使用Python按顺时针角度对二维坐标列表进行排序？

我有一组带有x和y坐标的点,如下图所示.9个点的坐标存储在列表中,如下所示:

L = [[5,2], [4,1], [3.5,1], [1,2], [2,1], [3,1], [3,3], [4,3] , [2,3]]

我们的想法是从原点顺时针对点进行排序.在这种情况下,原点是着色的点,并且具有指示排序方向的箭头.不要担心创建确定原点的方法,因为它已经解决了.

因此,在订购后,清单L应如下:

L = [[2,3], [3,3], [4,3], [5,2], [4,1], [3.5,1], [3,1], [2,1], [1,2]]

请注意,x和y坐标不会更改.存储顺序有哪些变化.

您是否知道python语言中针对此问题的算法,脚本或方法？

通过一点三角学,它并不那么难.也许你知道,但两个(标准化)矢量之间的角度是acos(vec1 * vec2).然而,这仅计算投影角度,但可以atan2用来计算方向感知角度.

对于这意味着计算它然后使用它作为key排序的函数将是一个好方法:

import math

pts = [[2,3], [5,2],[4,1],[3.5,1],[1,2],[2,1],[3,1],[3,3],[4,3]]
origin = [2, 3]
refvec = [0, 1]

def clockwiseangle_and_distance(point):
    # Vector between point and the origin: v = p - o
    vector = [point[0]-origin[0], point[1]-origin[1]]
    # Length of vector: ||v||
    lenvector = math.hypot(vector[0], vector[1])
    # If length is zero there is no angle
    if lenvector == 0:
        return -math.pi, 0
    # Normalize vector: v/||v||
    normalized = [vector[0]/lenvector, vector[1]/lenvector]
    dotprod  = normalized[0]*refvec[0] + normalized[1]*refvec[1]     # x1*x2 + y1*y2
    diffprod = refvec[1]*normalized[0] - refvec[0]*normalized[1]     # x1*y2 - y1*x2
    angle = math.atan2(diffprod, dotprod)
    # Negative angles represent counter-clockwise angles so we need to subtract them 
    # from 2*pi (360 degrees)
    if angle < 0:
        return 2*math.pi+angle, lenvector
    # I return first the angle because that's the primary sorting criterium
    # but if two vectors have the same angle then the shorter distance should come first.
    return angle, lenvector

一个sorted运行:

>>> sorted(pts, key=clockwiseangle_and_distance)
[[2, 3], [3, 3], [4, 3], [5, 2], [4, 1], [3.5, 1], [3, 1], [2, 1], [1, 2]]

并且在原点周围有一个矩形网格,这也按预期工作:

>>> origin = [2,3]
>>> refvec = [0, 1]
>>> pts = [[1,4],[2,4],[3,4],[1,3],[2,3],[3,3],[1,2],[2,2],[3,2]]
>>> sorted(pts, key=clockwiseangle_and_distance)
[[2, 3], [2, 4], [3, 4], [3, 3], [3, 2], [2, 2], [1, 2], [1, 3], [1, 4]]

即使您更改参考向量:

>>> origin = [2,3]
>>> refvec = [1,0]  # to the right instead of pointing up
>>> pts = [[1,4],[2,4],[3,4],[1,3],[2,3],[3,3],[1,2],[2,2],[3,2]]
>>> sorted(pts, key=clockwiseangle_and_distance)
[[2, 3], [3, 3], [3, 2], [2, 2], [1, 2], [1, 3], [1, 4], [2, 4], [3, 4]]

感谢@Scott Mermelstein好了函数名以及@f5r5e5d的atan2建议.