MATLAB实现拉丁超立方模拟？

发布网友 发布时间：2022-04-22 18:36

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共5个回答

热心网友 时间：2023-06-25 17:51

以上的回答都太复杂了，其中那个自己编MATLAB程序的，我看了看分布并不理想。建议楼主直接调用MATLAB里的lhsdesign函数进行抽样。以下是我举得一个例子：
%用lhsdesign(20,2)来生成20行2列的的矩阵，2列分别20个0～1之间的均匀样本点，你试试下面代码：
A=lhsdesign(20,2);
x1=A(:,1)*100;
x2=A(:,2)*110+20;
scatter(x1,x2,50,'filled')

热心网友 时间：2023-06-25 17:51

N=10;
UB=100;LB=-100;
D=2;
i=1;
for i=1:D
S(:, i) = ((randperm(N) -1 + rand(1, N)))' / N;
end
S=S*(UB-LB)+LB;
figure(6)
plot(S(:,1),S(:,2),'.')
以上是简单的拉丁超立法采样模拟，不为别的，就希望有人能看到，有批评指正就更好了

热心网友 时间：2023-06-25 17:52

function S=LHS(m,dist,mu,sigma,lowb,upb)
%code of Latin Hypercube Sampling%
%call S=lhs(m,dist,mu,sigma,lowb,upb)
%
%Input argument
%m: a scalar,the number of sample points
%dist: A row with distribution type flags of basic random variables;the
%value of the flag can be 1 (for uniform distribution, 2(for normal distribution), 3(for lognormal)
%and 4(for extreme type 1).
%mu: A row vector comprising the mean value of basic random variables.
%sigma: A row vector with its length equaligng to mu,including the standard
%deviation of basic random variables.
%lowb: a row vector with its elements are the lower bound of the sampling
%interval
%upb:a row vector with its elements are the upper bounds of the sampling
%interval
%dist,mu,sigma,lowb,upb must have the same length.
%Output argument
%S: sampling point matrix, of which each row is a sampling point.
%
%last update june 13, 2009
n=length(mu);
if length(dist)~=n|length(sigma)~=n|length(lowb)~=n|length(upb)~=n
error('dist,mu,sigma,lowb,upb must have the same length');
end
rvcom=[];
for j=1:n
rv=[];
if dist(j)==1
p_low(j)=unifcdf(lowb(j),lowb(j),upb(j));
p_up(j)=unifcdf(upb(j),lowb(j),upb(j));
p_bound(j)=p_up(j)-p_low(j);
p_subbound(j)=p_bound(j)./(m-1);
for i=0:m-1
rv=[rv;unifinv(p_low(j)+i.*p_subbound(j),lowb(j),upb(j))];
end
elseif dist(j)==2
p_low(j)=normcdf(lowb(j),mu(j),sigma(j));
p_up(j)=normcdf(upb(j),mu(j),sigma(j));
p_bound(j)=p_up(j)-p_low(j);
p_subbound(j)=p_bound(j)./(m-1);
for i=0:m-1
rv=[rv;norminv(p_low(j)+i.*p_subbound(j),mu(j),sigma(j))];
end
elseif dist(j)==3
p_low(j)=logncdf(lowb(j),mu(j),sigma(j));
p_up(j)=logncdf(upb(j),mu(j),sigma(j));
p_bound(j)=p_up(j)-p_low(j);
p_subbound(j)=p_bound(j)./(m-1);
for i=0:m-1
rv=[rv;logninv(p_low(j)+i.*p_subbound(j),mu(j),sigma(j))];
end
elseif dist(j)==4
p_low(j)=evcdf(lowb(j),mu(j),sigma(j));
p_up(j)=evcdf(upb(j),mu(j),sigma(j));
p_bound(j)=p_up(j)-p_low(j);
p_subbound(j)=p_bound(j)./(m-1);
for i=O:m-1
rv=[rv;evinv(p_low(j)+i.*p_subbound(j),mu(j),sigma(j))];
end
end
rvcom=[rvcom rv];
end
S=[];
for i=1:n
S=[S randsample(rvcom(:,i),m)];
end
if n==2
figure
plot(S(:,1),S(:,2),'k.');
elseif n==3
figure
plot3(S(:,1),S(:,2),S(:,3),'lc.');
end

热心网友 时间：2023-06-25 17:52

先说二维的举例，*你类推。
如你要在x，y两个变量中选k个点，先按概率分其区间为k分（或其他方法），选每一份的中点作为样本点，然后此时你有（x1,x2,…,xk)，(y1,y2,…,yk)，这时你在将他们两个随机组合（可以用randperm）。

热心网友 时间：2023-06-25 17:53

这中东西貌似大侠都不会说的，要自己去研究哦～