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), (560036.44957588764,6362071.8904932579), (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)]
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, 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, Width, Area 18.036, 10.392, 187.451
我的问题是：
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, 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）
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
