I want to use the command quiver3(X,Y,Z,M,N,O)
in order to get a vector field, where all matrices are 10x10x10 arrays with real entries. Now I wanted to plot it and got only 4 vectors instead of my expected 1000. Does anybody here know, what I could possibly have done wrong?
If you need further information, I am available for your comments.
Assuming the problem occurs when attempting to display the data (rather than with attempting to display unallowed numerical values such as Inf or NaN), I think the problem is the range of norms of the vectors, such that only 4 show up.
The importance of the norm of the vectors relative to the range of the coordinate system is important, as illustrated by the following example:
Here all 10 random vectors show up:
Na = 10;
[X Y Z M N O] = deal(rand(Na,1),rand(Na,1),rand(Na,1),rand(Na,1),rand(Na,1),rand(Na,1));
quiver3(X,Y,Z,M,N,O)
But if some of the vectors are made 100 x bigger, only the smaller ones show up:
mul = 1e+3;
[X Y Z M N O] = deal(rand(Na,1),rand(Na,1),rand(Na,1),...
[mul*rand(Na/2,1);rand(Na/2,1)],...
[mul*rand(Na/2,1);rand(Na/2,1)],...
[mul*rand(Na/2,1);rand(Na/2,1)]);
quiver3(X,Y,Z,M,N,O)
Similarly if some of the vectors are made too small they don't show up:
mul = 1e-3;
[X Y Z M N O] = deal(rand(Na,1),rand(Na,1),rand(Na,1),...
[mul*rand(Na/2,1);rand(Na/2,1)],...
[mul*rand(Na/2,1);rand(Na/2,1)],...
[mul*rand(Na/2,1);rand(Na/2,1)]);
quiver3(X,Y,Z,M,N,O)
Vectors too large or too small relative to the axis are not displayed at all.
If it is a problem with scaling you may want to inspect your function and see how to narrow the range of x,y,z so that the vectors have a narrower range of values, or change the scale (logarithm?) so that all the data can be displayed in one figure.
edit
As an alternative to the question of how to display your data, you may want to consider using isosurfaces. Here's an example, with each red sphere representing a different isosurface in a spherical potential:
Related
I have a problem to figure out how in a waterfall figure the x axis can correspond to the x values, and not their point number. This question seems rather simple but in my particular case (due to the size of vectors) it's not easy to get the correct figure. So i really need your help ... after several hours of unsatisfied results.
Assuming that two vectors x and y of the same length are recorded at a time t. This procedure is performed k times. I finally want to plot with waterfall y versus x for the different times.
I give you a script that corresponds to the experiment where xx is just added to get here two continuous functions x and y for the different times. The result is almost perfect but I would like the x-y values on the corresponding x, y axis instead of the point number.
xx=0:0.1:8;
for t=1:2:11
x(t,:)=sin(t*xx.^2);
y(t,:)=cos(t*xx.*4);
end
waterfall(x,y)
The problem comes probably from the different size of x, y with t. Thanks in advance for your advice.
Two comments:
waterfall takes either Z or X,Y,Z as coordinates. So it takes your x matrix as Z, and the other argument is mapped to the C input, which dictates color. You can see that the plot is the same if you do waterfall(x), except with different colors.
Your x is not monotonically increasing, so if you plot x(t,:) vs y(t,:) for any t, you'll get a web-like graph, not anything nice to look at.
So I'll plot xx vs y, and I'm modifying your y a bit so it looks nicer. I hope you can take this idea and modify it to do what you need.
The code below doesn't use waterfall at all, it simply calls plot3 once for each t. It might be possible to call plot3 with your full x and y matrices, but this is just as easy.
In the plot3 call, the x-coordinates are given by xx, the y-coordinates by t (simply repeated to match the expected size), and the z-coordinates by y:
xx = 0:0.1:8;
for t = 1:2:11
y = cos(t*xx/4);
plot3(xx,repmat(t,size(xx)),y)
hold on
end
xlabel('x')
ylabel('t')
zlabel('y=cos(tx/4)')
The title of this post may be a bit confusing. Please allow me to provide a bit of context and then elaborate on what I'm asking. For your reference, the question I'm asking is toward the end and is denoted by bold letters. I provide some code, outlining where I'm currently at in solving the problem, immediately beforehand.
Essentially what I'm trying to do is Kernel Regression, which is usually done using a single test point x and a set of training instances . A reference to this can be found on wikipedia here. The kernel I'm using is the RBF kernel, a Wikipedia reference for which can be found here.
Anyway, I have some code written in Matlab so that this can be done quickly for a single instance of x, which is 1 x p in size. What I'd like to do is make it so I can estimate for numerous points very quickly, say m x p.
For the sake of avoiding notational mixups, I'll let the training instances be denoted Train and the instances I want estimates for as Test: and . It also needs to be mentioned that I want to estimate a vector of numbers for each of the m points. For a single point this vector would be 1 x v in size. Now I need it to be m x v. Therefore, Train will also have a vector of these know values associated with it called TS: . Lastly, we need a vector of sigmas that is 1 x v in size. This is denoted as Sig.
Here's the code I have so far:
%First, we have to get the matrices to equivalent size so we can subtract Train from Test
tm0 = kron(ones(size(Train,1),1),Test) - kron(ones(size(Test,1),1),Train);
%Secondly, we apply the Euclidean norm sq by row and then multiply each of these results by each element (j) in Sig times 1/2j^2
tm3 = exp(-kron(sum((tm0).^2,2),1/2./(Sig.^2)));
Now, at this point tm3 is an (m*n) x v matrix. This is where my question is: I now need to multiply TS' (TS transpose) times each of the n x v-sized segments in tm3 (there are m of these segments), get the diagonal elements of each of these resulting segments (after multiplication one of the m segments will be v x v, so each chunk of diagonal elements will be 1 x v meaning the resulting matrix is m x v) and sum these diagonal elements together to produce an m x 1 sized matrix. Lastly, I will need to divide each entry i in this m x 1 matrix by each of the v elements in the ith row of the diagonal-holding m x v-sized matrix, producing an m x v-sized result matrix.
I hope all of that makes sense. I'm sure there's some kind of trick that can be employed, but I'm just not coming up with it. Any help is greatly appreciated.
Edit 1: I was asked to provide more of an example to help demonstrate what it is that I would like done. The following represent that two matrices I'm talking about, TS and tm3:
As you can see, TS'(TS transpose) is v x n and tm3 is mn x v. In tm3 there are blocks that are of size n x v -- there are m blocks of this size. Notice that the size of TS' is of size v x n. This means that I can multiply TS' by a single block of tm3, which again is of size n x v. This would result in a matrix that is v x v in size. I would like to do this operation -- individually multiplying TS' by each of the n x v-sized blocks of tm3, which would produce m v x v matrices.
From here, though, I would like to obtain the diagonal elements from each of these v x v matrices. So, for a single v x v matrix, denoted using a:
Ultimately, I would to do this for each of the m v x v matrices giving me something that looks like the following, where s is the mth v x v matrix:
If I denote this last matrix as Q, which is m x v in size, it is trivial to sum the elements across the rows to produce the m x 1 vector I was looking for. I will refer to this vector as C. However, I would then like to divide each of these m scalar values by the corresponding row of matrix Q, to produce another m x v matrix:
This is the final matrix I'm looking for. Hopefully this helps make it clear what I'm looking for. Thanks for taking the time to read this!
Thought: I'm pretty sure I could accomplish this by converting tm3 to a cell matrix by doing tc1 = mat2cell(tm3,repmat(length(Train),1,m),length(Sig)), and then put replicate TS m times in another cell matrix tc2 = mat2cell(TS',length(indirectSigma),repmat(length(Train),1,m))'. Finally, I could do operations like tc3 = cellfun(#(a,b) a*b, tc2,tc1,'UniformOutput',false), which would give me m cells filled with the v x v matrices I was looking for. I could proceed from there. However, I'm not sure how fast these cell operations are. Can anybody comment? I'm afraid they might be slow, so I would prefer operations be performed on normal matrices, which I know to be fast. Thanks!
I have multiple linear equations in the form of Zi=ai*Xi+bi*Yi for i = 1..30.
How can I calculate every pair of regression coefficient values, or those 30 values of a and b for each (Z,X,Y) combination using MATLAB?
I've tried the following code:
A=Z; B=[Xs Ys];
C = B \ A;
A are my Z points while B is a matrix of my X and Y points. However, I seem to only get one pair of regression coefficients for all of the points.
Thanks in advance!
What you have set up there is unfortunately not the right way to solve it if I understand your problem formulation. That assumption assumes that you are trying to fit all of the points on a single line. Each row of B would thus serve as one point on only one line that you are trying to find the linear regression of. If you want to solve for multiple lines simultaneously, you are going to need to change your formulation.
That is actually very simple. I'm going to assume that you have 30 (x,y) points where each point denotes one equation of a line. You have these set as Xs and Ys respectively. The outputs for each of these equations is also in Zs. I'm also going to assume these are column vectors, and therefore, you have a system set up such that:
a_i and b_i are the coefficients for each line. You know the (x,y) for each line and your goal is to solve for each corresponding a and b. As such, you would need to reformulate your system so that you're solve for a and b.
Rewriting that problem in matrix form, it can be done like so:
The right hand side vector of a_1, b_1, a_2, b_2, ... is what you are ultimately solving for. You can see that we have a matrix equation of Y = M*X where M and Y are known and X is what we need to solve for by doing X = M\Y. As such, all you need to rearrange your x and y values into a block matrix like the above. First we need to find the correct linear indices so that we can place our x and y values into this matrix, then solve the system by least squares with the ldivide operator. The matrix is a N x 2N matrix where N is the total number of equations or constraints that we have (so in your case, 30):
N = numel(Xs);
M = zeros(N, 2*N);
xind = sub2ind(size(M), 1:N, 1:2:2*N);
yind = sub2ind(size(M), 1:N, 2:2:2*N);
M(xind) = Xs;
M(yind) = Ys;
sub2ind allows you to place multiple values into a matrix with a single line of code. Specifically, sub2ind determine the linear indices from a set of row and column coordinates to access into a matrix. If you don't already know, you can access values (and set values) in a matrix using a single number instead of a pair of row and columns. sub2ind will allow you to set multiple values in a matrix at once by specifying a set of linear indices to access in the matrix with a corresponding vector.
In our case, we need two sets of linear indices - one for the x values and one for the y values. Note that the x values start from the first column and skip every other column. The same behaviour can be said for the y values but we start at the second column. Once we have those indices, we set the x and y values in this matrix and we now we simply solve for the coefficients:
coeff = M \ Z;
coeff will now be 2N x 1 vector, so if you want, you can reshape this into a matrix:
coeff = reshape(coeff, 2, []);
Now, coeff will be shaped such that each column will give you the pair of a,b for each equation that you had. As such, the first column denotes a_1, b_1, the second column denotes a_2, b_2 and so on. The first row of coeff is all of the a coefficients for each constraint while the second row is all of the b coefficients for each constraint.
I have just started working using CCA in Matlab. I have two vectors X and Y of dimension 60x1920 and 60x1536 with the number of samples being 60 and variables in the different set of vectors being 1920 and 1536 respectively. I want to know do CCA for reducing them to the subspace and then do feature matching.
I am using this commands.
%% DO CCA
[A,B,r,U,V] = canoncorr(X,Y);
The output I get is this :
Name Size Bytes Class Attributes
A 1920x58 890880 double
B 1536x58 712704 double
U 60x58 27840 double
V 60x58 27840 double
r 1x58 464 double
Can anyone please tell me what these variables mean. I have gone over the documentation several times and still is unclear about them. As I understand CCA finds two linear projection matrices Wx and Wy such that the projection of X and Y on Wx and Wy are maximally correlated.
1) Could anyone please tell me which of the following matrices are these?
2) Also how can I find the projected vectors in the learned subspace of CCA?
Any help will be appreciated. Thanks in advance.
As I understand it, with X and Y being your original data matrices, A and B are the sets of coefficients that perform a change of basis to maximally correlate your original data. Your data is represented in the new bases as the matrices U and V.
So to answer your questions:
The projection matrices you are looking for would be A and B since they transform X and Y into the new space.
The resulting projections of X and Y into the new space would be U and V, respectively. (The r vector represents the entries of the correlation matrix between U and V, which is a diagonal matrix.)
The The MATLAB documentation says this transformation can be done with the following formulae, where N is the number of observations:
U = (X-repmat(mean(X),N,1))*A
V = (Y-repmat(mean(Y),N,1))*B
This page lays out the process nicely so you can see what each coefficient means in the transformation process.
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.