在函数中实现脚本.有什么建议吗?

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

first of all, i am quite new in Python (an programming area) but i wish to learn and convert a function developed by jwpat7. Given a set of points derived from a convex hull

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

this script return a print of all possible area following this post problem. The code develop by jwpat7 is:

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

我希望创建一个单个函数，并返回最小矩形的Length，Width和Area.例如:

i wish to create a single function with the return of Length, Width, and Area of the smallest rectangle. Ex:

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

我的问题是:

1. 我需要创建一个功能还是两个功能.例如:def 最远和get_minimum_area_rectangle
2. hull是值列表.这是最好的格式吗?

1. do i need to create a single function or two function. ex: def mostfar and get_minimum_area_rectangle
2. hull is a list of value. Is it the best format?

1. 遵循一种功能方法，我在大多数情况下都无法集成

预先感谢

1)解决方案:一种功能 按照Scott Hunter提出的第一个解决方案，我在将mostfar()集成到get_minimum_area_rectangle()中时遇到了一个问题.任何建议或帮助都非常感激，因为我可以学习.

1) solution: one function following the first solution suggest by Scott Hunter, i have a problem to integrate mostfar() inside get_minimum_area_rectangle(). Any suggestion or help are really appreciate because i can learn.

#!/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

从这里跟随jwpat7的上述示例，我需要使用mostfar().我有一个问题要了解在这一点上大部分时候是如何整合的(对不起这个词，对不起)

from here following the above example of jwpat7 i need to use mostfar(). I have a problem to understand how integrate (sorry for the not right term) mostfar in this point