我正在制作一个程序，可以找到图像上两点之间的路径（很快就会成为视频帧）。我所做的是使用多边形对象来识别路径之间的障碍物，并使用该多边形对象根据需要绕过。我的代码如下：

simplepoint = tuple[int, int]

def getpolygonsinway(start: simplepoint, end: simplepoint, polygons: list[polygon]) -> list[polygon]:
    line = linestring([start, end])
    polygonsinway = [polygon for polygon in polygons if line.crosses(polygon)]  # make sure line.crosses() doesnt return true if the starting point is on the polygon
    return polygonsinway

def getpaths(start: simplepoint, end: simplepoint, polygons: list[polygon], prev_points: list[simplepoint] = []) -> list[list[simplepoint]]:
    polygonsinway = getpolygonsinway(start, end, polygons)
    if not polygonsinway:
        return [[start, end]]

    closestpolygon = closestpolygontopoint(start, polygons)
    xyvalsold = closestpolygon.exterior.xy
    xyvals = []

    for i in range(len(xyvalsold[0])):
        xyvals.append((xyvalsold[0][i], xyvalsold[1][i]))

    xyvals = list(set(xyvals)) # remove duplicates

    polxvals = [val[0] for val in xyvals]
    polyvals = [val[1] for val in xyvals]

    okpoints = []

    for i in range(len(polxvals)):
        if start == (polxvals[i], polyvals[i]):
            continue
        if int(polxvals[i]) != polxvals[i] or int(polyvals[i]) != polyvals[i]:
            continue
        if (int(polxvals[i]), int(polyvals[i])) in prev_points:
            continue
        if closestpolygon not in getpolygonsinway(start, (polxvals[i], polyvals[i]), polygons): #and closestpolygon not in getpolygonsinway((polxvals[i], polyvals[i]), end, polygons):
            okpoints.append((int(polxvals[i]), int(polyvals[i])))

    paths = [[start] for _ in range(len(okpoints))]

    for i in range(len(okpoints)):
        point = (okpoints[i][0], okpoints[i][1])
        paths[i].extend(getshortestpath(getpaths(point, end, polygons, prev_points + [point]), point, end))

    return paths

目前，我使用的唯一障碍是简单的形状，例如矩形、正方形、圆形等，一旦一切正常，我将改用视频。当两个部分是矩形的对角线时，函数 getpolygonsinway 中的列表理解中的 line.crosses(polygon) 返回 false 时，会出现问题。我已经查看了文档，但我不知道用什么来确保我正确地捕获这个案例，所以我在这里问了这个问题。

编辑：根据要求，示例多边形、起点和终点如下：

polygon = Polygon([[411, 182], [411, 335], [210, 335], [210, 182]])
start = (440, 35)
end = (90, 600)

使用上面的代码，运行 getshortestpath(getpaths(start, end, polygons, [])) 后得到以下路径： [(440, 35), (411, 182), (210, 335), (90, 600)]

正确答案

我找到了 line.within(polygon) 方法，该方法检查该线是否位于传递的多边形内部。我不知道为什么我以前没有看到它，但现在我看到了。通过在列表理解 if 条件中执行 line.crosses(polygon) 或 line.within(polygon) ，我设法获得了围绕矩形的路径。我现在将使用其他基本形状来测试它，然后无限地测试它的用途。