在函数中实现脚本有什么建议吗？

如何解决在函数中实现脚本有什么建议吗？

这是一个如何从代码中使其成为函子对象并使用它的示例，以及对一些我认为值得的其他事情的更改。函子是充当功能的实体，但可以像对象一样对其进行操作。

在Python中，由于函数已经是单例对象，因此两者之间的区别较小，但是有时为一个对象创建专用类很有用。在这种情况下，它允许将辅助函数做成私有类方法，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

解决方法

首先，我是Python（编程领域）的新手，但我希望学习和转换jwpat7开发的函数。给定一组从凸包得到的点

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,(560037.44957588764,6362070.3904932579),6362064.8904932579),6362061.3904932579),6362057.8904932579),(560025.44957588764,6362057.3904932579)]

这个脚本返回这个问题之后的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,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,1)
    if i==0: iR = iP
    xR,yR,iR = mostfar(iR,1,0)
    xL,yL,iL = mostfar(iL,-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

我希望创建一个单一函数，并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,Area
18.036,10.392,187.451

我的问题是：

我需要创建一个功能还是两个功能。例如：def最远和get_minimum_area_rectangle
船体是值的列表。是最好的格式吗？
1. followingint一种功能的方法，我有一个问题，要整合到大多数

提前致谢

1）解决方案：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（）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,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,math.cos(theta)
        yC = hull[i][0]*s + hull[i][1]*c

从上面的jwpat7示例开始，我需要使用mostfar（）。我有一个问题要了解在这一点上大多数情况下如何整合（抱歉，术语不正确）

在函数中实现脚本有什么建议吗？

如何解决在函数中实现脚本有什么建议吗？

解决方法

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