最小覆盖原算法 python 实现

百度一圈没有找到合适的博客，通过外网找到了python实现，所以整理记录一下。

最小圆问题

The smallest-circle problem (also known as minimum covering circle problem, bounding circle problem, smallest enclosing circle problem）
有多种算法求解实现，具体参见维基百科。
最容易理解，使用最多的是一些计算几何算法，大家可以参考这篇博文最小圆覆盖（经典算法【三点定圆）进行理论学习

代码实现

scipy.optimize.leastsq

转化将上述最小圆问题，转化为半径最小时的圆心定位问题，再通过scipy中的最小二乘算法进行求解即可。
展示一个例子

from numpy import *

# Coordinates of the 2D points

x = r_[  9, 35, -13,  10,  23,   0]
y = r_[ 34, 10,   6, -14,  27, -10]
basename = 'circle'
x_m = mean(x)
y_m = mean(y)

# Decorator to count functions calls
import functools
def countcalls(fn):
    "decorator function count function calls "

    @functools.wraps(fn)
    def wrapped(*args):
        wrapped.ncalls +=1
        return fn(*args)

    wrapped.ncalls = 0
    return wrapped

#  == METHOD 2 ==
# Basic usage of optimize.leastsq
from scipy      import optimize


method_2  = "leastsq"

def calc_R(xc, yc):
    """ calculate the distance of each 2D points from the center (xc, yc) """
    return sqrt((x-xc)**2 + (y-yc)**2)

@countcalls
def f_2(c):
    """ calculate the algebraic distance between the 2D points and the mean circle centered at c=(xc, yc) """
    Ri = calc_R(*c)
    return Ri - Ri.mean()

center_estimate = x_m, y_m
center_2, ier = optimize.leastsq(f_2, center_estimate)

xc_2, yc_2 = center_2
Ri_2       = calc_R(xc_2, yc_2)
R_2        = Ri_2.mean()
residu_2   = sum((Ri_2 - R_2)**2)
residu2_2  = sum((Ri_2**2-R_2**2)**2)
ncalls_2   = f_2.ncalls

# == METHOD 2b ==
# Advanced usage, with jacobian
method_2b  = "leastsq with jacobian"

def calc_R(xc, yc):
    """ calculate the distance of each 2D points from the center c=(xc, yc) """
    return sqrt((x-xc)**2 + (y-yc)**2)

@countcalls
def f_2b(c):
    """ calculate the algebraic distance between the 2D points and the mean circle centered at c=(xc, yc) """
    Ri = calc_R(*c)
    return Ri - Ri.mean()

@countcalls
def Df_2b(c):
    """ Jacobian of f_2b
    The axis corresponding to derivatives must be coherent with the col_deriv option of leastsq"""
    xc, yc     = c
    df2b_dc    = empty((len(c), x.size))

    Ri = calc_R(xc, yc)
    df2b_dc[ 0] = (xc - x)/Ri                   # dR/dxc
    df2b_dc[ 1] = (yc - y)/Ri                   # dR/dyc
    df2b_dc       = df2b_dc - df2b_dc.mean(axis=1)[:, newaxis]

    return df2b_dc

center_estimate = x_m, y_m
center_2b, ier = optimize.leastsq(f_2b, center_estimate, Dfun=Df_2b, col_deriv=True)

xc_2b, yc_2b = center_2b
Ri_2b        = calc_R(xc_2b, yc_2b)
R_2b         = Ri_2b.mean()
residu_2b    = sum((Ri_2b - R_2b)**2)
residu2_2b   = sum((Ri_2b**2-R_2b**2)**2)
ncalls_2b    = f_2b.ncalls

print ("""
Method 2b :
print "Functions calls : f_2b=%d Df_2b=%d""" % ( f_2b.ncalls, Df_2b.ncalls))

print  (method_2 , xc_2 , yc_2 , R_2 , ncalls_2 , Ri_2.std() , residu_2 , residu2_2 )

参考
https://scipy-cookbook.readthedocs.io/items/Least_Squares_Circle.html

Project Nayuki

网页提供了不同语言的算法源程序，可以供大家参考学习
算法实现参考
https://www.nayuki.io/page/smallest-enclosing-circle

标签：center,python,xc,yc,ncalls,最小,算法,2b,Ri
来源： https://blog.csdn.net/Forrest97/article/details/114366968