序正交多边形python

Question

如何排序正交多边形点的列表？在

例如，我有一个正交多边形点的列表

data = [(2, 0), (5, 0), (5, 7), (4, 7), (4, 5), (3, 5),(3, 3), (2, 3), (2, 2), (3, 2), (3, 7), (2, 7)]

不整齐。 我想按逆时针方向订购：

^{pr2}$

我已经尝试过使用deflate _hull，但没有正确。 有什么算法可以解决这个问题吗？在

我明白了：

但预计：

Answer 1

您可以使用以下递归函数：

def sort_ortho_poly(points, current=None, start=None, go_x=True):
    # initialize the starting point at the bottom left, which should have the least sum of x and y
    if not current:
        start = current = min(points, key=sum)
    # if we're going x-wards, v would be the y index (1), h would be the x index (0), and vice versa 
    v, h = go_x, not go_x
    # remove the current point from the list of points so the next recursion would be processing the remaining points
    remaining = points[:]
    remaining.remove(current)
    # if there is no more remaining point
    if not remaining:
        # we've found a path if we are able to connect back to the starting point, or else we don't
        return [current] if start[v] == current[v] else []
    # try each point in the remaining points that goes in the right direction from the current point
    for next in [p for p in remaining if p[v] == current[v]]:
        # recursively find a valid path from the remaining points after flipping the direction
        path = sort_ortho_poly(remaining, next, start, not go_x)
        # if we get a path that does go back to the starting point, we have to make sure the path is valid
        if path:
            # the current edge (e1, e2)
            e1, e2 = current, next
            # make sure e1 is lower than or left of e2
            if e1[h] > e2[h]:
                e1, e2 = e2, e1
            # for each edge (p1, p2) in the path, including the final edge connecting to the starting point
            for p1, p2 in zip(path, path[1:] + [start]):
                # make sure p1 is lower than or left of p2
                if p1[0] == p2[0] and p1[1] > p2[1] or p1[1] == p2[1] and p1[0] > p2[0]:
                    p1, p2 = p2, p1
                # if the edge is in the same line as the current edge
                if p1[v] == p2[v] == e1[v]:
                    # make sure the two edges don't overlap
                    if e1[h] < p1[h] < e2[h] or e1[h] < p2[h] < e2[h] or p1[h] < e1[h] < p2[h] or p1[h] < e2[h] < p2[h]:
                        break
                # if the edge is perpendicular to the current edge, make sure they don't cross over
                elif p1[h] == p2[h] and e1[h] < p1[h] < e2[h] and p1[v] < e1[v] < p2[v]:
                    break
            else:
                # the path is valid! we append the path to the current point and return
                return [current, *path]
    # return empty if it's a dead end
    return []

因此：

^{pr2}$

将输出：

[(2, 0), (5, 0), (5, 7), (4, 7), (4, 5), (3, 5), (3, 7), (2, 7), (2, 3), (3, 3), (3, 2), (2, 2)]