i in enumerate(walls):
        for j,w_j in enumerate(walls):
            (x0_i,x1_i),(y0_i,y1_i) = w_i
            (x0_j,x1_j),(y0_j,y1_j) = w_j
            if (x0_i > x1_j+eps or x1_i < x0_j-eps or
                y0_i > y1_j+eps or y1_i < y0_j-eps):
                idx_overlap[i,j] = False

    return idx_overlap


def get_orientations(walls):
    """
    0: horizontal,1: vertical
    """
    (x0,x1),(y0,y1) = walls.transpose(1,2,0)
    return ((x1 -x0) < (y1 - y0)).astype(int)


def cut_walls(walls,idx_overlap):
    """Move the walls to the most extreme boundary
     of all intersecting walls of different orientation."""
    for i,w_i in enumerate(walls):
        o_i = orientations[i]
        idx_temp = np.logical_and(idx_overlap[i],orientations != o_i)
        if idx_temp.sum() <= 0:  # Set to 1,to keep free ended walls
            continue
        walls[i,o_i,0] = np.min(walls[idx_temp,0])
        walls[i,1] = np.max(walls[idx_temp,1])

    return walls


walls = np.reshape(walls,(-1,2),order='F')
idx_overlap = walls2overlap(walls=walls,eps=10)
orientations = get_orientations(walls)
walls = cut_walls(walls=walls,idx_overlap=idx_overlap)

fig,ax = new_fig(aspect=1)
rects = [Rectangle(xy=w[:,0],width=w[0,1] - w[0,height=w[1,1] - w[1,0])
         for w in walls]
ax.add_collection(PatchCollection(rects,color='k',alpha=0.5))