如何在Python中获取多边形内网格点的坐标?

问题描述

我想输入一个多边形的4个顶点的坐标，指定点的数量来划分多边形的边(等分)，目标是生成一个矩阵，该矩阵带有多边形内部的网格点的坐标.

I want to input the coordinates of 4 vertices of a polygon, specify the number of points to divide the polygon's edges (in equal segments), and the goal is to generate a matrix with the coordinates of the grid points inside the polygon.

下图可能更好地解释了我的目标:

The following picture might explain better my goal:

所以在这种情况下，我将输入点 (P) 的坐标并指定我希望网格为 3 x 2.输出将是一个 3 x 2 矩阵，其坐标为网格的坐标 (x,y)点 (N).

So in that case I would input the coordinates of the points (P) and specify that I want grid to be 3 by 2. The output would be a 3 by 2 matrix with the coordinates (x,y) of the grid points (N).

我进行了很多搜索，但仍然找不到这样做的方法，老实说，我对 Python 一点经验都没有.我发现使用 numpy 的 meshgrid 结合 matplotlib.path 的 contains_points 在多边形内创建网格但我不知道如何获取网格点的坐标.我看到 shapely 在此类问题中被大量使用，但同样，我没有这方面的经验，因此不胜感激！

I have searched a lot but still couldn't find a way to do this, and honestly I am not at all experienced with Python. I found something using numpy's meshgrid in combination with matplotlib.path's contains_points to create a grid inside a polygon but I have no idea how to get the grid point's coordinates. I saw shapely being used a lot in problems like this but again, I'm not experience in this so some help would be appreciated!

提前谢谢大家！

推荐答案

为了解释这个方法，我们分为三个步骤:

To explain the approach, we have three steps:

1. 找出 4 边形每一边的刻度
2. 找到网格线
3. 找到网格线的交点

def det(a, b):
    return a[0] * b[1] - a[1] * b[0]

ticks = []
A = (1, 2)  # wlog., suppose the polygon is ABCD;
B = (3, 2)
C = (3, 4)
D = (1, 4)
polygon = [A, B, C, D]
n = 3  # number of parts on each side of the grid

# we first find ticks on each side of the polygon
for j in range(4):  # because we are talking about 4-gons
    temp_ticks = []
    for i in range(n-1):
        t = (i+1)*(1/n)
        Ex = polygon[j][0] * (1-t) + polygon[(j+1) % 4][0] * t
        Ey = polygon[j][1] * (1-t) + polygon[(j+1) % 4][1] * t
        temp_ticks.append((Ex, Ey))
    if j < 2:
        ticks.append(temp_ticks)
    else: # because you are moving backward in this part
        temp_ticks.reverse()
        ticks.append(temp_ticks)

# then we find lines of the grid
h_lines = []
v_lines = []
for i in range(n-1):
    h_lines.append((ticks[0][i], ticks[2][i]))
    v_lines.append((ticks[1][i], ticks[3][i]))
        
# then we find the intersection of grid lines
for i in range(len(h_lines)):
    for j in range(len(v_lines)):
        line1 = h_lines[i]
        line2 = v_lines[j]
        xdiff = (line1[0][0] - line1[1][0], line2[0][0] - line2[1][0])
        ydiff = (line1[0][1] - line1[1][1], line2[0][1] - line2[1][1])
        div = det(xdiff, ydiff)
        if div == 0:
            raise Exception('lines do not intersect')

        d = (det(*line1), det(*line2))
        x = det(d, xdiff) / div
        y = det(d, ydiff) / div
        print(x,y) # this is an intersection point that you want

注意:查找线交点的部分代码来自 这里

Note: the part of the code for finding the intersection of lines are from here

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