Question

给出Python中包含4个点的8个x，y坐标值（均为正数）的列表，如[x1, x2, x3, x4, y1, y2, y3, y4]（(xi, yi)是第i个点的x和y坐标），

如何排序新列表[a1, a2, a3, a4, b1, b2, b3, b4]使得1 2 3 4的坐标(ai, bi)顺时针排列，最接近xy平面的原点的是1，例如

          2--------3
          |        |
          |        |
          |        |
          1--------4

点将大致形成平行四边形。

当前，我正在考虑找到（x + y）最小值的点为1，然后通过剩余坐标中x最少的点找到2，通过（x + y）最大值得到3，剩下的为4点

Answer 1

您应该使用2项元组的列表作为数据结构，以有意义的方式表示可变数量的坐标。

from functools import reduce
import operator
import math
coords = [(0, 1), (1, 0), (1, 1), (0, 0)]
center = tuple(map(operator.truediv, reduce(lambda x, y: map(operator.add, x, y), coords), [len(coords)] * 2))
print(sorted(coords, key=lambda coord: (-135 - math.degrees(math.atan2(*tuple(map(operator.sub, coord, center))[::-1]))) % 360))

这将输出：

[(0, 0), (0, 1), (1, 1), (1, 0)]

Answer 2

# P4=8,10 P1=3,5   P2=8,5   P3=3,10
points=[8,3,8,3,10,5,5,10]
k=0
#we know these numbers are extreme and data won't be bigger than these
xmin=1000
xmax=-1000
ymin=1000
ymax=-1000
#finding min and max values of x and y
for i in points:
    if  k<4:
        if (xmin>i): xmin=i
        if (xmax<i): xmax=i        
    else:
        if (ymin>i): ymin=i
        if (ymax<i): ymax=i        
    k +=1

sortedlist=[xmin,xmin,xmax,xmax,ymin,ymax,ymax,ymin]
print(sortedlist)

输出：[3、3、8、8、5、10、10、5] 对于其他区域，您需要更改sortedlist行。如果中心位于框内，则将需要更多条件控制

Answer 3

import math

def centeroidpython(data):
    x, y = zip(*data)
    l = len(x)
    return sum(x) / l, sum(y) / l

xy = [405952.0, 408139.0, 407978.0, 405978.0, 6754659.0, 6752257.0, 6754740.0, 6752378.0]
xy_pairs = list(zip(xy[:int(len(xy)/2)], xy[int(len(xy)/2):]))

centroid_x, centroid_y = centeroidpython(xy_pairs)
xy_sorted = sorted(xy_pairs, key = lambda x: math.atan2((x[1]-centroid_y),(x[0]-centroid_x)))
xy_sorted_x_first_then_y = [coord for pair in list(zip(*xy_sorted)) for coord in pair]

Answer 4

我们要排序的是与起始坐标的夹角。我在这里使用了numpy来将起始坐标中的每个向量解释为一个复数，为此，有一种简便的方法来计算角度（沿单位球面逆时针）

def angle_with_start(coord, start):
    vec = coord - start
    return np.angle(np.complex(vec[0], vec[1]))

完整代码：

import itertools
import numpy as np


def angle_with_start(coord, start):
    vec = coord - start
    return np.angle(np.complex(vec[0], vec[1]))


def sort_clockwise(points):
    # convert into a coordinate system
    # (1, 1, 1, 2) -> (1, 1), (1, 2)
    coords = [np.array([points[i], points[i+4]]) for i in range(len(points) // 2)]

    # find the point closest to the origin,
    # this becomes our starting point
    coords = sorted(coords, key=lambda coord: np.linalg.norm(coord))
    start = coords[0]
    rest = coords[1:]

    # sort the remaining coordinates by angle
    # with reverse=True because we want to sort by clockwise angle
    rest = sorted(rest, key=lambda coord: angle_with_start(coord, start), reverse=True)

    # our first coordinate should be our starting point
    rest.insert(0, start)
    # convert into the proper coordinate format
    # (1, 1), (1, 2) -> (1, 1, 1, 2)
    return list(itertools.chain.from_iterable(zip(*rest)))

一些示例输入的行为：

In [1]: a
Out[1]: [1, 1, 2, 2, 1, 2, 1, 2]

In [2]: sort_clockwise(a)
Out[2]: [1, 1, 2, 2, 1, 2, 2, 1]

In [3]: b
Out[3]: [1, 2, 0, 2, 1, 2, 3, 1]

In [4]: sort_clockwise(b)
Out[4]: [1, 0, 2, 2, 1, 3, 2, 1]

Answer 5

根据IgnacioVazquez-Abrams的建议，我们还可以根据atan2角度进行排序：

代码：

import math
import copy
import matplotlib.pyplot as plt

a = [2, 4, 5, 1, 0.5, 4, 0, 4]
print(a)


def clock(a):
    angles = []
    (x0, y0) = ((a[0]+a[1]+a[2]+a[3])/4, (a[4]+ a[5] + a[6] + a[7])/4)  # centroid
    for j in range(4):
        (dx, dy) = (a[j] - x0, a[j+4] - y0)
        angles.append(math.degrees(math.atan2(float(dy), float(dx))))
    for k in range(4):
        angles.append(angles[k] + 800)
    # print(angles)

    z = [copy.copy(x) for (y,x) in sorted(zip(angles,a), key=lambda pair: pair[0])]
    print("z is: ", z)

plt.scatter(a[:4], a[4:8])
plt.show()

clock(a)

输出为：

[2, 4, 5, 1, 0.5, 4, 0, 4]
[-121.60750224624891, 61.92751306414704, -46.73570458892839, 136.8476102659946, 678.3924977537511, 861.9275130641471, 753.2642954110717, 936.8476102659946]
z is:  [2, 5, 4, 1, 0.5, 0, 4, 4]

Answer 6

基于BERA的回答，但作为一个类：

代码

import math

def class Sorter:
    @staticmethod    
    def centerXY(xylist):
        x, y = zip(*xylist)
        l = len(x)
        return sum(x) / l, sum(y) / l  

    @staticmethod    
    def sortPoints(xylist):  
        cx, cy = Sorter.centerXY(xylist)
        xy_sorted = sorted(xylist, key = lambda x: math.atan2((x[1]-cy),(x[0]-cx)))
        return xy_sorted

测试

def test_SortPoints():
    points=[(0,0),(0,1),(1,1),(1,0)]
    center=Sorter.centerXY(points)
    assert center==(0.5,0.5)
    sortedPoints=Sorter.sortPoints(points)
    assert sortedPoints==[(0, 0), (1, 0), (1, 1), (0, 1)]

Answer 7

尝试这一行代码

def sort_clockwise(pts):
    rect = np.zeros((4, 2), dtype="float32")
    s = pts.sum(axis=1)
    rect[0] = pts[np.argmin(s)]
    rect[2] = pts[np.argmax(s)]
    diff = np.diff(pts, axis=1)
    rect[1] = pts[np.argmin(diff)]
    rect[3] = pts[np.argmax(diff)]
    return rect

根据顺时针点坐标进行排序

7 个答案: