MATLAB最大熵法求概率密度的程序

2025-05-21 01:36:42
推荐回答(1个)
回答1:

样本熵代码
function [shang]=jss(xdate)
m=2;
n=length(xdate);
r=0.2*std(xdate);
cr=[];
gn=1;
gnmax=m;
while gn<=gnmax
x2m=zeros(n-m+1,m);%存放变换后的向量
d=zeros(n-m+1,n-m);% 存放距离结果的矩阵
cr1=zeros(1,n-m+1);%存放
k=1;
for i=1:n-m+1

for j=1:m

x2m(i,j)=xdate(i+j-1);

end
end
x2m;

for i=1:n-m+1

for j=1:n-m+1

if i~=j

d(i,k)=max(abs(x2m(i,-x2m(j,));%计算各个元素和响应元素的距离

k=k+1;

end
end

k=1;
end
d;

for i=1:n-m+1

[k,l]=size(find(d(i,
cr1(1,i)=l;
end
cr1;

cr1=(1/(n-m))*cr1;
sum1=0;
for i=1:n-m+1

if cr1(i)~=0

sum1=sum1+log(cr1(i));

end
end
cr1=1/(n-m+1)*sum1;
cr(1,gn)=cr1;
gn=gn+1;
m=m+1;
end
cr;

shang=cr(1,1)-cr(1,2);

function [shang]=ybs(xdate)
m=2;
n=length(xdate);
r=0.2*std(xdate);
cr=[];
gn=1;
gnmax=m;
while gn<=gnmax
x2m=zeros(n-m+1,m);%存放变换后的向量
d=zeros(n-m+1,n-m);% 存放距离结果的矩阵
cr1=zeros(1,n-m+1);%存放
k=1;
for i=1:n-m+1

for j=1:m

x2m(i,j)=xdate(i+j-1);

end
end
x2m;

for i=1:n-m+1

for j=1:n-m+1

if i~=j

d(i,k)=max(abs(x2m(i,-x2m(j,));%计算各个元素和响应元素的距离

k=k+1;

end
end

k=1;
end
d;

for i=1:n-m+1

[k,l]=size(find(d(i,
cr1(1,i)=l;
end
cr1;

cr1=(1/(n-m))*cr1;
sum1=0;
for i=1:n-m+1

sum1=sum1+cr1(i);

end
end
cr1=1/(n-m+1)*sum1;
cr(1,gn)=cr1;
gn=gn+1;
m=m+1;
end
cr;
shang=-log(cr(1,1)/cr(1,2));