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 mathdef 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：
- 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？
- 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/pythonimport 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）
回答：
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, sinclass 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
以上是 在函数中实现脚本。有什么建议吗？ 的全部内容， 来源链接： utcz.com/qa/406314.html
