logistic模型

y=[43.65 109.86 187.21 312.67 496.58 707.65 960.25 1238.75 1560.00 1824.29 2199.00 2438. 2737.71] 

y=L/(1+a*exp(-k*x)) 

利用线性回归模型所得到的a和k的估计值和L=3000作为Logistic模型的拟合初值，对Logistic模型做非线性回归。 

%第一步,线性回归模型得到a,k

%这里假定y=a*exp(k*x),对两边取ln(Matlab中，ln用log函数表示),有

%lny=lna+k*x

%即logy是x的线性函数，斜率为k*loge,截距为loga

x=0:1:12 ;

y=[43.65 109.86 187.21 312.67 496.58 707.65 960.25 1238.75 1560.00 1824.29 2199.00 2438. 2737.71] ;

line_A=polyfit(x,log(y),1);

k=line_A(1);

a=exp(line_A(2));

plot(x,y,'*',x,a*exp(k*x))

title('线性回归的参数曲线与已经点的关系')

%第二步，Logistic模型

%在Matlab下输入：edit，然后将下面两行百分号之间的内容，复制进去，保存 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

function y=zhidao_liziqiangde(A,x) 

%其中k=A(1),a=A(2)

k=A(1);

a=A(2);

L=3000;

y=L./(1+a*exp(-k*x));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

%返回Matlab，输入

[ABC,res]=lsqcurvefit('zhidao_liziqiangde',[k,a],x,y); 

kk=ABC(1)

aa=ABC(2)

y_logistic=zhidao_liziqiangde(ABC,x);

figure

plot(x,y,'*',x,y_logistic)

legend('实验数据点','Logistic模型')

==============================================================

总结以上

在Matlab下输入：edit，然后将下面两行百分号之间的内容，复制进去，保存 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

function y=zhidao_liziqiangde(A,x) 

%其中k=A(1),a=A(2)

k=A(1);

a=A(2);

L=3000;

y=L./(1+a*exp(-k*x));

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

返回Matlab下输入

x=0:1:12 ;

y=[43.65 109.86 187.21 312.67 496.58 707.65 960.25 1238.75 1560.00 1824.29 2199.00 2438. 2737.71] ;

line_A=polyfit(x,log(y),1);

k=line_A(1);

a=exp(line_A(2));

plot(x,y,'*',x,a*exp(k*x))

title('线性回归的参数曲线与已经点的关系')

[ABC,res]=lsqcurvefit('zhidao_liziqiangde',[k,a],x,y); 

kk=ABC(1)

aa=ABC(2)

y_logistic=zhidao_liziqiangde(ABC,x);

figure

plot(x,y,'*',x,y_logistic)

legend('实验数据点','Logistic模型')