Latin hypercube sampling of 5 indepentent distributions - matlab

Question

I have 5 variables A,V,h,l and b which all stem from different distributions. I'd like to make a 1,000 equally distributed samples from each distribution by the method of latin hypercube sampling. Is this a realistic request, ie is it really better than simple random sampling? Do you have any references of how I can do this in matlab? This page suggests I would need to transform the sample somehow...

Answer 1

UPDATE #2: solution using built-in function of Statistics Toolbox
The basic question is whether you want your samples on a regular grid or not. If not, you could use the built-in function lhsdesign:
p = 1000 % Number of points
N = 5 % Number of dimensions
lb = [1 1 1 1 1]; % lower bounds for A,V,h,l and b
ub = [10 10 10 10 10]; % upper bounds for A,V,h,l and b
X = lhsdesign(p,N,'criterion','correlation');
D = bsxfun(#plus,lb,bsxfun(#times,X,(ub-lb)));
'criterion','correlation' would give you the desired "equal distribution".
D then contains the irregular coordinate-distribution for your parameters.
First I thought you we're looking for samples on a regular grid, which really seems to be a tough task. I tried to modify the approach above D = round(bsxfun...), but it won't give you satisfying results. So for this case I still provide my initial idea here:
The following solution is far from fast and elegant, but it's at least a solution.
% For at least 1000 samples M=6 divisions are necessary
M = 6;
N = 5;
% the perfect LHC distribution would have 1296 samples for M=6 divisions
% and 5 dimensions
counter_max = M^(N-1); %=1296
% pre-allocation
D = zeros(6,6,6,6,6);
counter = 0;
while counter < 1000
c = randi(6,1,5);
if ( sum( D( c(1) , c(2) , c(3) , c(4) , : )) < 1 && ...
sum( D( c(1) , c(2) , c(3) , : , c(5) )) < 1 && ...
sum( D( c(1) , c(2) , : , c(4) , c(5) )) < 1 && ...
sum( D( c(1) , : , c(3) , c(4) , c(5) )) < 1 && ...
sum( D( : , c(2) , c(3) , c(4) , c(5) )) < 1 )
D(c(1),c(2),c(3),c(4),c(5)) = 1;
X(counter,:) = c;
counter = counter+1;
end
end
X is finally containing the coordinates of all your samples.
As you see I used a while-loop with an underlying if-condition. You wish 1000 samples, that's a realistic number and can be done in a reasonable time. I actually would recommend you to use a number of samples as close as possible to the maximum of 1296. That could take you a pretty while. But as you create the resulting matrix just once and use it again and again, don't hesitate to run it 24hrs. You could also implement an interuption code as described here: In MatLab, is it possible to terminate a script, but save all its internal variables to workspace? and see how many samples you got until then. (I got 900 samples in 20min when I was testing)
UPDATE: Example to show limitations of method:
The following example shall illustrate, what the asker could be willing to do and what the result is actually supposed to look like. Because I'm also very interested in a good solution, mine is limited and cannot provide the "100% result".
Imagine a cube (N=3) with M=10 divisions.
M = 10;
N = 3;
counter_max = M^(N-1); %=100 maximum number of placeable samples
% pre-allocations
D = zeros(10,10,10);
counter = 0;
while counter < counter_max
c = randi(10,1,3);
% if condition checks if there is already a sample in the same row,
% coloumn or z-coordinate,
if ( sum( D( c(1) , c(2) , : )) < 1 && ...
sum( D( c(1) , : , c(3) )) < 1 && ...
sum( D( : , c(2) , c(3) )) < 1 )
%if not a new sample is generated
D(c(1),c(2),c(3)) = 1;
counter = counter+1;
X(counter,:) = c;
end
end
After about 10000 iterations one gets the following distribution with 85 of 100 possible placed samples:
where the color indicates the normalized distance to the closest neighbor. For most of the points it's fine (1), but as there are 15 missing samples, some points are more distant from others.
The problem is: I doubt that it's possible to get all 100 samples in a reasonable time. When one plots the generated samples over the number of iterations you get:
...so the desired result seems hardly obtainable.
Please see this answer more as an encouragement than a solution.

Answer 2

By combining 1-D latin hypercube samples (LHS), you can make a full set of LHS for regular grid in higher order dimension. For example, imagine 3X3 LHS (ie 2-D and 3 divisions). First, you just make 1-D LHS for regular grid. (1,0,0), (0, 1, 0), (0, 0, 1) for 1-D. And then, combine the 1-D LHS to make 2-D LHS.
1, 0, 0
0, 1, 0
0, 0, 1
or
0, 1, 0
1, 0, 0
0, 0, 1
... etc.
LHS for 3-D can also be created using the same method(by combining 2-D LHS).
There are 12 possible LHS for 3X3. Generally, the number of possible LHS is N x ((M-1)!)^(M-1).
(N=divisions, M=dimensions)
The following code shows LHS for 3-D and 10 divisions.
This code generates only one LHS.
result is random.
it takes 0.001288 sec for 100% result
clear;
clc;
M = 3; % dimension
N = 10; % division
Sel2 = ':,';
stop = 0;
P_matrix = repmat([num2str(N),','],1,M);
P_matrix = P_matrix(1:end-1);
eval(['P = zeros(', P_matrix, ');']);
P(1,1) = 1;
tic
while stop == 0
for i = 1 : M-1
for j = 2:N
if i == 1
P(end , j, 1) = P(1 , j-1, 1);
P(1:end-1, j, 1) = P(2:end, j-1, 1);
else
Sel_2 = repmat(Sel2,1,i-1);
Sel_2 = Sel_2(1:end-1);
eval(['P(', Sel_2, ',end , j, 1) = P(', Sel_2 , ', 1 , j-1, 1);']);
eval(['P(', Sel_2, ',1:end-1 , j, 1) = P(', Sel_2 , ', 2:end, j-1, 1);']);
end
end
if i == 1
P(:,:,1) = P(randperm(N),:,1);
elseif i <M-1
Sel_2 = repmat(Sel2,1,i);
Sel_2 = Sel_2(1:end-1);
eval(['P(',Sel_2,',:,1) = P(',Sel_2,',randperm(N),1);']);
else
Sel_2 = repmat(Sel2,1,i);
Sel_2 = Sel_2(1:end-1);
eval(['P(',Sel_2,',:) = P(',Sel_2,',randperm(N));']);
end
end
% you can add stop condition
stop = 1;
end
toc
[x, y, z] = ind2sub(size(P),find(P == 1));
scatter3(x,y,z);
xlabel('X');
ylabel('Y');
zlabel('Z');
Result

Answer 3

This code gives the same result as the accepted one in this discussion.
Check it out!
n=1000; p=5;
distr=lhsdesign(n,p); %creates LHS of n samples on each of your p dimensions
% Then, you can choose any inverted distribution. in this case it is the Discrete Uniform Distribution
Parameters=[unidinv(distr(:,1),UB1) unidinv(distr(:,2),UB2) ...
unidinv(distr(:,3),UB3) unidinv(distr(:,4),UB4) ...
unidinv(distr(:,5),UB5) ];
%At the end, you'll do a simple work of indexing.