【发布时间】:2015-03-31 14:10:18
【问题描述】:
这是我第一次使用curve_fit,我还没有找到与我的问题相匹配的示例。我的问题是,我是否正确使用了 curve_fit 数据格式?如果是,那么我的问题是数学问题,我必须纠正它(我还没有发现任何错误......)。 这是我的代码:
import numpy as np
import math as m
import scipy.optimize as sci
def f4(X,Tx,Ty,Tz,i,j,k):
res=[]
#X looks like [x1,y1,z1,x2,y2,z2....]
for n in range(np.shape(X)[0]/3):
xr=X[n*3]+Tx+m.cos(j)*m.cos(k)*X[n*3]-m.cos(j)*m.sin(k)*X[n*3+1]+m.sin(j)*X[n*3+2]
yr=X[n*3+1]+Ty+(m.sin(i)*m.sin(j)*m.cos(k)+m.cos(i)*m.sin(k))*X[n*3]+(-m.sin(i)*m.sin(j)*m.sin(k)+m.cos(i)*m.cos(k))*X[n*3+1]-m.sin(i)*m.cos(j)*X[n*3+2]
zr=X[n*3+2]+Tz+(-m.cos(i)*m.sin(j)*m.cos(k)+m.sin(i)*m.sin(k))*X[n*3]+(m.cos(i)*m.sin(j)*m.sin(k)+m.sin(i)*m.cos(k))*X[n*3+1]+m.cos(i)*m.cos(j)*X[n*3+2]
res.append(xr)
res.append(yr)
res.append(zr)
return res
xdata2=[998.362,5000.052,99.4,997.862,5000.052,99.4,998.362,5000.052,98.9,997.862,5000.052,98.9]
ydata2=[999.46555,4999.801,99.4,999.112,5000.15455,99.4,999.46555,4999.801,98.9,999.112,5000.15455,98.9]
p0=[1,0,0,0,0,0.8]
popt,pcov=sci.curve_fit(f4,xdata2,ydata2,p0)
它引发:RuntimeError:未找到最佳参数:函数调用次数已达到 maxfev = 1400。
根据我的计算,p0 非常接近解。 我的代码还可以吗,还是应该继续寻找数学错误?
如果有人好奇,我试图找到 3 个平移和 3 个旋转,我必须将其应用于一组点 Xp 以获得一组点 Xr。
感谢任何建议
编辑 1/04:
我尝试了 unutbu 的方式来改变我的功能:
def f2(X,Tx,Ty,Tz,i,j,k):
res=[]
for n in range(np.shape(X)[0]):
xr=X[n][0]+Tx+m.cos(j)*m.cos(k)*X[n][0]-m.cos(j)*m.sin(k)*X[n][1]+m.sin(j)*X[n][2]
yr=X[n][1]+Ty+(m.sin(i)*m.sin(j)*m.cos(k)+m.cos(i)*m.sin(k))*X[n][0]+(-m.sin(i)*m.sin(j)*m.sin(k)+m.cos(i)*m.cos(k))*X[n][1]-m.sin(i)*m.cos(j)*X[n][2]
zr=X[n][2]+Tz+(-m.cos(i)*m.sin(j)*m.cos(k)+m.sin(i)*m.sin(k))*X[n][0]+(m.cos(i)*m.sin(j)*m.sin(k)+m.sin(i)*m.cos(k))*X[n][1]+m.cos(i)*m.cos(j)*X[n][2]
aux=[xr,yr,zr]
res.append(aux)
res=np.array(res)
return res
我在数组中添加了两个点:
xdata3=np.array([[998.362,5000.052,99.4],[997.862,5000.052,99.4],[998.362,5000.052,98.9],[997.862,5000.052,98.9],[999.112,4999.801,98.9],[999.112,4999.801,99.4]])
ydata3=np.array([[999.46555,4999.801,99.4],[999.112,5000.15455,99.4],[999.46555,4999.801,98.9],[999.112,5000.15455,98.9],[1000.112,4999.801,98.9],[1000.112,4999.801,99.4]])
我试用了这个功能:
test=f2(xdata3,1,0.00001,0.00001,0.00001,0.00001,0.8)
In [17]:test
Out[17]:
array([[-1891.88925911, 9199.80185261, 198.87092002],
[-1892.73761247, 9199.44317456, 198.87091992],
[-1891.88926411, 9199.80185761, 197.87092002],
[-1892.73761747, 9199.44317956, 197.87091992],
[-1890.4366777 , 9199.91400129, 197.87091663],
[-1890.4366727 , 9199.91399629, 198.87091663]])
也改变了 p0:
p0=[1,0.00001,0.00001,0.00001,0.00001,0.8]
然后我尝试了curve_fit:
test=f2(xdata3,1,0.00001,0.00001,0.00001,0.00001,0.8)
我有一个不同的错误:
error: Result from function call is not a proper array of floats.
不确定发生了什么,因为我的测试变量看起来不错。
【问题讨论】:
标签: python scipy least-squares