The commands
a = magic(3);
b = pascal(3);
c = cat(4,a,b);
produce a 3-by-3-by-1-by-2 array.
Why is the result 3-3-1-2 when the dimension is 4?
Both a and b are two-dimensional matrices of size 3-by-3. When you concatenate them along a fourth dimension, the intervening third dimension is singleton (i.e. 1). So c(:,:,1,1) will be your matrix a and c(:,:,1,2) will be your matrix b.
Here's a link to some documentation that may help with understanding multidimensional arrays.
EDIT:
Perhaps it will help to think of these four dimensions in terms that us humans can more easily relate to...
Let's assume that the four dimensions in the example represent three dimensions in space (x, y, and z) plus a fourth dimension time. Imagine that I'm sampling the temperature in the air at a number of points in space at one given time. I can sample the air temperature in a grid that comprises all combinations of three x positions, three y positions, and one z position. That will give me a 3-by-3-by-1 grid. Normally, we'd probably just say that the data is in a 3-by-3 grid, ignoring the trailing singleton dimension.
However, let's say that I now take another set of samples at these points at a later time. I therefore get another 3-by-3-by-1 grid at a second time point. If I concatenate these sets of data together along the time dimension I get a 3-by-3-by-1-by-2 matrix. The third dimension is singleton because I only sampled at one z value.
So, in the example c=cat(4,a,b), we are concatenating two matrices along the fourth dimension. The two matrices are 3-by-3, with the third dimension implicitly assumed to be singleton. However, when concatenating along the fourth dimension we end up having to explicitly show that the third dimension is still there by listing its size as 1.
Related
Okay, so I'm working on a problem related to quantum chaos and one of the things I need to do is to map the unit cube in n-dimensions to a parallelepiped in n-dimensions and find all integer points in the interior of this parallelepiped. I have been trying to do this using the following scheme:
Given the linear map B and the dimension of the cube n, we find the coordinates of the corners of the unit hypercube by converting numbers j from 0 to (2^n -1) into their binary representation and turning them into vectors that describe the vertices of the cube.
The next step was to apply the map B to each of these vectors, which gives me a set of 2^n vectors describing the coordinates of the vertices of the parallelepiped in n dimensions
Now, we take the maximum and minimum value attained by any of these vertices in each coordinate direction, i.e the first element of my vectors might have a maximum value of 4 across all of the vertices and a minimum value of -3 etc. This gives me an n-dimensional rectangular prism that contains my parallelepiped and some extra unwanted space.
I now find all points with integer coordinates in this bounding rectangular prism described as vectors in n dimensions
Finally, I apply the inverse of the map B to each of the points and throw away any points that have any coefficients greater than 1 as they must originally have lain outside my unit hypercube.
My issue arises in step 4, I'm struggling to come up with a way of generating all vectors with integer coordinates in my rectangular hyper-prism such that I can change the number of dimensions n on the fly. Ideally, i'd like to be able to increase n at will until it becomes too computationally heavy to do so, but every method of finding all integer points in the prism i've tried so far has relied on n for loops to permute each element and thus I need to rewrite the code every time.
So I guess my question is this, is there any way to code this up so that I can change n on the fly? Also, any thoughts on the idea of the algorithm itself would be appreciated :) It wouldn't surprise me if i've overcomplicated things massively...
EDIT:
Of course as soon as I post the question I see a lovely little link in the side-bar where a clever method has been given already for how to do this: Generate a matrix containing all combinations of elements taken from n vectors
I'll leave this up for the moment just in case anyone has any comments on the method in general, but otherwise (since I can't upvote yet I'll just say it here) Luis Mendo, you are a hero!
I want to create a 3D matrix of (MxNx1) dimension. where third dimension is equal to 1.
I am trying something like zeros(M,N,1) but it is returning a matrix of (MxN) dimension instead of (MxNx1) dimension. Is there any way of creating a 3D matrix where third dimension equal to 1.
How do you process a MxNxD image? I could only image that the useage of the third dimension is:
[M,N,D] = size(your3DMatrix);
Then you use D for further coding. Another possible useage is
permute(your3DMatrix,[n1 n2 n3])
Anyway, these both work fine for matrix whose third dimension is 1.
If you detail your code dealing with the MxNx1 matrix, it would be easier to help you.
I have a 3d matrix of 100x100x100. Each point of that matrix has assigned a value that corresponds to a certain signal strength. If I plot all the points the result is incomprehensible and requires horsepower to compute, due to the large amount of points that are painted.
The next picture examplify the problem (in that case the matrix was 50x50x50 for reducing the computation time):
[x,y,z] = meshgrid(1:50,1:50,1:50);
scatter3(x(:),y(:),z(:),5,strength(:),'filled')
I would like to plot only the highest values (for example, the top 10). How can I do it?
One simple solution that came up in my mind is to asign "nan" to the values higher than the treshold.
Even the results are nice I think that it must be a most elegant solution to fix it.
Reshape it into an nx1 vector. Sort that vector and take the first ten values.
num_of_rows = size(M,1)
V = reshape(M,num_of_rows,1);
sorted_V = sort(V,'descend');
ind = sorted_V(1:10)
I am assuming that M is your 3D matrix. This will give you your top ten values in your matrix and the respective index. The you can use ind2sub() to get the x,y,z.
I have to two evenly sized very large vectors (columns) A and B. I would like to divide vector A by vector B. This will give me a large matrix AxB filled with zeros, except the last column. This column contains the values I'm interested in. When I simple divide the vectors in a Matlab script, I run out of memory. Probably because the matrix AxB becomes very large. Probably I can prevent this from happening by repeating the following:
calculating the first row of matrix AxB
filter the last value and put it into another vector C.
delete the used row of matrix AxB
redo step 1-4 for all rows in vector A
How can I make a loop which does this?
You're question doesn't make it clear what you are trying to do, although it sounds like you want to do an element wise division.
Try:
C = A./B
"Matrix product AxB" and "dividing vectors" are distinct operations.
If we understood this correctly, what you do want to calculate is "C = last column from AxB", such that:
lastcolsel=zeros(size(B,2),1)
C=(A*B)*lastcolsel
If that code breaks your memory limit, recall that matrix product is associative (MxN)xP = Mx(NxP). Simplifying your example, we get:
lastcolsel=zeros(size(B,2),1)
simplifier=B*lastcolsel
C=A*simplifier
I have multiple vectors of varying lengths that I would like to plot next to each other in 3D space in Matlab.
As an example:
Say I have three vectors:
X is a 5x2 vector,
Y is a 10x2 vector and
Z is a 15x2 vector.
Each element of every vector has the format:
x value, y value
but the x values of the various vectors do not match.
I would like to plot these vectors in 3D space, next to each other. The reason why I don't want to plot them using "hold" is because most of the data have the same values, but I would like to see how many of the plots have the same value at a specific time.
I hope my questions makes sense. Please just ask if anyone is unsure.
I think you are looking for the function ribbon.
Documentation: http://www.mathworks.fr/help/techdoc/ref/ribbon.html
EDIT:
if your x's do not have the same length, you can combine it with interp1 as follow:
x1=0:0.1:1;
x2=0:0.02:1.5;
y1=x1.^2;
y2=sqrt(x2);
y2=interp1(x2,y2,x1);
ribbon(x1',[y1;y2]')