首先，我对Python(一个编程领域)很陌生，但我希望学习并转换一个由我自己开发的函数。jwpat7. 给定一组由凸型壳导出的点
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)]
该脚本返回打印所有可能的区域，如下图所示。岗位问题. 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
我想创建一个函数，返回最小矩形的长度，宽度和面积。例：我想创建一个函数，返回最小矩形的长度、宽度和面积。
Length, Width, Area = get_minimum_area_rectangle(hull)
print Length, Width, Area
18.036, 10.392, 187.451
我的问题是：
先谢谢你
1)解决方案：一个函数按照Scott Hunter建议的第一个解决方案，我有一个问题，要把mostfar()整合到get_minimum_area_rectangle()里面。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
在这里按照上面jwpat7的例子，我需要使用mostfar()。我有一个问题要了解如何在这一点上整合（对不起，这个词用得不对）mostfar。
下面是一个例子，说明如何使它成为一个 向量 对象从你的代码中移除并使用它--同时对其他一些我认为值得的东西做了一些修改。触发器是一个实体，它扮演着函数的角色，但可以像对象一样被操作。
在 Python 中，由于函数已经是单人对象，所以两者之间的区别不大，但有时为一个函数创建一个专门的类是很有用的。在这种情况下，它允许帮助函数成为一个私有类方法，而不是像你反对的那样成为全局的或嵌套的。
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
你可以使用一个函数或两个函数 但使用两个函数可能会更干净和容易一些。你可以把 mostfar 函数的原样。然后，只需通过添加函数定义行将代码的后半部分转换成函数。
def get_minimum_area_rectangle(hull):
...然后将代码的其余部分缩进（以... ...为起点）。n = len(hull))来构成函数的主体。你还想改变函数来返回你想得到的值（长度、宽度和面积）。这将使你的代码保持模块化和简洁，并且只需要很少的改动。
使用值的列表来获取 hull 似乎可以满足这个目的。另一种选择是使用一个数组（比如NumPy数组），但在这种情况下，你是在反复检查数据，一次一个项目，而不是同时对多个数据点进行任何计算。所以，一个列表应该是可以的。在列表中访问项目的速度很快，和你要做的数学相比，这应该不会是一个瓶颈。
EDIT: (我忽略了一些代码是在mostfar之外的)我会把 "脚本 "部分(mostfar之后的代码)包装成一个函数，然后按照上面的描述修改那个函数。 然后，你的 "脚本 "就会调用那个函数，或者，如果使用第二个修改，从返回的列表中找到最小值。
我在贴出另一个答案，说明如何按照我（和其他人）的建议来做，这只是嵌套帮助函数--------。mostfar() 内的主函数。这在 Python 中是很容易做到的，因为嵌套函数可以访问其外层作用域的局部变量 (如 hull 的情况下）。) 我还把这个函数改名为 _mostfar() 遵循惯例来表示某些东西是私有的，但这并不是严格意义上的必要条件（从来没有，这里也绝对没有）。
正如你所看到的，大部分代码与我的另一个答案中的代码非常相似，尽管我简化了一些与函数嵌套无关的东西（所以它们可能会被整合到你选择的任何答案中）。
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))