I have been trying to use the multidimensional array function to store NxN vectors of length n in each (i,j) entry of a 3d matrix of NxNxn dimensions.
My code looks like:
a=zeros(N,N,n);
a(i,j,:)=v_ij; %here v is a vector of length n, which differs for each i,j
However when I try to extract each vector by typing e.g. a(1,1,:) I don't get a nice vector. Rather I get something like:
ans(:,:,1) = ..
ans(:,:,2) = ..
...
I don't understand why it says ans(:,:)...
That's because each vector v_ij is stored along the 3rd dimension of your matrix, so when you access a(1,1,:), you are looking for a multidimensional array consisting of every value at the location (1,1) along the 3rd dimension of a.
Let's consider a simple example:
N = 2;
n = 3;
a = zeros(N,N,n);
for k = 1:N
for l = 1:N
v_kl = randi([0 10],1,n);
a(k,l,:) = v_kl;
end
end
The randomly-generated matrix a is a 2x2x3 matrix that looks like this:
a(:,:,1) =
4 1
4 10
a(:,:,2) =
0 2
0 5
a(:,:,3) =
2 2
9 5
Therefore, using a(1,1,:) is equivalent to getting the element located at (1,1) for all 3 dimensions, here it would be 4,0 and 2.
Indeed, calling a(1,1,:) yields:
ans(:,:,1) =
4
ans(:,:,2) =
0
ans(:,:,3) =
2
Benoit_11's answer plus squeeze should work, but I would like to propose a different solution.
Rather than creating a NxNxn array, why not make it nxNxN?
N = 2;
n = 3;
a = zeros(n,N,N);
for p = 1:N
for q = 1:N
v_pq = randi([0 10],1,n);
a(:,p,q) = v_pq;
if (p == 1) && (q == 1)
v_pq %// display vector at a(:,1,1)
end
end
end
a
v_pq =
3 4 8
a =
ans(:,:,1) =
3 3
4 9
8 7
ans(:,:,2) =
5 6
10 1
9 5
Now the vectors are stored along the first dimension, so [3 4 8] shows up as the first column of ans(:,:,1). To access it you would use a(:,p,q) rather than a(p,q,:):
a(:,1,1)
ans =
3
4
8
Related
I want to find all possible variations (combinations) of a vector, choosing various numbers of elements from that vector.
For example, suppose I have the vector:
x = [1 2 3 4 5];
I can determine the number of combinations for each number of chosen elements:
x = [1 2 3 4 5]';
n = numel(x);
for k = 1:n
combs(k) = nchoosek(n,k);
end
sum(combs)
This results in:
combs = 5 10 10 5 1
sum(combs) = 31
I want a way to store all 31 of these combinations in an array, for example a cell array, with n cells, within each is an array in which each row is a vector combination of the elements.
e.g. at k = 4:
combs{4} =
1 2 3 4
1 2 3 5
1 2 4 5
1 3 4 5
2 3 4 5
Is there an existing function that does this, or what would be the most simple approach to this?
Call nchoosek with a vector as first input, using arrayfun (or equivalently for) to loop over the number of picked elements:
n = 5;
combs = arrayfun(#(k) nchoosek(1:n,k), 1:n, 'UniformOutput', false);
Here is an approach using dec2bin , find and accumarray:
x = [1 2 3 4 5];
[a b] = find(dec2bin(1:2^numel(x)-1)=='1');
combs = accumarray(a,x(b),[],#(c){c});
I am looking for a matrix operation of the form: B = M*A*N where A is some general square matrix and M and N are the matrices I want to find.
Such that the columns of B are the diagonals of A. The first column the main diagonal, the second the diagonal shifted by 1 from the main and so on.
e.g. In MATLAB syntax:
A = [1, 2, 3
4, 5, 6
7, 8, 9]
and
B = [1, 2, 3
5, 6, 4
9, 7, 8]
Edit:
It seems a pure linear algebra solution doesn't exist. So I'll be more precise about what I was trying to do:
For some vector v of size 1 x m. Then define C = repmat(v,m,1). My matrix is A = C-C.';.
Therefore, A is essentially all differences of values in v but I'm only interested in the difference up to some distance between values.
Those are the diagonals of A; but m is so large that the construction of such m x m matrices causes out-of-memory issues.
I'm looking for a way to extract those diagonals in a way that is as efficient as possible (in MATLAB).
Thanks!
If you're not actually looking for a linear algebra solution, then I would argue that constructing three additional matrices the same size as A using two matrix multiplications is very inefficient in both time and space complexity. I'm not sure it's even possible to find a matrix solution, given my limited knowledge of linear algebra, but even if it is it's sure to be messy.
Since you say you only need the values along some diagonals, I'd construct only those diagonals using diag:
A = [1 2 3;
4 5 6;
7 8 9];
m = size(A, 1); % assume A is square
k = 1; % let's get the k'th diagonal
kdiag = [diag(A, k); diag(A, k-m)];
kdiag =
2
6
7
Diagonal 0 is the main diagonal, diagonal m-1 (for an mxm matrix) is the last. So if you wanted all of B you could easily loop:
B = zeros(size(A));
for k = 0:m-1
B(:,k+1) = [diag(A, k); diag(A, k-m)];
end
B =
1 2 3
5 6 4
9 7 8
From the comments:
For v some vector of size 1xm. Then B=repmat(v,m,1). My matrix is A=B-B.'; A is essentially all differences of values in v but I'm only interested in the difference up to some distance between values.
Let's say
m = 4;
v = [1 3 7 11];
If you construct the entire matrix,
B = repmat(v, m, 1);
A = B - B.';
A =
0 2 6 10
-2 0 4 8
-6 -4 0 4
-10 -8 -4 0
The main diagonal is zeros, so that's not very interesting. The next diagonal, which I'll call k = 1 is
[2 4 4 -10].'
You can construct this diagonal without constructing A or even B by shifting the elements of v:
k = 1;
diag1 = circshift(v, m-k, 2) - v;
diag1 =
2 4 4 -10
The main diagonal is given by k = 0, the last diagonal by k = m-1.
You can do this using the function toeplitz to create column indices for the reshuffling, then convert those to a linear index to use for reordering A, like so:
>> A = [1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
>> n = size(A, 1);
>> index = repmat((1:n).', 1, n)+n*(toeplitz([1 n:-1:2], 1:n)-1);
>> B = zeros(n);
>> B(index) = A
B =
1 2 3
5 6 4
9 7 8
This will generalize to any size square matrix A.
I have a 4x5 matrix called A from which I want to select randomly 3 rows, then 4 random columns and then select those elements which coincide in those selected rows and columns so that I have 12 selected elements.Then I want to create a diagonal matrix called B which will have entries either 1 or 0 so that multiplication of that B matrix with reshaped A matrix (20x1) will give me those selected 12 elements of A.
How can I create that B matrix? Here is my code:
A=1:20;
A=reshape(A,4,5);
Mr=4;
Ma=3;
Na=4;
Nr=5;
M=Ma*Mr;
[S1,S2]=size(A);
N=S1*S2;
y2=zeros(size(A));
k1=randperm(S1);
k1=k1(1:Ma);
k2=randperm(S2);
k2=k2(1:Mr);
y2(k1,k2)=A(k1,k2);
It's a little hard to understand what you want and your code isn't much help but I think I've a solution for you.
I create a matrix (vector) of zeros of the same size as A and then use bsxfun to determine the indexes in this vector (which will be the diagonal of B) that should be 1.
>> A = reshape(1:20, 4, 5);
>> R = [1 2 3]; % Random rows
>> C = [2 3 4 5]; % Random columns
>> B = zeros(size(A));
>> B(bsxfun(#plus, C, size(A, 1)*(R-1).')) = 1;
>> B = diag(B(:));
>> V = B*A(:);
>> V = V(V ~= 0)
V =
2
3
4
5
6
7
8
9
10
11
12
13
Note: There is no need for B = diag(B(:)); we could have simply used element by element multiplication in Matlab.
>> V = B(:).*A(:);
>> V = V(V ~= 0)
Note: This may be overly complex or very poorly put together and there is probably a better way of doing it. It's my first real attempt at using bsxfun on my own.
Here is a hack but since you are creating y2 you might as well just use it instead of creating the useless B matrix. The bsxfun answer is much better.
A=1:20;
A=reshape(A,4,5);
Mr=4;
Ma=3;
Na=4;
Nr=5;
M=Ma*Mr;
[S1,S2]=size(A);
N=S1*S2;
y2=zeros(size(A));
k1=randperm(S1);
k1=k1(1:Ma);
k2=randperm(S2);
k2=k2(1:Mr);
y2(k1,k2)=A(k1,k2);
idx = reshape(y2 ~= 0, numel(y2), []);
B = diag(idx);
% "diagonal" matrix 12x20
B = B(all(B==0,2),:) = [];
output = B * A(:)
output =
1
3
4
9
11
12
13
15
16
17
19
20
y2 from example.
y2 =
1 0 9 13 17
0 0 0 0 0
3 0 11 15 19
4 0 12 16 20
Let's say I have a matrix,
>> m = magic(3)
m =
8 1 6
3 5 7
4 9 2
and suppose I have a vector containing a subset of the first elements of the columns,
>> v = [3 4]
v =
3 4
is there any function that I can use to find the index of the row from the first element of the columns? what I mean --
>> rows = row_index_from_col(m, 1, v)
rows =
2 3
??
if not, what is the best way to do this?
Looks like I got it --
m = magic(3)
v = [3 4]
[~,c] = size(m)
i = 1:c
i(ismember(m(:,1),v))
there might be a better way, I am not sure.
I have 2 vectors (n and t) eg:
n t
1 5
5 3
5 2
2 6
2 9
Once I sample from vector n through randsample(n,1), I want to sample from the vector t but only from the values corresponding to that same one in vector n.
eg. If I drew a value of 2 from n, I then want to draw the value 6 or 9 from t. But how do I tell matlab to do that?
You could potentially do this:
out = t(n == randsample(n, 1))
This will create a filter based on whether n = its own random sample, ie if
randsample(n, 1) = 2
(n == randsample(n, 1)) = [0
0
0
1
1]
and applying this to t ie:
t(n == randsample(n, 1)) = [6
9]
which are the two corresponding values to 2 in n, but in t.
Hope this helps.
PS if you need just one value from t then you can randsample the output that this function gives you.
Simple one-liner, assuming you have them stored in a Nx2 matrix
nt = [
1 5;
5 3;
5 2;
2 6;
2 9];
meaning:
n = nt(:,1);
t = nt(:,2);
you can sample nSamples with replacement by randmonly indexing the matrix row-wise, i.e.:
nSamples = 5;
keepSamples = nt(randi(length(nt),nSamples,1),:);