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首先,我对 Python(一个编程领域)还很陌生,但我想学习和转换一个由jwpat7开发的函数。给定一组从凸包导出的点

hull= [(560023.44957588764,6362057.3904932579), 
      (560023.44957588764,6362060.3904932579), 
      (560024.44957588764,6362063.3904932579), 
      (560026.94957588764,6362068.3904932579), 
      (560028.44957588764,6362069.8904932579), 
      (560034.94957588764,6362071.8904932579), 
      (560036.44957588764,6362071.8904932579), 
      (560037.44957588764,6362070.3904932579), 
      (560037.44957588764,6362064.8904932579), 
      (560036.44957588764,6362063.3904932579), 
      (560034.94957588764,6362061.3904932579), 
      (560026.94957588764,6362057.8904932579), 
      (560025.44957588764,6362057.3904932579), 
      (560023.44957588764,6362057.3904932579)]

此脚本在此帖子问题之后返回所有可能区域的打印。jwpat7开发的代码是:

import math

def mostfar(j, n, s, c, mx, my): # advance j to extreme point
    xn, yn = hull[j][0], hull[j][1]
    rx, ry = xn*c - yn*s, xn*s + yn*c
    best = mx*rx + my*ry
    while True:
        x, y = rx, ry
        xn, yn = hull[(j+1)%n][0], hull[(j+1)%n][1]
        rx, ry = xn*c - yn*s, xn*s + yn*c
        if mx*rx + my*ry >= best:
            j = (j+1)%n
            best = mx*rx + my*ry
        else:
            return (x, y, j)

n = len(hull)
iL = iR = iP = 1                # indexes left, right, opposite
pi = 4*math.atan(1)
for i in range(n-1):
    dx = hull[i+1][0] - hull[i][0]
    dy = hull[i+1][1] - hull[i][1]
    theta = pi-math.atan2(dy, dx)
    s, c = math.sin(theta), math.cos(theta)
    yC = hull[i][0]*s + hull[i][1]*c    
    xP, yP, iP = mostfar(iP, n, s, c, 0, 1)
    if i==0: iR = iP
    xR, yR, iR = mostfar(iR, n, s, c,  1, 0)
    xL, yL, iL = mostfar(iL, n, s, c, -1, 0)
    area = (yP-yC)*(xR-xL) 
    print '    {:2d} {:2d} {:2d} {:2d} {:9.3f}'.format(i, iL, iP, iR, area)

结果是:

i iL iP iR    Area
 0  6  8  0   203.000
 1  6  8  0   211.875
 2  6  8  0   205.800
 3  6 10  0   206.250
 4  7 12  0   190.362
 5  8  0  1   203.000
 6 10  0  4   201.385
 7  0  1  6   203.000
 8  0  3  6   205.827
 9  0  3  6   205.640
10  0  4  7   187.451
11  0  4  7   189.750
12  1  6  8   203.000

我希望创建一个返回最小矩形的长度、宽度和面积的函数。前任:

Length, Width, Area = get_minimum_area_rectangle(hull)
print Length, Width, Area
18.036, 10.392, 187.451

我的问题是:

  1. 我需要创建一个函数还是两个函数。例如:def mostfar 和 get_minimum_area_rectangle
  2. hull 是一个价值列表。它是最好的格式吗?
    1. 遵循一种功能方法,我有一个问题要在里面集成 mostfar

提前致谢

1) 解决方案:在 Scott Hunter 建议的第一个解决方案之后的一个函数,我在将 mostfar() 集成到 get_minimum_area_rectangle() 中时遇到问题。任何建议或帮助都非常感谢,因为我可以学习。

#!/usr/bin/python
import math

def get_minimum_area_rectangle(hull):
    # get pi greek
    pi = 4*math.atan(1)
    # number of points
    n = len(hull)
     # indexes left, right, opposite
    iL = iR = iP = 1
    # work clockwise direction
    for i in range(n-1):
        # distance on x axis
        dx = hull[i+1][0] - hull[i][0]
        # distance on y axis
        dy = hull[i+1][1] - hull[i][1]
        # get orientation angle of the edge
        theta = pi-math.atan2(dy, dx)
        s, c = math.sin(theta), math.cos(theta)
        yC = hull[i][0]*s + hull[i][1]*c

从这里开始,按照上面的 jwpat7 示例,我需要使用 mostfar()。在这一点上,我很难理解如何整合(抱歉,这个术语不正确)

4

4 回答 4

2
  1. 您可以使用一个函数或两个函数,但使用两个函数可能更简洁、更容易。您可以按原样保留该mostfar功能。然后,只需通过添加函数定义行将代码的后半部分转换为函数:

    def get_minimum_area_rectangle(hull):
    

    …然后缩进其余代码(以 开头n = len(hull))以形成函数的主体。您还需要更改函数以返回您想要获取的值(长度、宽度和面积)。这将使您的代码保持模块化和干净,并且只需要很少的更改。

  2. hull为此目的,使用值列表似乎很好。另一种方法是使用数组(如 NumPy 数组),但在这种情况下,您将迭代地遍历数据,一次一个项目,而不是同时跨多个数据点进行任何计算。所以列表应该没问题。访问列表中的项目很快,与您必须做的数学相比,它不应该是一个瓶颈。

于 2012-11-25T18:30:51.207 回答
1
  1. 您当然可以将其作为单个功能进行:稍微修改 mostfar,而不是打印找到的区域,跟踪最小的区域和随之而来的信息。或者你可以让它收集它正在打印的值到一个 lst 中,然后 GEAR 可以使用它来找到最小值。

编辑:(我错过了一些代码在最远之外)我将“脚本”部分(最远之后的代码)包装到一个函数中,并如上所述修改它。然后,您的“脚本”将只调用该函数,或者,如果使用第二个修改,则从返回的列表中找到最小值。

  1. 我认为您对船体的表示没有任何问题。
于 2012-11-25T18:17:40.710 回答
1

下面是一个示例,说明如何从您的代码中使其成为仿函数对象并使用它——以及对我认为值得的其他一些事情的一些更改。函子是充当函数角色但可以像对象一样操作的实体。

在 Python 中,两者之间的区别不大,因为函数已经是单例对象,但有时为一个对象创建一个专门的类很有用。在这种情况下,它允许将辅助函数制作为私有类方法,而不是您似乎反对这样做的全局或嵌套方法。

from math import atan2, cos, pi, sin

class GetMinimumAreaRectangle(object):
    """ functor to find length, width, and area of the smallest rectangular
        area of the given convex hull """
    def __call__(self, hull):
        self.hull = hull
        mostfar = self._mostfar  # local reference
        n = len(hull)
        min_area = 10**100  # huge value
        iL = iR = iP = 1  # indexes left, right, opposite
#        print '    {:>2s} {:>2s} {:>2s} {:>2s} {:>9s}'.format(
#                   'i', 'iL', 'iP', 'iR', 'area')
        for i in xrange(n-1):
            dx = hull[i+1][0] - hull[i][0]  # distance on x axis
            dy = hull[i+1][1] - hull[i][1]  # distance on y axis
            theta = pi-atan2(dy, dx)   # get orientation angle of the edge
            s, c = sin(theta), cos(theta)
            yC = hull[i][0]*s + hull[i][1]*c
            xP, yP, iP = mostfar(iP, n, s, c, 0, 1)
            if i==0: iR = iP
            xR, yR, iR = mostfar(iR, n, s, c,  1, 0)
            xL, yL, iL = mostfar(iL, n, s, c, -1, 0)
            l, w = (yP-yC), (xR-xL)
            area = l*w
#            print '    {:2d} {:2d} {:2d} {:2d} {:9.3f}'.format(i, iL, iP, iR, area)
            if area < min_area:
                min_area, min_length, min_width = area, l, w
        return (min_length, min_width, min_area)

    def _mostfar(self, j, n, s, c, mx, my):
        """ advance j to extreme point """
        hull = self.hull  # local reference
        xn, yn = hull[j][0], hull[j][1]
        rx, ry = xn*c - yn*s, xn*s + yn*c
        best = mx*rx + my*ry
        while True:
            x, y = rx, ry
            xn, yn = hull[(j+1)%n][0], hull[(j+1)%n][1]
            rx, ry = xn*c - yn*s, xn*s + yn*c
            if mx*rx + my*ry >= best:
                j = (j+1)%n
                best = mx*rx + my*ry
            else:
                return (x, y, j)

if __name__ == '__main__':

    hull= [(560023.44957588764, 6362057.3904932579),
           (560023.44957588764, 6362060.3904932579),
           (560024.44957588764, 6362063.3904932579),
           (560026.94957588764, 6362068.3904932579),
           (560028.44957588764, 6362069.8904932579),
           (560034.94957588764, 6362071.8904932579),
           (560036.44957588764, 6362071.8904932579),
           (560037.44957588764, 6362070.3904932579),
           (560037.44957588764, 6362064.8904932579),
           (560036.44957588764, 6362063.3904932579),
           (560034.94957588764, 6362061.3904932579),
           (560026.94957588764, 6362057.8904932579),
           (560025.44957588764, 6362057.3904932579),
           (560023.44957588764, 6362057.3904932579)]

    gmar = GetMinimumAreaRectangle()  # create functor object
    print "dimensions and area of smallest enclosing rectangular area:"
    print "  {:.3f}(L) x {:.3f}(W) = {:.3f} area".format(*gmar(hull))  # use it

输出:

dimensions and area of smallest enclosing rectangular area:
  10.393(L) x 18.037(W) = 187.451 area
于 2012-11-26T03:32:15.913 回答
1

我正在发布另一个答案,显示如何按照我(和其他人)的建议进行操作,这只是将辅助函数嵌套在mostfar()被调用的主函数中。这在 Python 中很容易做到,因为嵌套函数可以访问其封闭范围的局部变量(就像hull在案例中一样)。我还_mostfar()按照约定重命名了该函数,以表明某些东西是私有的,但这并不是绝对必要的(永远,而且绝对不是这里)。

如您所见,尽管我确实简化了一些与嵌套函数无关的事情(因此它们可能会集成到您选择的任何答案中),但大多数代码与我的其他答案中的代码非常相似。

from math import atan2, cos, pi, sin

def get_minimum_area_rectangle(hull):
    """ find length, width, and area of the smallest rectangular
        area of the given convex hull """

    def _mostfar(j, n, s, c, mx, my):
        """ advance j to extreme point """
        xn, yn = hull[j]
        rx, ry = xn*c - yn*s, xn*s + yn*c
        best = mx*rx + my*ry
        k = j + 1
        while True:
            x, y = rx, ry
            xn, yn = hull[k % n]
            rx, ry = xn*c - yn*s, xn*s + yn*c
            if mx*rx + my*ry < best:
                return (x, y, j)
            else:
                j, k = k % n, j + 1
                best = mx*rx + my*ry

    n = len(hull)
    min_area = 10**100
    iL = iR = iP = 1  # indexes left, right, opposite
#   print '    {:>2s} {:>2s} {:>2s} {:>2s} {:>9s}'.format(
#              'i', 'iL', 'iP', 'iR', 'area')
    for i in xrange(n-1):
        dx = hull[i+1][0] - hull[i][0]  # distance on x axis
        dy = hull[i+1][1] - hull[i][1]  # distance on y axis
        theta = pi-atan2(dy, dx)   # get orientation angle of the edge
        s, c = sin(theta), cos(theta)
        yC = hull[i][0]*s + hull[i][1]*c
        xP, yP, iP = _mostfar(iP, n, s, c, 0, 1)
        if i==0: iR = iP
        xR, yR, iR = _mostfar(iR, n, s, c,  1, 0)
        xL, yL, iL = _mostfar(iL, n, s, c, -1, 0)
        l, w = (yP-yC), (xR-xL)
        area = l*w
#       print '    {:2d} {:2d} {:2d} {:2d} {:9.3f}'.format(i, iL, iP, iR, area)
        if area < min_area:
            min_area, min_length, min_width = area, l, w
    return (min_length, min_width, min_area)

if __name__ == '__main__':

    hull= [(560023.44957588764, 6362057.3904932579),
           (560023.44957588764, 6362060.3904932579),
           (560024.44957588764, 6362063.3904932579),
           (560026.94957588764, 6362068.3904932579),
           (560028.44957588764, 6362069.8904932579),
           (560034.94957588764, 6362071.8904932579),
           (560036.44957588764, 6362071.8904932579),
           (560037.44957588764, 6362070.3904932579),
           (560037.44957588764, 6362064.8904932579),
           (560036.44957588764, 6362063.3904932579),
           (560034.94957588764, 6362061.3904932579),
           (560026.94957588764, 6362057.8904932579),
           (560025.44957588764, 6362057.3904932579),
           (560023.44957588764, 6362057.3904932579)]

    print "dimensions and area of smallest enclosing rectangular area:"
    print "  {:.3f}(L) x {:.3f}(W) = {:.3f} area".format(
             *get_minimum_area_rectangle(hull))
于 2012-11-26T15:56:54.670 回答