[讨论]曲线拟合的方法 - 编程论坛

#2

abingchem2006-11-30 10:43

来个最简单的：

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【拟合　目标】：多项式
【函数｜方法】：polyfit,polyval
【具体　实例】：

x=0:0.1:5;
y=exp(-5*x)+x.*x;
subplot(2,1,1),plot(x,y,'r+'),hold on
p=polyfit(x,y,2)
yp=polyval(p,x);
plot(x,yp,'bo')
subplot(2,1,2),plot(yp-y,'*')

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#3

尘埃1222006-11-30 12:38

呵呵，我也说说吧

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【拟合　目标】：模型参数拟合
【函数｜方法】：beta=lsqcurvefit(@fun,beta0,xdata,ydata) 返回求解参数值
beta= nlinfit(xdata,ydata,@fun,beta0)

【具体　实例】：

根据给定数据集合，模型，参数初值，拟合模型中需要确定的参数值

具体实例太长，就不列出来了，

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[此贴子已经被作者于2006-11-30 12:43:37编辑过]

#4

abingchem2006-11-30 14:10

奖励
　　尘埃122

金币　：10
经验　：10
魅力　：10
金钱　：20

#5

ydgsl2006-11-30 23:39

matlab曲线拟合的能力有限。下为一简例：
模型：y=b1+b2*x.^b3+b4*x.^b5+b6*x.^b7;
数据：
x=[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
y=[8.2, 4.6, 4.3, 4.6, 5.1, 5.5, 5.7, 5.5, 5, 3.8]

#6

abingchem2006-12-01 09:03

ydgsl是这方面的大牛，给大家系统说说非线性拟合的情况啊

#7

abingchem2006-12-06 09:31

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【拟合　目标】：多元线性拟合y=a1*x1+a2*x2+..+am*xm
【函数｜方法】：pinv,\

【具体　实例】：数据点{x1i,x2i,…xmi,yi}，

A=|x12,x22,…xm2|
|…………… |
|x1n,x2n,…xmn|

Y={y1,y2,y3,…,yn}'

则系数{a1,a2,…,am}'=pinv(A)*Y
在matlab中使用
coeff=A\Y
则可以得到最小二乘意义上的拟合系数

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#8

abingchem2006-12-06 10:18

【转】Matlab中如何作圆回归
:#Peter Boettcher (boettcher@ll.mit.edu),2002/5/16, comp.soft-sys.matlab#

Q5.5: How can I fit a circle to a set of XY data?
=================================================

An elegant chunk of code to perform least-squares circle fitting was
written by Bucher Izhak and has been floating around the newgroup for
some time. The first reference to it that I can find is in:

function [xc,yc,R,a] = circfit(x,y)
%CIRCFIT Fits a circle in x,y plane
%
% [XC, YC, R, A] = CIRCFIT(X,Y)
% Result is center point (yc,xc) and radius R.A is an optional
% output describing the circle's equation:
%
% x^2+y^2+a(1)*x+a(2)*y+a(3)=0

% by Bucher izhak 25/oct/1991

n=length(x); xx=x.*x; yy=y.*y; xy=x.*y;
A=[sum(x) sum(y) n;sum(xy) sum(yy) sum(y);sum(xx) sum(xy) sum(x)];
B=[-sum(xx+yy) ; -sum(xx.*y+yy.*y) ; -sum(xx.*x+xy.*y)];
a=A\B;
xc = -.5*a(1);
yc = -.5*a(2);
R = sqrt((a(1)^2+a(2)^2)/4-a(3));

Tom Davis provided a more sophisticated approach that works for more
cases in and Code included.

#9

hitzhang2006-12-08 17:29

介绍一种能以人已精度逼近任意函数的方法：神经网络

1.BP网络





>> a=1:10;

b=rand(1,10);
net=newff(minmax(a),[6,1],{'tansig','purelin'},'trainlm');
>> net=train(net,a,b);
TRAINLM, Epoch 0/100, MSE 2.4582/0, Gradient 28.1399/1e-010 TRAINLM, Epoch 25/100, MSE 0.0100718/0, Gradient 0.0645639/1e-010 TRAINLM, Epoch 50/100, MSE 0.00279072/0, Gradient 0.0371558/1e-010 TRAINLM, Epoch 75/100, MSE 0.000259069/0, Gradient 0.000950231/1e-010 TRAINLM, Epoch 100/100, MSE 0.000246122/0, Gradient 0.00282375/1e-010 TRAINLM, Maximum epoch reached, performance goal was not met.
>> b1=sim(net,a);%预测
>> b-b1;%误差
ans = Columns 1 through 7 -0.0000 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0.0000 Columns 8 through 10 0.0351 -0.0351 0.0003

2.径向基函数网络

>> net=newrb(a,b,1);
NEWRB, neurons = 0, SSE = 0.587343
>> b2=sim(net,a);%预测
>> b-b2
ans = Columns 1 through 7 0.0891 0.0611 -0.2402 0.2367 -0.1718 -0.1205 -0.0118 Columns 8 through 10 0.2286 -0.1139 0.0427

3.零误差径向基函数网络

>> net=newrbe(a,b,1);
>> b2=sim(net,a);%预测
>> b-b2
ans = 1.0e-015 * Columns 1 through 7 -0.1110 0.3331 0.1665 -0.6661 -0.1249 -0.0555 0.2220 Columns 8 through 10 0 -0.2776 0.3331

#10

hitzhang2006-12-08 17:32

当然，真正的预测并不如此，还需做区间估计这一部分比较难