Question

首先，我是Python（编程领域）的新手，但我希望学习并转换由jwpat7开发的函数。给出一组来自凸壳的点

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)]

此脚本返回此post problem之后所有可能区域的打印件。代码开发 jwpat7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

我希望创建一个函数，返回最小矩形的长度，宽度和面积。例如：

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

我的问题是：

我需要创建一个函数或两个函数。例如：def mostfar和get_minimum_area_rectangle
船体是一个价值清单。它是最好的格式吗？
1. 关注一种功能方法，我有一个问题要整合到大多数里面

提前致谢

1）解决方案：一个功能按照Scott Hunter建议的第一个解决方案，我有一个问题是在get_minimum_area_rectangle（）中集成mostfar（）。任何建议或帮助都非常感激，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

从这里按照上面的jwpat7示例我需要使用mostfar（）。我有一个问题需要了解如何整合（抱歉不正确的术语）至少在这一点上

Answer 1

您可以使用单个功能或两个功能，但它可能更简洁，更容易使用两个功能。您可以按原样保留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):
```
...然后缩进代码的其余部分（以n = len(hull)开头）以形成函数体。您还需要更改函数以返回要获取的值（长度，宽度和面积）。这将使您的代码保持模块化和清洁，并且只需要很少的更改。
使用hull的值列表似乎没有用于此目的。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，与您必须的数学相比，它不应该成为瓶颈。

Answer 2

您当然可以将其作为单一功能进行操作：稍微修改一下，而不是打印找到的区域，跟踪最小的＆amp;与之相关的信息。或者你可以让它将它打印的值收集到lst中，G.E.A.R。然后可以用来找到最小值。

我认为你的船体没有任何问题。

Answer 3

这是一个如何从代码中创建仿函数对象并使用它的示例 - 以及对我认为值得的其他一些事情的一些更改。仿函数是一个实现函数角色但可以像对象一样操作的实体。

在Python中，两者之间的区别较小，因为函数已经是单例对象，但有时为一个对象创建一个专用类很有用。在这种情况下，它允许将辅助函数设置为私有类方法，而不是它是全局的或嵌套的，您似乎反对这样做。

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

Answer 4

我正在发布另一个答案，显示我和其他人建议的方法，这只是将辅助函数mostfar()嵌套在被调用的主函数中。这在Python中很容易实现，因为嵌套函数可以访问其封闭范围的局部变量（例如hull）。我还按照惯例重命名了函数_mostfar()，以表明某些内容是私有的，但这并非绝对必要（永远，绝对不是在这里）。

正如您所看到的，大多数代码与我的其他答案中的代码非常相似，尽管我确实简化了与嵌套函数无关的一些事情（因此它们可能会集成到您选择的任何答案中）。

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))

Python：在函数中实现脚本。一些建议

4 个答案: