用极差变换法把X归一化,得到归一矩阵Z作者: 匿名 时间: 2012-8-10 11:41
当你写这一行程序作者: gaoxian 时间: 2012-8-10 11:43
for j=5:6 Z3(:,j)=(X(:,j)-min(X(:,j)))/(max(X(:,j))-min(X(:,j)));已经默认Z3是一个四行六列的矩阵只不过前面四列都默认为0后面两列存储了数据,而后面Z=[Z1,Z2,Z3];这时候你的Z是一个四行十列的矩阵,所以中间会有四列的0作者: gaoxian 时间: 2012-8-10 11:48
for j=1:3
Z1(:,j)=(X(:,j)-min(X(:,j)))/(max(X(:,j))-min(X(:,j)));
for j=1
Z2=(min(X(:,j+3))-X(:,j+3))/(max(X(:,j+3))-min(X(:,j+3)));
for j=1:2
Z3(:,j)=(X(:,j+4)-min(X(:,j+4)))/(max(X(:,j+4))-min(X(:,j+4)));
Z=[Z1,Z2,Z3];
end
end
end
Z
这样就可以了作者: gaoxian 时间: 2012-8-10 11:51
for j=1:3
Z1(:,j)=(X(:,j)-min(X(:,j)))/(max(X(:,j))-min(X(:,j)));
end
for j=1
Z2=(min(X(:,j+3))-X(:,j+3))/(max(X(:,j+3))-min(X(:,j+3)));
end
for j=1:2
Z3(:,j)=(X(:,j+4)-min(X(:,j+4)))/(max(X(:,j+4))-min(X(:,j+4)));
end
Z=[Z1,Z2,Z3]
不过我这样写和你那样写是同样的作用,不明白你为什么要嵌套进去,其实这样更好理解吧作者: HNzhangjie 时间: 2012-8-10 14:53
gaoxian 发表于 2012-8-10 11:51
for j=1:3
Z1(:,j)=(X(:,j)-min(X(:,j)))/(max(X(:,j))-min(X(:,j)));
end