(完整版)云模型matlab程序

1、1. 绘制云图Ex=18En=2He=0.2hold onfor i=1:1000Enn=randn(1)*He+En;x(i)=randn(1)*Enn+Ex;y(i)=exp(-(x(i)-Ex)2/(2*Enn2);plot(x(i),y(i),*)endEx=48.7En=9.1He=0.39hold onfor i=1:1000Enn=randn(1)*He+En;x(i)=randn(1)*Enn+Ex;y(i)=exp(-(x(i)-Ex)2/(2*Enn2);plot(x(i),y(i),*)end2. 求期望、熵及超熵X1=51.93 52.51 54.70 43.14 43

2、.85 44.48 44.61 52.08;Y1=0.911692415730.9218750.960323033710.757373595510.76983848315 0.7808988764 0.78318117978 0.9143258427;m=8;Ex=mean(X1)En1=zeros(1,m);for i=1:mEn1(1,i)=abs(X1(1,i)-Ex)/sqrt(-2*log(Y1(1,i);endEn=mean(En1);He=0;for i=1:mHe=He+(En1(1,i)-En)2;endEn=mean(En1)He=sqrt(He/(m-1)3. 平顶山 s

3、o2 环境 :X1=0.013 0.04 0.054 0.065 0.07 0.067 0.058 0.055 0.045;Y1=0.1756756760.5405405410.729729730.8783783780.945945946 0.905405405 0.783783784 0.743243243 0.608108108;m=9;Ex=mean(X1)En1=zeros(1,m);for i=1:mEn1(1,i)=abs(X1(1,i)-Ex)/sqrt(-2*log(Y1(1,i);endEn=mean(En1);He=0;for i=1:mHe=He+(En1(1,i)-En

4、)2;endEn=mean(En1)He=sqrt(He/(m-1)1. 绘制正向云图Ex=18En=2He=0.2hold on for i=1:1000Enn=randn(1)*He+En;x(i)=randn(1)*Enn+Ex;y(i)=exp(-(x(i)-Ex)2/(2*Enn2);plot(x(i),y(i),*)endEx=48.7En=9.1He=0.39hold onfor i=1:1000Enn=randn(1)*He+En;x(i)=randn(1)*Enn+Ex;y(i)=exp(-(x(i)-Ex)2/(2*Enn2);plot(x(i),y(i),*)end2.

5、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;6983848315,0.7808988764,0.78318117978,0.9143258427;X1=x;Y1=y;i=1;n=9;flag=0;m=0;while i0.9999Y1(:,i)=;X1(:,i)=;flag=flag+1;elsei=i+1;m=m+1;endendmX1Y1输出：m=8X1=51.93 52.51 54.70 43.14 43.85 44.48 44.61 52.0

6、8;%掉的 56.96 得到 Y1，云模型在水资源供求预测中的应用Y1=0.911692415730.9218750.960323033710.757373595510.76983848315 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=0.911692415730.9218750.960323033710.757373595510.76983848315 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:mEn1(1,i)=abs(X1(1,i)-Ex)/sqr