Is there a simpler way to create a matrix of matrixes:
A first matrix (9 x 32) that contain value (through a loop) for each item. I'll have 8 item, so 8 matrix of (9 x 32). I'll have also 2 condition so 2 * 8 of matrix of (9 x 32).
What is the simplest way to create it?
According to Documentation you can use various techniques to create each matrix. If you have a specific matrix you need to copy you can use the repmat(M, v, h) function to repeatedly create it. Otherwise to create a multidimensional array you can do:
B = repmat(0, [r c 8 2])
That should give you the matrices you need as a 4 dimensional array, where r is the number of rows, c the columns, 8 the number of repetitions, and 2 conditions. Hopefully that helps you
Just do
A = zeros(m,n,8,2)
or
A(m,n,8,2) = 0;
Related
I have a 2 dimensional array L, and I am trying to create a vector of linear indices ind for each row of this array.
L=
1 5 25 4 0 0
2 3 3 45 5 6
45 5 6 0 0 0
I am using lenr to store the number of non zero elements in each row (starting from column 1).
lenr=
4
6
3
Then I have 1x45 array RULES. Indices stored in L refer to elements in RULES. Since I want to vectorize the code, I decided to create linear indices and then run RULES(ind).
This works perfectly:
ind=sub2ind(size(L),1,lenr(1));
while this doesn't work:
ind=sub2ind(size(L),1:3,1:lenr(1:3));
Any ideas?
UPDATE:
This is what I initially tried to vectorize the code, but it did not works and that's why I checked linear indices:
rul=repmat(RULES,3);
result = rul((L(1:J,1:lenr(1:J))));
If I correctly interpret your edit, you want to create a variable result that contains the elements of RULES indicated by the non-zero elements of L. Note that in general, the best way to vectorize sub2ind is to not use sub2ind.
If you want result to be a linear array, you can simply write
%// transpose so that result is ordered by row
L_transposed = L.';
result = RULES(L_transposed(L_transposed>0));
If, instead, you want result to be an array of the same size as L, with all numbers in L replaced by the corresponding element in RULES, it's even simpler:
result = L;
result(result>0) = RULES(result>0);
I simplify my problem, let says I have three matrices.
I want to extract the red-boxed sub-matrices. I define
S = [1 4;
2 5]
that are the linear indices of the above matrices. So, A(S), B(S) and C(S) can extract the entries of the three matrices.
I pack them into vector by V = [ A(S)(:); B(S)(:); C(S)(:) ]. Let says after some manipulations, I obtain a new vector
V_new = [12 9 8 12 21 8 7 5 3 12 11 10]'
Here comes to my problem:
E.g for matrix A, I want to obtain
2->12, 5->9, 4->8 and 6->12
which are the first four entries of my V_new.
Since I have around 200 matrices, I have no idea to swap along the 200 matrices and the updated vector, V_new at the same time. Is writing a for-loop best way to do this purpose?
Thanks in advance.
Assuming that your A, B and C matrices have the same dimensions, rather work with a 3D matrix.
e.g. assuming your example matrices
M = cat(3,A,B,C)
No to extract those 4 upper left elements:
M_subset = M(1:2,1:2,:)
And then to reshape them into the vector you had:
V = M_subset(:)
then manipulate it to get V_new and finally put it back in the original:
M(1:2,1:2,:) = reshape(V_new,2,2,[])
What I intend to do is very simple but yet I haven't found a proper way to do it. I have a function handle which depends on two variables, for example:
f = #(i,j) i+j
(mine is quite more complicated, though)
What I'd like to do is to create a matrix M such that
M(i,j) = f(i,j)
Of course I could use a nested loop but I'm trying to avoid those. I've already managed to do this in Maple in a quite simple way:
f:=(i,j)->i+j;
M:=Matrix(N,f);
(Where N is the dimension of the matrix) But I need to use MATLAB for this. For now I'm sticking to the nested loops but I'd really appreciate your help!
Use bsxfun:
>> [ii jj] = ndgrid(1:4 ,1:5); %// change i and j limits as needed
>> M = bsxfun(f, ii, jj)
M =
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
If your function f satisfies the following condition:
C = fun(A,B) accepts arrays A and B of arbitrary, but equal size and returns output of the same size. Each element in the output array C is the result of an operation on the corresponding elements of A and B only. fun must also support scalar expansion, such that if A or B is a scalar, C is the result of applying the scalar to every element in the other input array.
you can dispose of ndgrid. Just add a transpose (.') to the first (i) vector:
>> M = bsxfun(f, (1:4).', 1:5)
Function handles can accept matrices as inputs. Simply pass a square matrix of size N where the values corresponds to the row number for i, and a square matrix of size N where the values correspond to the column number for j.
N = 5;
f = #(i,j) i+j;
M = f(meshgrid(1:N+1), meshgrid(1:N+1)')
In MATLAB I have a very large matrix (matrix A). Now I would like to find the row-index of the row which contain certain values in the second column. These values - which I'm looking for in Matrix A - are stored in anonther Matrix (Matrix B) with consists out of a row (800 numbers).
Besides I would like to redo this calculation for the same matrix A, but for ten different matrices, with different sizes (which contain the values I'm looking for in different columns of matrix A).
Because of the sizes of the matrix I think i need a loop to extract the row in matrix A which contain te value of Matrix B. How can I do this?
regards,
V
Thanks for the quick response! Indeed the problem is maybe a bit complex to answer without an example, and indeed duplicate entries cause some problems. Therefore hereby an example
For example I have a -simplified- matrix A:
1 2 3 4
9 9 9 9
4 3 2 1
And a -simplified- matrix (row) B: [9 3]
And a -simplified- matrix (row) C: [9 2]
Then I would like to get matrix D and matrix E.These matrices should contain in the first column the numbers from the original matrix D(or E) and in the second column the corresponding row-location of this value in matrix A.
So, matrix D =
9 2
3 3
matrix E =
9 2
2 3
As represented in this example matrix B and matrix C can contain data which is present in several column of matrix A (like the nine). However, martix B should "search" in column 2 of matrix A. Likewise, should matrix C "search" in column 3 of Matrix A, resulting in matrix D and E as given in the example.
As mentionned by Shai in his comment, your question is quite vague and a lot of special case could arise (duplicate entries, relative size of A and B, etc.). But in all generality I tried a small piece of code that seems to do what you want. There are certainly quicker ways of doing it, and certainly more information on your problem could help optimize this.
colA=2;
% Example
nmax=10;
nA=5;
A=randi(nmax,[nA nA]);
nB=3;
B=randi(nmax,[1 nB]);
% Find rows
rows=cell(size(B));
for i=1:numel(B)
rows(i)={find(A(:,colA)==B(i))};
end
The input / output was:
A =
3 7 8 5 4
9 7 3 7 5
8 2 9 9 8
9 5 9 7 9
3 3 4 6 8
B =
1 7 5
rows =
[0x1 double] [1;2] [4]
Assuming you have two vectors, largeDataIndex (the second column of your matrix) and interestingIndex (your b) and you want the following:
For each value of interestingIndex , find the position in largeDataIndex
Then an easy method would be this:
result = zeros(size(interestingIndex))
for i = 1:length(result)
result(i) = find(interestingIndex(i) == largeDataIndex)
end
Note that this assumes there is always just one entry that matches, otherwise you should define result as a cell array rather than a vector.
I have an m by n matrix in MATLAB, say M. I have an n-element row vector, i.e. a one by n column matrix, say X.
I know X is a row somewhere in M. How can I find the index in M?
EDIT:
gnovice's suggestion is even simpler than mine:
[~,indx]=ismember(X,M,'rows')
indx =
3
FIRST SOLUTION:
You can easily do it using find and ismember. Here's an example:
M=magic(4); %#your matrix
M =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
X=[9 7 6 12]; %#your row vector
find(ismember(M,X),1)
ans =
3
Before I learned of ismember, I used to do:
index = find(all(bsxfun(#eq, M, X), 2));
But using ismember(X, M, 'rows') is definitely preferable.
Another solution that returns a row index for each occurrence of X is
find(sum(abs(M-ones(rows(M),1)*X),2)==0)
Also, this solution can be easily adapted to find rows that are within threshold of X as follows (if numerical noise is an issue)
tolerance = 1e-16; %setting the desired tolerance
find(sum(abs(M-ones(rows(M),1)*X),2)<tolerance)
This is a non-loop version. It is only suitable, if M (your matrix) is not very large, ie. n and m are small. X is your row:
function ind = findRow(M,X)
tmp = M - repmat(X,size(M,1),1);
ind = find(tmp,1);
end
If M is too large, it might be faster, to iterate the rows of M and compare every row with your vector.
#Edit: renamed variables to match the names used in the question.