1.绘制云图
Ex=18
En=2
He=0.2
hold on
for i=1:1000
Enn=randn(1)*He+En;
x(i)=randn(1)*Enn+Ex;
y(i)=exp(-(x(i)-Ex)^2/(2*Enn^2));
plot(x(i),y(i),'*')
end
Ex=48.7
En=9.1
He=0.39
hold on
for i=1:1000
Enn=randn(1)*He+En;
x(i)=randn(1)*Enn+Ex;
y(i)=exp(-(x(i)-Ex)^2/(2*Enn^2));
plot(x(i),y(i),'*')
end
2.求期望、熵及超熵
X1=[51.93 52.51 54.70 43.14 43.85 44.48 44.61 52.08];
Y1=[0.91169241573
0.921875
0.96032303371
0.75737359551
0.76983848315 0.7808988764 0.78318117978 0.9143258427];
m=8;
Ex=mean(X1)
En1=zeros(1,m);
for i=1:m
En1(1,i)=abs(X1(1,i)-Ex)/sqrt(-2*log(Y1(1,i)));
end
En=mean(En1);
He=0;
for i=1:m
He=He+(En1(1,i)-En)^2;
end
En=mean(En1)
He=sqrt(He/(m-1))
3.平顶山 so2 环境:
X1=[0.013 0.04 0.054 0.065 0.07 0.067 0.058 0.055 0.045];
Y1=[0.175675676
0.540540541
0.72972973
0.878378378
0.945945946 0.905405405 0.783783784 0.743243243 0.608108108];
m=9;
Ex=mean(X1)
En1=zeros(1,m);
for i=1:m
En1(1,i)=abs(X1(1,i)-Ex)/sqrt(-2*log(Y1(1,i)));
end
En=mean(En1);
He=0;
for i=1:m
He=He+(En1(1,i)-En)^2;
end
En=mean(En1)
He=sqrt(He/(m-1))
1.绘制正向云图
Ex=18
En=2
He=0.2
hold on
for i=1:1000
Enn=randn(1)*He+En;
x(i)=randn(1)*Enn+Ex;
y(i)=exp(-(x(i)-Ex)^2/(2*Enn^2));
plot(x(i),y(i),'*')
end
Ex=48.7
En=9.1
He=0.39
hold on
for i=1:1000
Enn=randn(1)*He+En;
x(i)=randn(1)*Enn+Ex;
y(i)=exp(-(x(i)-Ex)^2/(2*Enn^2));
plot(x(i),y(i),'*')
end
2.逆向云发生器中需要剔除隶属度大于 0. 9999 的云滴,剩
下 个云滴。代码如下:
x=[51.93,52.51,54.7,56.96,43.14,43.85,44.48,44.61,52.08];
y=[0.91169241573,0.921875,0.96032303371,1,0.75737359551,0.7
6983848315,0.7808988764,0.78318117978,0.9143258427];
X1=x;
Y1=y;
i=1;n=9;flag=0;m=0;
while i<=(n-flag)
if Y1(1,i)>0.9999
Y1(:,i)=[];
X1(:,i)=[];
flag=flag+1;
else
i=i+1;
m=m+1;
end
end
m
X1
Y1
输出:
m=8
X1=[51.93 52.51 54.70 43.14 43.85 44.48 44.61 52.08];%除以去
掉的 56.96 得到 Y1,云模型在水资源供求预测中的应用
Y1=[0.91169241573
0.921875
0.96032303371
0.75737359551
0.76983848315 0.7808988764 0.78318117978 0.9143258427];%确定
度或者隶属度
求期望、熵及超熵
X1=[51.93 52.51 54.70 43.14 43.85 44.48 44.61 52.08];%除以去
掉的 56.96 得到 Y1,云模型在水资源供求预测中的应用
Y1=[0.91169241573
0.921875
0.96032303371
0.75737359551
0.76983848315 0.7808988764 0.78318117978 0.9143258427];%确定
度或者隶属度
m=8;
Ex=mean(X1)
En1=zeros(1,m);
for i=1:m
En1(1,i)=abs(X1(1,i)-Ex)/sqrt(-2*log(Y1(1,i)));
end
En=mean(En1);
He=0;
for i=1:m
He=He+(En1(1,i)-En)^2;
end
En=mean(En1)
He=sqrt(He/(m-1))