I am trying to find the Mahalanobis distance of some points from the origin.The MATLAB command for that is mahal(Y,X)
But if I use this I get NaN as the matrix X =0 as the distance needs to be found from the origin.Can someone please help me with this.How should it be done
I think you are a bit confused about what mahal() is doing. First, computation of the Mahalanobis distance requires a population of points, from which the covariance will be calculated.
In the Matlab docs for this function it makes it clear that the distance being computed is:
d(I) = (Y(I,:)-mu)*inv(SIGMA)*(Y(I,:)-mu)'
where mu is the population average of X and SIGMA is the population covariance matrix of X. Since your population consists of a single point (the origin), it has no covariance, and so the SIGMA matrix is not invertible, hence the error where you get NaN/Inf values in the distances.
If you know the covariance structure that you want to use for the Mahalanobis distance, then you can just use the formula above to compute it for yourself. Let's say that the covariance you care about is stored in a matrix S. You want the distance w.r.t. the origin, so you don't need to subtract anything from the values in Y, all you need to compute is:
for ii = 1:size(Y,1)
d(ii) = Y(ii,:)*inv(S)*Y(ii,:)'; % Where Y(ii,:) is assumed to be a row vector.'
end
Related
Suppose I have a continuous probability distribution, e.g., Normal, on a support A. Suppose that there is a Matlab code that allows me to draw random numbers from such a distribution, e.g., this.
I want to build a Matlab code to "approximate" this continuous probability distribution with a probability mass function spanning over r points.
This means that I want to write a Matlab code to:
(1) Select r points from A. Let us call these points a1,a2,...,ar. These points will constitute the new discretised support.
(2) Construct a probability mass function over a1,a2,...,ar. This probability mass function should "well" approximate the original continuous probability distribution.
Could you help by providing also an example? This is a similar question asked for Julia.
Here some of my thoughts. Suppose that the continuous probability distribution of interest is one-dimensional. One way to go could be:
(1) Draw 10^6 random numbers from the continuous probability distribution of interest and store them in a column vector D.
(2) Suppose that r=10. Compute the 10-th, 20-th,..., 90-th quantiles of D. Find the median point falling in each of the 10 bins obtained. Call these median points a1,...,ar.
How can I construct the probability mass function from here?
Also, how can I generalise this procedure to more than one dimension?
Update using histcounts: I thought about using histcounts. Do you think it is a valid option? For many dimensions I can use this.
clear
rng default
%(1) Draw P random numbers for standard normal distribution
P=10^6;
X = randn(P,1);
%(2) Apply histcounts
[N,edges] = histcounts(X);
%(3) Construct the new discrete random variable
%(3.1) The support of the discrete random variable is the collection of the mean values of each bin
supp=zeros(size(N,2),1);
for j=2:size(N,2)+1
supp(j-1)=(edges(j)-edges(j-1))/2+edges(j-1);
end
%(3.2) The probability mass function of the discrete random variable is the
%number of X within each bin divided by P
pmass=N/P;
%(4) Check if the approximation is OK
%(4.1) Find the CDF of the discrete random variable
CDF_discrete=zeros(size(N,2),1);
for h=2:size(N,2)+1
CDF_discrete(h-1)=sum(X<=edges(h))/P;
end
%(4.2) Plot empirical CDF of the original random variable and CDF_discrete
ecdf(X)
hold on
scatter(supp, CDF_discrete)
I don't know if this is what you're after but maybe it can help you. You know, P(X = x) = 0 for any point in a continuous probability distribution, that is the pointwise probability of X mapping to x is infinitesimal small, and thus regarded as 0.
What you could do instead, in order to approximate it to a discrete probability space, is to define some points (x_1, x_2, ..., x_n), and let their discrete probabilities be the integral of some range of the PDF (from your continuous probability distribution), that is
P(x_1) = P(X \in (-infty, x_1_end)), P(x_2) = P(X \in (x_1_end, x_2_end)), ..., P(x_n) = P(X \in (x_(n-1)_end, +infty))
:-)
I am trying to find the point that is at a minimum distance from the candidate set. Z is a matrix where the rows are the dimension and columns indicate points. Computing the inter-point distances, and then recording the point with minimum distance and its distance as well. Below is the code snippet. The code works fine for a small dimension and small set of points. But, it takes a long time for large data set (N = 1 million data points and dimension is also high). Is there an efficient way?
I suggest that you use pdist to do the heavy lifting for you. This function will compute the pairwise distance between every two points in your array. The resulting vector has to be put into matrix form using squareform in order to find the minimal value for each pair:
N = 100;
Z = rand(2,N); % each column is a 2-dimensional point
% pdist assumes that the second index corresponds to dimensions
% so we need to transpose inside pdist()
distmatrix = squareform(pdist(Z.','euclidean')); % output is [N, N] in size
% set diagonal values to infinity to avoid getting 0 self-distance as minimum
distmatrix = distmatrix + diag(inf(1,size(distmatrix,1)));
mindists = min(distmatrix,[],2); % find the minimum for each row
sum_dist = sum(mindists); % sum of minimal distance between each pair of points
This computes every pair twice, but I think this is true for your original implementation.
The idea is that pdist computes the pairwise distance between the columns of its input. So we put the transpose of Z into pdist. Since the full output is always a square matrix with zero diagonal, pdist is implemented such that it only returns the values above the diagonal, in a vector. So a call to squareform is needed to get the proper distance matrix. Then, the row-wise minimum of this matrix have to be found, but first we have to exclude the zero in the diagonals. I was lazy so I put inf into the diagonals, to make sure that the minimum is elsewhere. In the end we just have to sum up the minimal distances.
I've got an arbitrary probability density function discretized as a matrix in Matlab, that means that for every pair x,y the probability is stored in the matrix:
A(x,y) = probability
This is a 100x100 matrix, and I would like to be able to generate random samples of two dimensions (x,y) out of this matrix and also, if possible, to be able to calculate the mean and other moments of the PDF. I want to do this because after resampling, I want to fit the samples to an approximated Gaussian Mixture Model.
I've been looking everywhere but I haven't found anything as specific as this. I hope you may be able to help me.
Thank you.
If you really have a discrete probably density function defined by A (as opposed to a continuous probability density function that is merely described by A), you can "cheat" by turning your 2D problem into a 1D problem.
%define the possible values for the (x,y) pair
row_vals = [1:size(A,1)]'*ones(1,size(A,2)); %all x values
col_vals = ones(size(A,1),1)*[1:size(A,2)]; %all y values
%convert your 2D problem into a 1D problem
A = A(:);
row_vals = row_vals(:);
col_vals = col_vals(:);
%calculate your fake 1D CDF, assumes sum(A(:))==1
CDF = cumsum(A); %remember, first term out of of cumsum is not zero
%because of the operation we're doing below (interp1 followed by ceil)
%we need the CDF to start at zero
CDF = [0; CDF(:)];
%generate random values
N_vals = 1000; %give me 1000 values
rand_vals = rand(N_vals,1); %spans zero to one
%look into CDF to see which index the rand val corresponds to
out_val = interp1(CDF,[0:1/(length(CDF)-1):1],rand_vals); %spans zero to one
ind = ceil(out_val*length(A));
%using the inds, you can lookup each pair of values
xy_values = [row_vals(ind) col_vals(ind)];
I hope that this helps!
Chip
I don't believe matlab has built-in functionality for generating multivariate random variables with arbitrary distribution. As a matter of fact, the same is true for univariate random numbers. But while the latter can be easily generated based on the cumulative distribution function, the CDF does not exist for multivariate distributions, so generating such numbers is much more messy (the main problem is the fact that 2 or more variables have correlation). So this part of your question is far beyond the scope of this site.
Since half an answer is better than no answer, here's how you can compute the mean and higher moments numerically using matlab:
%generate some dummy input
xv=linspace(-50,50,101);
yv=linspace(-30,30,100);
[x y]=meshgrid(xv,yv);
%define a discretized two-hump Gaussian distribution
A=floor(15*exp(-((x-10).^2+y.^2)/100)+15*exp(-((x+25).^2+y.^2)/100));
A=A/sum(A(:)); %normalized to sum to 1
%plot it if you like
%figure;
%surf(x,y,A)
%actual half-answer starts here
%get normalized pdf
weight=trapz(xv,trapz(yv,A));
A=A/weight; %A normalized to 1 according to trapz^2
%mean
mean_x=trapz(xv,trapz(yv,A.*x));
mean_y=trapz(xv,trapz(yv,A.*y));
So, the point is that you can perform a double integral on a rectangular mesh using two consecutive calls to trapz. This allows you to compute the integral of any quantity that has the same shape as your mesh, but a drawback is that vector components have to be computed independently. If you only wish to compute things which can be parametrized with x and y (which are naturally the same size as you mesh), then you can get along without having to do any additional thinking.
You could also define a function for the integration:
function res=trapz2(xv,yv,A,arg)
if ~isscalar(arg) && any(size(arg)~=size(A))
error('Size of A and var must be the same!')
end
res=trapz(xv,trapz(yv,A.*arg));
end
This way you can compute stuff like
weight=trapz2(xv,yv,A,1);
mean_x=trapz2(xv,yv,A,x);
NOTE: the reason I used a 101x100 mesh in the example is that the double call to trapz should be performed in the proper order. If you interchange xv and yv in the calls, you get the wrong answer due to inconsistency with the definition of A, but this will not be evident if A is square. I suggest avoiding symmetric quantities during the development stage.
I am using pdist2(x(i), y(j), 'euclidean') formula to find euclidean distance between x and y instead of the manual method
sqrt((x(i)-y(i))^2).
And to find co relation coefficient I am using corrcoeff( x(i), y(j)) . Is this the correct way to find correlation coefficient and euclidean distance between x and y matrix?
I get different answers when I use formula and manual method.
I think it is not correct.
I suppose x and y are matrixes, so you are using pdist2 to compute the distance between the observation i (from x(i)) and j (from y(j)). In the case of the manual method you are using the i index in both of them. Maybe the differences are due to the incorrect use of the indexes i and j. Show us your code so we can confirm it.
By the way, as #Luis pointed out, it is better to use pdist2 to compute all the distances at the same time (it is much faster). So, if you have two matrixes x and y, use: pdist2(x,y).
The same for the correlation.
The correlation between the two matrices can be calculated as:
r = corr2(x,y)
Now, if you are after the element-wise distance, how about:
dist=gsqrt((x-y).^2);
I have two matrices X and Y. Both represent a number of positions in 3D-space. X is a 50*3 matrix, Y is a 60*3 matrix.
My question: why does applying the mean-function over the output of pdist2() in combination with 'Mahalanobis' not give the result obtained with mahal()?
More details on what I'm trying to do below, as well as the code I used to test this.
Let's suppose the 60 observations in matrix Y are obtained after an experimental manipulation of some kind. I'm trying to assess whether this manipulation had a significant effect on the positions observed in Y. Therefore, I used pdist2(X,X,'Mahalanobis') to compare X to X to obtain a baseline, and later, X to Y (with X the reference matrix: pdist2(X,Y,'Mahalanobis')), and I plotted both distributions to have a look at the overlap.
Subsequently, I calculated the mean Mahalanobis distance for both distributions and the 95% CI and did a t-test and Kolmogorov-Smirnoff test to asses if the difference between the distributions was significant. This seemed very intuitive to me, however, when testing with mahal(), I get different values, although the reference matrix is the same. I don't get what the difference between both ways of calculating mahalanobis distance is exactly.
Comment that is too long #3lectrologos:
You mean this: d(I) = (Y(I,:)-mu)inv(SIGMA)(Y(I,:)-mu)'? This is just the formula for calculating mahalanobis, so should be the same for pdist2() and mahal() functions. I think mu is a scalar and SIGMA is a matrix based on the reference distribution as a whole in both pdist2() and mahal(). Only in mahal you are comparing each point of your sample set to the points of the reference distribution, while in pdist2 you are making pairwise comparisons based on a reference distribution. Actually, with my purpose in my mind, I think I should go for mahal() instead of pdist2(). I can interpret a pairwise distance based on a reference distribution, but I don't think it's what I need here.
% test pdist2 vs. mahal in matlab
% the purpose of this script is to see whether the average over the rows of E equals the values in d...
% data
X = []; % 50*3 matrix, data omitted
Y = []; % 60*3 matrix, data omitted
% calculations
S = nancov(X);
% mahal()
d = mahal(Y,X); % gives an 60*1 matrix with a value for each Cartesian element in Y (second matrix is always the reference matrix)
% pairwise mahalanobis distance with pdist2()
E = pdist2(X,Y,'mahalanobis',S); % outputs an 50*60 matrix with each ij-th element the pairwise distance between element X(i,:) and Y(j,:) based on the covariance matrix of X: nancov(X)
%{
so this is harder to interpret than mahal(), as elements of Y are not just compared to the "mahalanobis-centroid" based on X,
% but to each individual element of X
% so the purpose of this script is to see whether the average over the rows of E equals the values in d...
%}
F = mean(E); % now I averaged over the rows, which means, over all values of X, the reference matrix
mean(d)
mean(E(:)) % not equal to mean(d)
d-F' % not zero
% plot output
figure(1)
plot(d,'bo'), hold on
plot(mean(E),'ro')
legend('mahal()','avaraged over all x values pdist2()')
ylabel('Mahalanobis distance')
figure(2)
plot(d,'bo'), hold on
plot(E','ro')
plot(d,'bo','MarkerFaceColor','b')
xlabel('values in matrix Y (Yi) ... or ... pairwise comparison Yi. (Yi vs. all Xi values)')
ylabel('Mahalanobis distance')
legend('mahal()','pdist2()')
One immediate difference between the two is that mahal subtracts the sample mean of X from each point in Y before computing distances.
Try something like E = pdist2(X,Y-mean(X),'mahalanobis',S); to see if that gives you the same results as mahal.
Note that
mahal(X,Y)
is equivalent to
pdist2(X,mean(Y),'mahalanobis',cov(Y)).^2
Well, I guess there are two different ways to calculate mahalanobis distance between two clusters of data like you explain above:
1) you compare each data point from your sample set to mu and sigma matrices calculated from your reference distribution (although labeling one cluster sample set and the other reference distribution may be arbitrary), thereby calculating the distance from each point to this so called mahalanobis-centroid of the reference distribution.
2) you compare each datapoint from matrix Y to each datapoint of matrix X, with, X the reference distribution (mu and sigma are calculated from X only)
The values of the distances will be different, but I guess the ordinal order of dissimilarity between clusters is preserved when using either method 1 or 2? I actually wonder when comparing 10 different clusters to a reference matrix X, or to each other, if the order of the dissimilarities would differ using method 1 or method 2? Also, I can't imagine a situation where one method would be wrong and the other method not. Although method 1 seems more intuitive in some situations, like mine.