创建一个周期从散点 [英] Create a cycle out of scattered points

问题描述

我知道这听起​​来微不足道，但我的头被拒绝给出一个算法这一点。

I know this sounds trivial, but my head is refusing to give an algorithm for this.

我有一堆散落在一个2-D平面上的点，并希望将它们存储在一个列表中，这样他们创造一个环。点不属于上一周期。

I have a bunch of points scattered on a 2-D plane and want to store them in a list such that they create a ring. The points do not belong on a cycle.

开始从该列表中的第一个点（红在这个图），并依次基于其距离其余添加

Start from the first point in the list (red in this pic) and sequentially add the rest based on their distance.

因为我不能回答我的问题，我会在这里发表一个可能的答案。

Since I cannot answer my question I will post here a possible answer.

这是似乎做这项工作的方法。 V.pos保持节点和距离的位置（）仅仅是一个函数，返回两个点之间的欧几里得距离。更快的方法也将其附加到环后删除next_node，这样你就不必去通过已连接点

This is an approach that seems to do the job. V.pos holds the positions of the nodes and distance() is just a function that returns the Euclidean distance between two points. A faster approach would also delete the next_node after appending it to the ring so that you don't have to go through the already connected points

圈= [节点[0]     而LEN（圈）＆LT;的len（节点）：         MINL = 99999         因为我在范围内（LEN（节点））：             DIST =距离（V.pos [圈[-1]，V.pos [节点[I]）             如果DIST＆LT; MINL和节点[我]不响：                 MINL = DIST                 next_node =节点[I]         ring.append（next_node）

ring = [nodes[0]] while len(ring) < len(nodes): minl=99999 for i in range(len(nodes)): dist = distance(V.pos[ring[-1]],V.pos[nodes[i]]) if dist<minl and nodes[i] not in ring: minl = dist next_node = nodes[i] ring.append(next_node)

推荐答案

下面有一个想法，这将使好吗杂交的结果，如果你的点云已经环形喜欢你的例子：

Here's an idea that will give okay-ish results if your point cloud is already ring-shaped like your example:

• 确定中心点;这既可以是中心的所有点的重心或边界框的中心。
• 重新present径向坐标（半径，角度）参照所有点到中心
• 排序角度

这当然会，产生锯齿星为随机的云彩，但目前尚不清楚，到底什么是环是。你也许可以用这个作为一个初稿，并开始交换节点，如果说给你一个较短的总距离。也许这种简单的code是所有你需要实现短的最小距离过图的所有节点。

This will, of course, produce jagged stars for random clouds, but it is not clear, what exactly a "ring" is. You could probably use this as a first draft and start swapping nodes if that gives you a shorter overall distance. Maybe this simple code is all you need short of implementing the minimum distance over all nodes of a graph.

Anayway，这里有云：

Anayway, here goes:

import math

points = [(0, 4), (2, 2), ...]     # original points in Cartesian coords    
radial = []                        # list of tuples(index, angle)

# find centre point (centre of gravity)
x0, y0 = 0, 0
for x, y in points:
    x0 += x
    y0 += y

x0 = 1.0 * x0 / len(points)
y0 = 1.0 * y0 / len(points)

# calculate angles
for i, p in enumerate(points):
    x, y = p
    phi = math.atan2(y - y0, x - x0)

    radial += [(i, phi)]

# sort by angle
def rsort(a, b):
    """Sorting criterion for angles"""
    return cmp(a[1], b[1])

radial.sort(rsort)

# extract indices
ring = [a[0] for a in radial]

这篇关于创建一个周期从散点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！